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https://www.physicsforums.com/threads/dot-product-of-vectors.571244/ | # Dot product of vectors
1. Jan 26, 2012
### anigeo
say for example for 2 vectors A.B=0 and for another pair A.C=0.is it necessary that B is parallel to C.If yes how?
2. Jan 27, 2012
### Rap
As long as A and B are not zero, A.B=0 means A and B are perpendicular. A.C means A and C are perpendicular. That does not mean that B and C are parallel. Think of the three cartesian axes vectors X, Y, and Z: X.Y=0 and X.Z=0 but Y and Z are not parallel.
3. Jan 27, 2012
### mathfeel
Only in 2D.
Similar Discussions: Dot product of vectors | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8801280856132507, "perplexity": 2216.143459373912}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818689572.75/warc/CC-MAIN-20170923070853-20170923090853-00538.warc.gz"} |
https://par.nsf.gov/biblio/10053915-agent-based-model-construction-using-inverse-reinforcement-learning | Agent-based model construction using inverse reinforcement learning
Agent-based modeling (ABM) assumes that behavioral rules affecting an agent's states and actions are known. However, discovering these rules is often challenging and requires deep insight about an agent's behaviors. Inverse reinforcement learning (IRL) can complement ABM by providing a systematic way to find behavioral rules from data. IRL frames learning behavioral rules as a problem of recovering motivations from observed behavior and generating rules consistent with these motivations. In this paper, we propose a method to construct an agent-based model directly from data using IRL. We explain each step of the proposed method and describe challenges that may occur during implementation. Our experimental results show that the proposed method can extract rules and construct an agent-based model with rich but concise behavioral rules for agents while still maintaining aggregate-level properties.
Authors:
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10053915
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2017 Winter Simulation Conference (WSC)
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1264-1275 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8506724834442139, "perplexity": 1876.5997160609734}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335362.18/warc/CC-MAIN-20220929163117-20220929193117-00388.warc.gz"} |
http://math.stackexchange.com/questions/159026/curious-laplacian-inequality-delta-f0-rightarrow-f-has-no-maxima-additio | # Curious Laplacian inequality ($\Delta f>0 \Rightarrow f$ has no maxima). Additional premises just to confuse students?
In a homework question, we've been given the following things:
• Let $B=\overline{B(0,1)} \subseteq \mathbb R^n$ be the closed unit ball, $f : B\to \mathbb R^n$ a $\mathcal C^2$ function and let some $a_i : B \to \mathbb R$ be continuous for $i=1,\ldots,n$.
• For all $x\in B$, we have $$\Delta f(x)+\sum_{i=1}^n a_i(x)\frac{\partial f}{\partial x_i}(x) > 0$$ where $\Delta f$ is the Laplacian of $f$.
Then show that $f$ has no local maximum in $Int(B)$.
I'm kind of confused by the question. We're used to having questions where all the premises matter, but this doesn't seem to be the case here, or is it?
I mean, let $x_0$ be the maximum whose existence we want to contradict. Then $x_0$ has to be a critical point and the sum in the inequality vanishes anyway. But $\Delta f$ needs to be negative in maxima, right? Hence contradiction.
So, what's the point of the $a_i$, why make them continuous? Do we really need the contraint on the unit ball, or if not, what property of $B$ matters? Convexity, compactness?
Are all these premises just to confuse the students and we actually don't need them? Or am I severely overlooking something? If so, could you give a hint please on what to keep in mind for the proof?
Thank you.
-
The Laplacian should be non positive. And yes, continuity of $a_i$ is irrelevant and domain can be arbitrary. Perhaps the course is about classical solutions, so... – Andrew Jun 16 '12 at 11:25
The $a_i$ are not additional assumptions, but additional constraints. The statement is, of course, true, if the $a_i$ are all identically zero. The point is that you may even have this additional term in the equation and the statement still holds true.
And the Laplacian is not necessarily negative, only nonpositive, in $x_0$ such that $f(x_0)$ is maximal.
And, finally, if the $a_i$ were not continuous you'd run into problems assuming $f\in C^2$ (usually one considers solutions of the type $\Delta f + \sum a_i \frac{\partial f}{\partial x_i} = g(x)$) But you are right, to just prove the above statement the assumption can be weakened.
(Edit: note though, that it is not clear that the statement remains true if the $a_i$ have, say, a pole at $x_0$. If then, e.g., $\lim_{x\rightarrow x_0}a_i \frac{\partial f}{\partial x_i} < 0$, how would you reason in that case? For this reason I'd say it is, for an exercise, safe and reasonable to assume the $a_i$ are continuous.)
-
Thanks. Regarding your comment on the continuity of the $a_i$: As I see it, for infering $x_0$ is not a maximum, all we need is the local information that $\Delta f(x_0) > 0$, not anything from an environment of $x_0$, right? Could you elaborate please why I ran into problems if $a_i$ weren't continuous? – H.Kroeger Jun 16 '12 at 14:16
@H.Kroeger It is, in general, not sufficient to look at this pointwise. If, as I suggested, $a_i$ is unbounded and $a_i\frac{\partial f}{\partial x_i} > M >0$, say, when $x$ tends to a critical point of $f$, then you will not be able to conclude that $\Delta f >0$, but only $\Delta F > - M$, if at all. So, if $a_i$ is not continuous you need at least to know that $a_i \frac{\partial f}{\partial x_i}\rightarrow 0$ if $x$ tends to a critical point of $f$ (e.g. cause $a_i$ is assumed to be bounded). (Or you need additional reasoning, which I, right now, could not provide). – user20266 Jun 16 '12 at 14:31
Ah okay, I'll have a try with this approach, thank you. – H.Kroeger Jun 16 '12 at 14:40
@H.Kroeger: maybe Andrew can comment on that (he who commented on your question). As pointed out in my answer, though, I don't think that your exercise is aiming at a discussion like this. It's intention, for sure, is rather that you figure out the reasoning you actually found, so, for simplicity's sake, they just assumed $a_$ to be continuous. – user20266 Jun 16 '12 at 14:53 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9389814138412476, "perplexity": 254.56271368601602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119658961.52/warc/CC-MAIN-20141024030058-00255-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/138152/characterization-of-convex-space-curve | # Characterization of convex space curve
In general, a smooth curve $C$ in $R^3$ that lies entirely on the boundary of its convex hull, $\mathcal{H}(C)$, is defined to be convex.
1. Does any one know of a characterization of a curve in space as convex depending on the sign of its torsion $\tau$?
2. Is the projection of a non-singular "short" curve in space to any plane always convex?
Where a "short" curve in space is such that there is a 1-1 correspondence between the space curve and the corresponding projected curve in a plane.
Thanks.
-
Any set $A$ lies entirely on its convex hull by definition: $A\subset CH(A)$. Or did I missunderstood something in your question? – Luc Jul 30 '13 at 9:22
I think it means "on the boundary of its convex hull". – alvarezpaiva Jul 30 '13 at 9:40
Sorry about not being precise. I am referring to the definition of a convex curve as provided by Sedykh in link.springer.com/article/10.1007%2FBF01077070#page-1 – Sunayana Ghosh Jul 30 '13 at 9:42
I cannot make sense of the 2nd question. Perhaps this is what is meant: If some curve $C$ in space has the property that every projection of $C$ to a plane is simple, i.e., has no self-intersections, must then $C$ lie on its convex hull in 3D? – Joseph O'Rourke Jul 30 '13 at 11:53
In any case it is impossible that all projections be simple: join two points of the curve by a segment and project in the direction perpendicular to it. The most you can ask is that almost all projections be simple. I guess this only says the curve is planar. – alvarezpaiva Jul 30 '13 at 12:38 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8790066838264465, "perplexity": 258.02316956669245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701153585.76/warc/CC-MAIN-20160205193913-00244-ip-10-236-182-209.ec2.internal.warc.gz"} |
https://terrytao.wordpress.com/2008/10/24/distinguished-lecture-series-iii-elias-stein-algebras-of-operators/ | In the third of the Distinguished Lecture Series given by Eli Stein here at UCLA, Eli presented a slightly different topic, which is work in preparation with Alex Nagel, Fulvio Ricci, and Steve Wainger, on algebras of singular integral operators which are sensitive to multiple different geometries in a nilpotent Lie group.
For sake of discussion, let us begin by working on the Heisenberg group ${\Bbb H}^n = \{ (z,t) \in {\Bbb C}^n \times {\Bbb R} \}$ with group law $(z,t) (z',t') = (z+z',t+t'+2\hbox{Im}(z\overline{z'}))$ (this notation differs slightly from that in the previous lectures, as we have decremented n by 1). We have two classical algebras of singular integral operators on this space. On the one hand, one can view ${\Bbb H}^n$ as a 2n+1-dimensional Euclidean space ${\Bbb R}^{2n+1}$ and consider the algebra of pseudodifferential operators on this space. On the other hand, one can look at convolution operators $T: f \mapsto f*K$ using the group multiplication (thus $Tf(x) = \int_{{\Bbb H}^n} K(y) f(xy^{-1})\ dy$), where K is a Calderòn-Zygmund kernel adapted to the scaling structure of the Heisenberg group. Both algebras of operators are indeed algebras, and are bounded on every $L^p$ space for $1 < p < \infty$. It is then a natural question to ask whether there is some larger algebra that contains both types of operators. This will naturally create operators with “hybrid kernels” that detect both the Euclidean geometric structure and the Heisenberg geometric structure. A model example of this is the convolution operator $Tf := f * K$ where K is the distribution given by a suitable principal value interpretation of the function
$\displaystyle K( z, t ) := \frac{z_j z_k}{(|z|^2+|t|^2)^{n+1}} \frac{1}{|z|^2+it}$ (1)
where $1 \leq j,k \leq n$ are fixed indices. More generally, one could consider convolution kernels $f\mapsto f*K$ where K is smooth for non-zero (z,t) and obeys the derivative estimates
$\displaystyle |\partial_t^\beta \partial_{z,\overline{z}}^\alpha K(x)| \leq C_{\alpha,\beta} |x|_E^{-2n-|\alpha|} |x|_H^{-2-2\beta}$ (2)
where $x = (z,t)$, $|x|_E := (|z|^2+|t|^2)^{1/2}$, and $|x|_H := |z|+|t|^{1/2}$, and also obeys the cancellation condition $\langle K, \phi_{r,R} \rangle| \leq C$ for all $0 < R^2 \leq r^2 \leq R$, where $\phi$ is a non-trivial bump function and $\phi_{r,R}(z,t) = \phi(z/r,t/R)$. [Note that (2) just barely prevents K from being absolutely integrable, in analogy with other singular integral operator kernels.] There is also a new condition that appears in this hybrid setting that does not occur in the previous settings: one needs the “marginal distributions” $\int_{\Bbb R} K(z,t) \phi(t/R)\ dt$ and $\int_{{\Bbb C}^n} K(z,t) \phi(z/R)\ dz$ to themselves be Calderòn-Zygmund kernels on ${\Bbb C}^n$ and ${\Bbb R}$ respectively, uniformly in $R > 0$.
The first main result of Nagel et al. is that the class of such operators is an algebra, and is bounded on $L^p({\Bbb H}^n)$ for all $1 < p < \infty$. Furthermore, this algebra of convolution operators can be extended to a larger algebra of “hybrid pseudodifferential operators”
$\displaystyle Tf(x) := \int_{{\Bbb H}^n} K(x,y) f(xy^{-1})\ dy$ (3)
where for each fixed x, $K(x,y)$ (and more generally $\partial_x^\alpha K(x,y)$) is a kernel of the above form, uniformly in x. This algebra contains both standard pseudodifferential operators and Calderòn-Zygmund operators on the Heisenberg group. The former turn out to be “almost central” in the sense that the commutator between a pseudodifferential operator and any other operator in this algebra is a smoothing operator (more precisely, it gains 1/2 of a Euclidean derivative in $L^2$, and a proportionally lesser amount of regularity in other $L^p$ spaces). Similar results exist for other nilpotent groups than the Heisenberg group, as discussed a little later.
One major difficulty here is that these kernels have a more complicated singularity than just being singular along the diagonal $x=y$, in that the region where the kernel is large also concentrates along various higher-dimensional spaces, such as the space $\{ ((z,t), (z',t')): x=x' \}$. Because of this, the second part of the Calderòn-Zygmund paradigm – i.e. leveraging $L^2$ boundedness to obtain $L^p$ bounds – does not work well any more. Instead, what Eli and his coauthors did was to go back to an older paradigm, the product paradigm, in which results in multidimensional (or multiparameter) harmonic analysis were obtained by iterating results in one-dimensional (or one-parameter) harmonic analysis.
To explain this paradigm, Eli went back to one of the very first applications of this paradigm, namely the Marcinkiewicz multiplier theorem, proven in 1933, and used among other things to show that the Riesz-like transforms $\partial_{x_i} \partial_{x_j} \Delta^{-1}$ were bounded on $L^p({\Bbb R}^n)$ for $1 < p < \infty$. The precise statement of the theorem is slightly technical and was not given here, but the main ingredient of this theorem was the multidimensional Littlewood-Paley inequality, which in turn followed from iterations of the one-dimensional Littlewood-Paley inequality. To describe this in more detail, Eli then discussed an improvement of the Marcinkiewicz multiplier theorem due to R. Fefferman and Stein, concerning product singular integral operators, which serves as a partial model for the more recent results of Nagel et al.. For simplicity, let us consider convolution operators $T: f \mapsto f * K$ on the plane ${\Bbb R}^2 \equiv {\Bbb R} \times {\Bbb R}$, although the theory extends to arbitrary products of finitely many Euclidean spaces. We define a product kernel to be a distribution K, smooth away from the coordinate axes $\{ (x_1,x_2): x_1=0 \hbox{ or } x_2 = 0 \}$, that obeys the kernel bounds
$|\partial_{x_1}^{\alpha_1} \partial_{x_2}^{\alpha_2} K(x_1,x_2)| \leq C_{\alpha_1, \alpha_2} |x_1|^{-1-|\alpha_1|} |x_2|^{-1-|\alpha_2|}$
for all $\alpha_1, \alpha_2 \geq 0$ and all non-zero $x_1, x_2$. We also require two cancellation conditions: firstly, that $|\langle K, \phi_{R_1,R_2} \rangle| \leq C$ for all $R_1,R_2 > 0$, where $\phi_{R_1,R_2}(x_1,x_2) := \phi(x_1/R_1,x_2/R_2)$ and $\phi$ is a non-trivial bump function, and secondly that the marginal kernels $\int_{{\Bbb R}} K(x_1,x_2) \phi(x_1/R_1)\ dx_1$ and $\int_{{\Bbb R}} K(x_1,x_2) \phi(x_2/R_2)\ dx_2$ are one-dimensional Calderòn-Zygmund operators. [Examples of such kernels include the classical two-dimensional Calderòn-Zygmund kernels on ${\Bbb R}^2$, as well as tensor products $K_1(x_1) K_2(x_2)$ of one-dimensional Calderòn-Zygmund kernels.]
The main result of R. Fefferman and Stein was that such convolution operators are bounded on $L^p({\Bbb R}^2)$ for all $1 < p < \infty$ and form a (commutative) algebra. The proof proceeds by the product paradigm. Firstly, one observes that kernels in the above class can be characterised in two different manners, much as in the classical case. On the Fourier-analytic side, the Fourier transform $m = \hat K$ of a product kernel obeys the product symbol estimates $|\partial_{\xi_1}^{\alpha_1} \partial_{\xi_2}^{\alpha_2} m(\xi_1,\xi_2)| \leq C_{\alpha_1,\alpha_2} |\xi_1|^{-|\alpha_1|} |\xi_2|^{-|\alpha_2|}$ for all non-zero $\alpha_1,\alpha_2$. This is enough to establish $L^p$ boundedness, but the non-local nature of the singularity means that the usual Calderòn-Zygmund argument to boost the $L^2$ bound to $L^p$ bounds no longer works well. Instead, one uses the third type of characterisation of these kernels, namely that they can be decomposed as $K = \sum_I \phi_I$, where $I = (i_1,i_2) \in {\Bbb Z}^2$, $\phi_{i_1,i_2}(x_1,x_2) = 2^{i_1+i_2} \phi^{(i_1,i_2)}( 2^{i_1} x_1, 2^{i_2} x_2)$, and the $\phi^{(i_1,i_2)}$ are bump functions uniformly in $i_1,i_2$ that satisfy the marginal cancellation conditions $\int_{\Bbb R} \phi^{(i_1,i_2)}(x_1,x_2)\ dx_2 = 0$ and $\int_{\Bbb R} \phi^{(i_1,i_2)}(x_1,x_2)\ dx_1 = 0$.
This characterisation can be combined with the Littlewood-Paley square function inequality to establish the theorem of R. Fefferman and Stein. To explain this, Eli returned to the one-dimensional (or one-parameter) theory. If $T: f \mapsto f*K$ is a classical Calderòn-Zygmund convolution operator, then by decomposing K into dyadic pieces it is not difficult to show a pointwise estimate of the form
$|P_j ( Tf )| \leq C M |\tilde P_j f|$ (4)
for all j, where $P_j: f \mapsto f * \phi_j$ is a Littlewood-Paley type projection (thus $\phi_j(x) = 2^j \phi(2^j x)$ and $\phi$ has mean zero), M is the Hardy-Littlewood maximal function, and $\tilde P_j$ is a slightly larger variant of $P_j$ (chosen so that $P_j = P_j \tilde P_j$, and hence $P_j(Tf) = P_j(T \tilde P_j f)$ by commutativity). The $L^p$ boundedness of T can then be deduced from the Littlewood-Paley square function inequality
$\| (\sum_j |P_j(f)|^2)^{1/2} \|_{L^p} \sim \|f\|_{L^p}$ (5)
and the variant
$\| (\sum_j |M \tilde P_j(f)|^2)^{1/2} \|_{L^p} \sim \|f\|_{L^p}$ (6)
(which follows by combining the square function inequality with the vector-valued maximal inequality of C. Fefferman and Stein).
The same method can be applied in the product setting, using the product version of the Littlewood-Paley inequality. On ${\Bbb R}^2$, for instance, one can introduce the Littlewood-Paley projections $P_{j_1,j_2} := P^1_{j_1} P^2_{j_2}$ for $(j_1,j_2) \in {\Bbb Z}^2$, defined by first applying a one-dimensional Littlewood-Paley projection $P^1_{j_1}$ in the $x_1$ variable (keeping $x_2$ frozen), and then applying a one-dimensional Littlewood-Paley projection $P^2_{j_2}$ in the $x_2$ variable (keeping $x_1$ frozen). For the product operators T considered by R. Fefferman and Stein, there is a product analogue of (4), namely
$|P_J ( Tf )| \leq C {\mathcal M} |\tilde P_J f|$ (7)
for all $J = (j_1,j_2) \in {\Bbb Z}^2$, where $\tilde P_J$ is again a variant of $P_J$, and ${\mathcal M} =M_1 M_2$ is the strong maximal function, defined by composing the two one-dimensional Hardy-Littlewood maximal functions. There are analogues of the Littlewood-Paley inequalities (5), (6), which can be proven by concatenating various one-dimensional inequalities (basically, one needs certain “vector-valued” extensions of (5), (6)). From these facts, the proof of the theorem of R. Fefferman and Stein is relatively straightforward.
The moral of the above story is this: the product operator T was not itself a tensor product of lower-dimensional operators, but it was nevertheless controlled by objects (such as the Littlewood-Paley square function and the strong maximal function) which were products of lower-dimensional objects, thus allowing this operator to be controlled by the lower-dimensional theory.
Eli now returned to the Heisenberg group ${\Bbb H}^n$ to apply the product paradigm to convolution operators $f \mapsto f*K$ such as convolution with the kernel (1). Note that kernels like (1) are already almost of product form; morally we have
$\displaystyle K(z,t) = \frac{z_j z_k}{(|z|^2+|t|^2)^{n+1}} \frac{1}{|z|^2+it} \leq \frac{z_j z_k}{|z|^{2n+2}} \frac{1}{it}$
which suggests that K is “dominated” by the tensor product of two classical kernels. If we were dealing with a Euclidean convolution rather than a Heisenberg convolution, the product theory of R. Fefferman and Stein would then presumably allow us to conclude the bounds one wants for this operator. So the main new difficulty is how to deal with the “twisted” nature of Heisenberg convolution.
It turns out that the correct way to deal with this is to extend this class of hybrid kernels to a larger class of kernels, which were singular on a flag of subspaces rather than at a point. In the Heisenberg case, the relevant flag is $\{0\} \subset \{ (0,t): t \in {\Bbb R} \} \subset {\Bbb H}^n$. The corresponding kernels are those that are smooth on $\{ (x,t): x \neq 0 \}$ and obey the estimates
$|\partial_t^\alpha \partial_{z,\overline{z}}^\beta K(z,t)| \leq C_{\alpha,\beta} |z|^{-2n-|\alpha|} |(z,t)|_H^{-2-2|\beta|}$
for all multi-indices $\alpha,\beta$, where $|(z,t)|_H := |z|+|t|^{1/2}$. We also need some cancellation conditions; firstly, that $|\langle K, \phi_{r,R} \rangle| \leq C$ whenever $0 < R^2 \leq r^2 \leq R$, where $\phi_{r,R}(z,t) := \phi(z/r,t/R)$, and secondly that the marginal distributions of K are themselves Calderòn-Zygmund kernels.
Examples of such operators includes multipliers of the form $m( {\mathcal L}, iT)$, where ${\mathcal L} :=- \sum_{i=1}^n X_i^2+Y_i^2$ is the sub-Laplacian (which commutes with T), and m is a product symbol of order 0. (In fact, every convolution operator whose kernel obeys the properties of the previous paragraph and is invariant under rotations of the z variable, is of this form.)
The work of Nagel et al. shows that these flag kernel convolution operators are bounded on $L^p$ and form an algebra; furthermore, there is a “variable-coefficient” extension of this algebra involving operators such as (3). To establish this claim, it is in fact convenient to work in a more general setting than a Heisenberg group, namely that of a graded simply connected nilpotent Lie group G. This group can be described as the exponential $G = \exp{\mathfrak g}$ of its Lie algebra ${\mathfrak g}$, which in turn can be split as a direct sum ${\mathfrak g} = \bigoplus_{j=1}^k {\mathfrak h}_j$, where the ${\mathfrak h}_j$ are subspaces obeying the commutation relations ${}[{\mathfrak h}_j, {\mathfrak h}_l] \subset {\mathfrak h}_{j+l}$ (with the convention that ${\mathfrak h}_j$ is trivial for $j > k$). This grading leads to a filtration (or flag)
${\mathfrak g} = {\mathfrak g}_1 \supset \ldots \supset {\mathfrak g}_k \supset \{0\}$
defined by ${\mathfrak g}_j := {\mathfrak h}_j \oplus \ldots \oplus {\mathfrak h}_k$. Exponentiating this leads to a corresponding filtration of Lie groups
$G = G_1 \geq G_2 \geq \ldots \geq G_k \geq \{\hbox{id}\}.$
We parameterise elements of ${\mathfrak g} \equiv G$ by $(x_1,\ldots,x_k)$, where $x_j \in {\mathfrak g}_j$. We then have a natural scaling
$(x_1,\ldots,x_k) \mapsto (\delta x_1, \ldots, \delta^k x_k)$
that respects all the previous structures, and gives the Lie algebra a homogeneous dimension of $Q := \sum_{j=1}^k j \hbox{dim}({\mathfrak g}_j)$. We also have some “semi-norms”
$N_j(x_1,\ldots,x_k) := |x_1|^j + |x_2|^{j/2} + \ldots + |x_j|$
which can be viewed as a variant of $|x_j|$ (for instance, they behave the same way with respect to scaling). We then define a flag kernel to be a function K on ${\mathfrak g} \equiv G$ that is smooth on the set $x_1 \neq 0$, obeys the derivative bounds
$|\partial_{x_1}^{\alpha_1} \ldots \partial_{x_k}^{\alpha_k} K(x_1,\ldots,x_k)| \leq C_{\alpha_1,\ldots,\alpha_k} \prod_{j=1}^k N_j(x_j)^{-\hbox{dim}({\mathfrak g}_j)-|\alpha_j|}$,
is such that $|\langle K, \phi_R \rangle| \leq C$ for all “acceptable” radii $R = (r_1,\ldots,r_k)$, where $\phi_R(x_1,\ldots,x_k) = \phi(x_1/r_1,\ldots,x_k/r_k)$ and “acceptable” means that $r_1^k \leq r_2^{k/2} \leq \ldots \leq r_k$, and such that every marginal integral of K is itself a flag kernel of lower dimension (so this is a recursive definition). The main theorem here is, once again, that convolution operators with flag kernels are bounded on $L^p(G)$ for every $1 < p < \infty$ and form a (non-commutative) algebra. Once one has this result (as well as a variable coefficient generalisation of this result), all the other results claimed earlier follow fairly easily.
[The main reason we restrict attention to “acceptable” radii scales is that the corresponding bump functions $\phi_R$ are stable under twisted convolution: the convolution of two such $\phi_R$ is essentially a multiple of another $\phi_R$.]
Eli then briefly sketched the main ingredients of the proof of the above theorem. The first step is to introduce Littlewood-Paley operators $P^{(m)}_j$ for each subgroup $G_m$, which one can then lift to act also on the whole group G (by foliating G into cosets of the normal subgroup $G_m$). One can then create multiparameter Littlewood-Paley operators $P_J = P_{j_k}^{(k)} \ldots P_{j_1}^{(1)}$ for $J = (j_1,\ldots,j_k) \in {\Bbb Z}^k$. (One has to be a little careful here because these operators don’t commute with each other, but as long as one orders things consistently, things turn out to be OK.) One also has a strong maximal function ${\mathcal M}$ associated to acceptable boxes. The main proposition is then a variant of (7), namely that
$|P_J T P_I^* f| \leq C 2^{-c|I-J|} {\mathcal M} f$ (8)
for all $I, J \in {\Bbb Z}^k$, where $c > 0$ is an absolute constant. One can use this inequality to deduce everything one needs using appropriate variants of the Littlewood-Paley square function inequality as before.
The proof of (8) is somewhat complicated by the twisted nature of convolution on nilpotent Lie groups. A key technical step in the proof of (8) (and one which, apparently, held up the project for quite some time) was to show that the class of flag kernels was stable under various “acceptable” changes of variables, such as the change from canonical coordinates $\exp(\sum_i a_i X_i ) \to (a_1,\ldots,a_n)$ of the first kind for the Lie group G, to canonical coordinates $\prod_i \exp(a_i X_i) \to (a_1,\ldots,a_n)$ of the second kind. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 147, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9512866139411926, "perplexity": 379.6040244019248}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218193716.70/warc/CC-MAIN-20170322212953-00429-ip-10-233-31-227.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-algebra/160222-question-rings-print.html | # Question in rings
• October 19th 2010, 05:04 AM
raed
Question in rings
Hi All:
I need the solution of the following question:
Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.
Best wishes
• October 19th 2010, 06:03 AM
tonio
Quote:
Originally Posted by raed
Hi All:
I need the solution of the following question:
Let (R,+,.) satisfy all conditions of a ring except the condition of a+b=b+a for all a ,b in R ( abelian with +); If 1 in R (R is arig with unity) prove that a+b=b+a for all a,b in R.
Best wishes
Existence of additive inverse: $(a+b)+(-(a+b))=0$
But also $a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0$ , so by uniqueness of inverse
$-b-a = -(a+b)=-a-b$ .
Justify each step above and end the argument.
Tonio
• October 19th 2010, 06:30 AM
raed
Quote:
Originally Posted by tonio
$a+b+(-b-a)=a+(b+(-b))+(-a)=a+0+(-a)=a+(-a)=0$ , so by uniqueness of inverse
$-b-a = -(a+b)=-a-b$ .
Tonio
Thank for your reply but this ring is not a belian under additive is it true that $-(a+b)=-a-b=-b-a$
• October 19th 2010, 08:56 AM
tonio
Quote:
Originally Posted by raed
Thank for your reply but this ring is not a belian under additive is it true that $-(a+b)=-a-b=-b-a$
I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:
$-(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b$ . That this equals $-b-a$ follows from uniqueness of the additive inverse.
Tonio
• October 20th 2010, 12:35 AM
raed
Quote:
Originally Posted by tonio
I didn't say it was: that's what you must prove! But you said that "ring" fulfills all the other ring axioms, so there's distributivity:
$-(a+b)=(-1)(a+b)=(-1)a+(-1)b=-a-b$ . That this equals $-b-a$ follows from uniqueness of the additive inverse. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8252274990081787, "perplexity": 2238.8702999586044}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398448227.60/warc/CC-MAIN-20151124205408-00308-ip-10-71-132-137.ec2.internal.warc.gz"} |
https://math.libretexts.org/Bookshelves/Algebra/Book%3A_College_Algebra_and_Trigonometry_(Beveridge)/03%3A_Exponents_and_Logarithms/3.03%3A_Solving_Exponential_Equations | $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.3: Solving Exponential Equations
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Because of the fact that logarithms are exponents, the rules for working with logarithms are similar to those that govern exponential expressions. One very helpful rule of equality for working with logarithms is related to the exponential rule for raising a power to a power. We recall one of the rules of exponents as:
$\left(b^{x}\right)^{y}=b^{x * y}$
in other words
$\left(5^{2}\right)^{4}=\left(5^{2}\right)\left(5^{2}\right)\left(5^{2}\right)\left(5^{2}\right)=5^{2 * 4}=5^{8}$
In logarithmic notation, this rule works out as:
$\log _{b} M^{p}=p * \log _{b} M$
The reason for this comes from the rule for exponents. Let's say that $$\log _{b} M=x$$
Then:
$\log _{b} M=x$
this means that
$b^{x}=M$
and
$\begin{array}{c} \left(b^{x}\right)^{p}=(M)^{p} \\ \text { so } \\ b^{p * x}=M^{p} \end{array}$
Now, we come back to the question of $$\log _{b} M^{p}=?$$. This expression $$\left(\log _{b} M^{p}\right)$$ is asking the question "What power do we raise $$b$$ to in order to get an answer of
$$M^{p} ?$$ The result on the previous page shows that:
$b^{p x}=M^{p}$
This means that we must raise $$b$$ to the $$p x$$ power to get an answer of $$M^{p}$$. Remember that $$x=\log _{b} M .$$ This means that:
$\begin{array}{c} b^{p x}=M^{p} \\ \text { so } \\ \log _{b} M^{p}=p x=p * \log _{b} M \end{array}$
This statement of equality is useful if we are trying to solve equations in which the variable is an exponent.
Example
Solve for $$x$$
$4^{x}=53$
We start by taking a logarithm on both of the equation. Just as we can add to both sides of an equation, or multiply on both sides of an equation, or raise both sides of an equation to a power, we can also take the logarithm of both sides. So long as two quantities are equal, then their logarithms will also be equal.
\begin{aligned} 4^{x} &=53 \\ \log 4^{x} &=\log 53 \\ x \log 4 &=\log 53 \\ x &=\frac{\log 53}{\log 4} \approx 2.864 \end{aligned}
since the log base 10 and log base $$e$$ are both programmed into most calculators, these are the most commonly used bases for logarithms.
Example
Solve for $$x$$
$5^{2 x+3}=17$
We start this problem in the same fashion, but this time we will use a logarithm to the base $$e:$$
$\begin{array}{c} 5^{2 x+3}=17 \\ \ln 5^{2 x+3}=\ln 17 \\ (2 x+3) \ln 5=\ln 17 \end{array}$
There are several possibilities for finishing the problem from this point. We will focus on two of them that are the most useful for solving more complex problems. First, we will distribute the $$\ln 5$$ into the parentheses and then get the $$x$$ by itself.
\begin{aligned} (2 x+3) \ln 5 &=\ln 17 \\ x * 2 \ln 5+3 \ln 5 &=\ln 17 \\ -3 \ln 5 &=-3 \ln 5 \\ x * 2 \ln 5 &=\ln 17-3 \ln 5 \\ x &=\frac{\ln 17-3 \ln 5}{2 \ln 5} \approx-0.620 \end{aligned}
And we can check the answer by plugging it back in:
$5^{2 *(-0.0620)+3} \approx 5^{1.760} \approx 16.9897 \approx 17$
We can also approximate the logarithms in the problem and solve for an approximate answer:
\begin{aligned} (2 x+3) \ln 5 &=\ln 17 \\ x * 2 \ln 5+3 \ln 5 &=\ln 17 \\ 3.2189 x+4.8283 & \approx 2.8332 \\ -4.8283 & \approx-4.8283 \\ 3.2189 x & \approx-1.9951 \\ x & \approx-0.620 \end{aligned}
If you use the method of approximating, it's important to make a good approximation. At least $$4-5$$ decimal places are necessary for an accurate answer.
Let's look at an example that has variables on both sides of the equation:
Example
Solve for $$x$$
$4^{3 x}=9^{2 x-1}$
We'll use log base 10 in this problem.
\begin{aligned} 4^{3 x} &=9^{2 x-1} \\ \log 4^{3 x} &=\log 9^{2 x-1} \\ 3 x * \log 4 &=(2 x-1) \log 9 \\ x * 3 \log 4 &=x * 2 \log 9-\log 9 \end{aligned}
If we collect like terms, we'll end up with:
\begin{aligned} x * 3 \log 4 &=x * 2 \log 9-\log 9 \\ \log 9 &=x * 2 \log 9-x * 3 \log 4 \end{aligned}
At this point, if we want to get the $$x$$ by itself, we need to factor out the $$x$$ on the right-hand side:
$\begin{array}{l} \log 9=x * 2 \log 9-x * 3 \log 4 \\ \log 9=x(2 \log 9-3 \log 4) \end{array}$
Then divide on both sides by the coefficient in parentheses:
$\frac{\log 9}{2 \log 9-3 \log 4}=\frac{x\cancel{(2 \log 9-3 \log 4)}}{\cancel{2 \log 9-3 \log 4}} \\ \frac{\log 9}{2 \log 9-3 \log 4}=x \\ 9.327 \approx x$
Again, we can check our answer by plugging it back into the equation:
$4^{3 * 9.327} \approx 4^{27.981} \approx 7.0184 * 10^{16}$
$9^{2 * 9.327-1} \approx 9^{17.654} \approx 7.0177 * 10^{16}$
We could also have solved this equation by approximating the logarithms in the beginning.
\begin{aligned} 4^{3 x} &=9^{2 x-1} \\ \log 4^{3 x} &=\log 9^{2 x-1} \\ 3 x * \log 4 &=(2 x-1) \log 9 \\ 3 x(0.60206) & \approx(2 x-1) 0.95424 \\ 1.80618 x & \approx 1.9085 x-0.95424 \\ 0.95424 & \approx 0.10232 x \\ 9.326 & \approx x \end{aligned}
This answer is less accurate than the other approximation $$(9.326036 \text { vs. } 9.327424)$$ The accuracy of an answer depends upon the original approximations for the logarithms.
Solve for the indicated variable.
1) $$\quad 2^{x}=5$$
2) $$\quad 2^{x}=9$$
3) $$\quad 3^{x}=7$$
4) $$\quad 3^{x}=20$$
5) $$\quad 2^{x+1}=6$$
6) $$\quad 7^{x+1}=41$$
7) $$\quad 5^{x+1}=36$$
8) $$\quad 8^{x-2}=6$$
9) $$\quad 4^{2 x+3}=50$$
10) $$\quad 4^{x+2}=5^{x}$$
11) $$\quad 5^{2 x+1}=9$$
12) $$\quad 6^{x+4}=10^{x}$$
13) $$\quad 7^{y+1}=3^{y}$$
14) $$\quad 2^{x+1}=3^{x-2}$$
15) $$\quad 6^{y+2}=5^{y}$$
16) $$\quad 7^{x-3}=3^{x+1}$$
17) $$\quad 6^{2 x+1}=5^{x+2}$$
18) $$\quad 9^{1-x}=12^{x+1}$$
19) $$\quad 5^{2 x-1}=3^{x-3}$$
20) $$\quad 3^{x-2}=4^{2 x+1}$$
21) $$\quad 8^{3 x-2}=9^{x+2}$$
22) $$\quad 2^{2 x-3}=5^{-x-1}$$
23) $$\quad 10^{3 x+2}=5^{x+3}$$
24) $$\quad 5^{3 x}=3^{x+4}$$
25) $$\quad 3^{x+4}=2^{1-3 x}$$
26) $$\quad 4^{2 x+3}=5^{x-2}$$
27) $$\quad 3^{2-3 x}=4^{2 x+1}$$
28) $$\quad 2^{2 x-3}=5^{x-2}$$
3.3: Solving Exponential Equations is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Richard W. Beveridge. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9999847412109375, "perplexity": 486.1300880687055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662510138.6/warc/CC-MAIN-20220516140911-20220516170911-00016.warc.gz"} |
http://math.stackexchange.com/questions/69810/what-is-the-origin-of-the-prefix-logic-notation-used-in-wff-n-proof?answertab=active | What is the origin of the prefix logic notation used in WFF 'N PROOF?
The classic "modern logic" game of WFF 'N PROOF uses a set of symbols to represent logical relations that I've seen used nowhere else: $C$ for then; $A$ for or; $K$ for and; $E$ for if and only if; and $N$ for not. These are used in prefix notation so that, for example $$CNpq$$ means $$\neg p\Rightarrow q,$$ and $$EpAqNr$$ means $$p\Leftrightarrow \left({q\vee\neg r}\right).$$
What is the origin of these symbols and have they been used widely or elsewhere? Are they still in use?
-
Your link contains a recent forum thread debating that very question. – Henning Makholm Oct 4 '11 at 18:52
It's polish notation. For example, $K$ stands for koniunkcja, from the same root as "conjunction".
-
It was called "Polish notation" because of people who could not remember the name Łukasiewicz. – GEdgar Oct 4 '11 at 18:47
Some examples of the use of this notation. – raxacoricofallapatorius Oct 5 '11 at 12:48
The "Polish" notation was introduced by my academic grandfather, Jan Łukasiewicz (1878-1956).
A famous joint paper with Tarski, "Untersuchungen über den Aussagenkalkül," (1930) contains the phrase "Łukasiewicz employs the formulas Cpq and Np." A footnote indicates that Łukasiewicz used the notation in a paper from 1929 and in his famous text Elementy logiki matematycznej (1929) which contains the contents of a course given "in the autumn trimester of the academic year 1928/9". In a paper by Łukasiewicz from 1931 a footnote reports that "I came upon the idea of a parenthesis-free notation in 1924." Thus 1924 seems to be the most likely date of the creation of Łukasiewicz's parenthese free notation.
The notation has been, and continues to be, widely used in the study of propositional calculi.
-
This answer deserves to get accepted. By "academic grandfather" do you mean that your thesis adviser had Łukasiewicz as his/her thesis adviser (or someone further down the line)? Also, and I guess this wanders a bit... I've read that C. A. Meredith developed condensed detachment building on Łukasiewicz's work. I'd like to know how he did this, but I don't have access to his work. Did Meredith basically get the idea or build on the way in Łukasiewicz notated his proofs? – Doug Spoonwood Jun 15 '13 at 2:55
On the work of C. A. Meredith, see "In memoriam: Carew Arthur Meredith (1904--1976)" Notre Dame Journal of Formal Logic, Volume 18, Number 4 (1977), 513-516, by his cousin David Meredith. Google – Fred Rickey Jun 15 '13 at 3:52
The paper is on line: projecteuclid.org/euclid.ndjfl/1093888116. Yes, my advisor, Boleslaw Sobocinski was a PhD student of Łukasiewicz. – Fred Rickey Jun 15 '13 at 4:03 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8569793701171875, "perplexity": 1939.3322618454727}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413558067467.26/warc/CC-MAIN-20141017150107-00123-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/radial-free-fall-proper-time.232916/ | # Radial Free Fall Proper Time
• Start date
• #1
11
0
Foster and Nightingale give an equation for the proper time of a radially falling clock between any two distances. (Short Course in General Relativity, 1979, p. 128.) As they state, the equation is the "same as its Newtonian counterpart." So there are no factors representing any slowing of the clocks. This is confusing to me because it seemingly implies that such a clock suffers no time dilation. But that doesn't seem right.
Consider an experiment such as that of Vessot-Levine, but suppose the clock is simply dropped from rest at "apogee" over a large non-rotating planet with no atmosphere. Suppose the time on the clock is set to zero at the moment of release. At rigidly built observing stations alongside the path and at the surface the elapsed time on the falling clock is recorded. The clock's frequency is affected by its speed and its location in the gravitational field. (The frequency of the Vessot-Levine clock was monitored from a distance using a clever first-order Doppler canceling system. But they did not measure elapsed proper time.) My question is, how would these idealized proper time observations differ from the predictions of Newton's theory, in which all clocks always tick at the same rate? Or is it that the combined effects of acquired falling speed and spacetime curvature (i.e, the difference between proper and coordinate distance) "conspire" to make the Newtonian equation match the General Relativistic prediction?
• #2
3,962
20
Foster and Nightingale give an equation for the proper time of a radially falling clock between any two distances. (Short Course in General Relativity, 1979, p. 128.) As they state, the equation is the "same as its Newtonian counterpart." So there are no factors representing any slowing of the clocks. This is confusing to me because it seemingly implies that such a clock suffers no time dilation. But that doesn't seem right.
Hi Stoonroon,
This is an interesting question. Could you post the Foster and Nightingale equation so that we can discuss it?
Consider an experiment such as that of Vessot-Levine, but suppose the clock is simply dropped from rest at "apogee" over a large non-rotating planet with no atmosphere. Suppose the time on the clock is set to zero at the moment of release. At rigidly built observing stations alongside the path and at the surface the elapsed time on the falling clock is recorded. The clock's frequency is affected by its speed and its location in the gravitational field. ....
I am not sure but I think the frequency of the falling clock is affected by its frequency but not its location in the gravitational field. This is because the clock is experiencing no proper acceleration as it falls. As I said, I'm not sure and I hope the experts on this forum can give a more definitive answer.
There seems to scant information on the details of the Vessot-Levine experiment (Gravity Probe A) freely available on the internet. It seems ironic that the billions of dollars to fund scientific research on a big scale is ultimately paid for by the general public and yet we have to pay more to be able to read the original papers. :(
Getting off my soap box and back to the original question, there are a number of obvious factors to take into account (and I am sure a greater number of not obvious factors). First, if the observing stations are spaced at regular intervals on a sort of ladder built in space, then when the ladder is lowered into the gravitational field it will be length contracted with the lower parts being contracted to a greater extent. The clocks on the ladder will each be time dilated according to their location in the gravitational field. The falling clock will tick like a clock in flat space far from any gravitational field but it will also be knematically time dilated according to its falling velocity which will obviously be increasing as it falls. Signals sent from the falling clock to a ground station will be doppler shifted and this shift will have to be corrected to obtain the proper time of the falling clock. I think the first order doppler self correction that you mention applies to the functioning of the maser clock itself rather than doppler shift of the signals sent from it, as far as I can ascertain.
I have not done the calculations, but I think the Schwarzschild equation tells us something about the proper time of a falling clock compared to coordinate time, but I am not sure if there is an unambiguous solution in this case.
• #3
11
0
The Foster and Nightingale equation is attached as a jpg. The initial distance is given by r_0 and the final distance by r.
The General Relativity equation for the clock frequency has in its argument one term for the location ("potential") and one for the speed (squared). They are simply added to give the total effect.
Although I'd still like to see an explicit authoritative assessment, as I contemplate the problem, it seems that use of the Newtonian equation makes sense at least insofar as it assures that both the falling clock and ground-based observers should both measure the same speed. Whereas, if the clock were falling from infinity (and I think also from 2R) the coordinate speed would be diminished by the metric coefficient, 1-2GM/rc^2. (As per Hobson, Efstathiou and Lasenby, General Relativity, p. 210.)
In the Vessot-Levine experiment, the natural clock frequency was not altered; the first order Doppler cancelation was achieved by having a separate out-back link that facilitated the subtraction. (I recommend their original papers, which you may have to go to a university physics library to find.)
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https://brilliant.org/discussions/thread/could-it-be-possible/?sort=new | Could it be Possible ? Maybe Confusion
I was scribbling in my note yesterday, when I thought to write something about infinity algebraic expressions. After many tried sums, I had found out a very strange expression in my note.
As I tried to solve it, I saw something very keen and interesting which is in the image.
Is it true ? I too don't think so, but think maybe could be true Like if agree. . Share if you find confusing and interesting or disagree
Note by Anish Harsha
5 years, 6 months ago
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Thanks for getting me right !
- 5 years, 1 month ago
I agree with the others - infinity is a concept, not a number.
But aren't you contradicting your own statement? If [(infinity)^2 = infinity] and [-2infinity2 = infinity], then isn't infinity-2 = infinity??
- 5 years, 3 months ago
Infinity is used to indicate very large number which we can't even imagine. It is a theoretical concept.
- 5 years, 4 months ago
$\infty - \infty \neq 0$
so you can't do like that !!!
- 5 years, 4 months ago
infinity -infinity is not 0
- 5 years, 5 months ago
If you do what is in the brackets first you get ∞^2... which is still ∞
- 5 years, 5 months ago
Here lies the fallacy... infinity is a relative concept...its not an absolute number...hence it fails to satisfy laws of algebra...
- 5 years, 5 months ago
your 2 infinites are not equal in the second line {unless infintiy is 4, which clearly it isnt as it isnt a fixed number}
- 5 years, 5 months ago
infinity is something , which is taken sometimes for FOR DOES NOT EXIST condition or sometimes when we take a very large no.
- 5 years, 5 months ago
This reminds me of myself back then. No, infinity is an idea/concept. There's no way you can treat it as a number, thus you can't solve it.
Cheers!
- 5 years, 5 months ago
infinity is concept of numbers but it can't be defined so it's not true
- 5 years, 5 months ago
You have done wrong calculation.
- 5 years, 5 months ago
I don't exactly know how Infinity works but my teacher once told me that infinity minus infinity is not 0
- 5 years, 5 months ago
Lets consider infinite as n for simpler conventions to facilitate operations ...so n is large no. n^2 is of even larger order we cannot add them both as they are of diff orders /or subtract them.
- 5 years, 5 months ago
$\infty$ ^2 is not $\infty$
- 5 years, 5 months ago
It is
- 5 years, 1 month ago
OK I didn't know.
- 5 years, 1 month ago
No
- 5 years, 5 months ago
Here two things are wrong ..
1. The laws of algebra hold on real number system , $\infty$ and $- \infty$ don't belong to real number system , they can be thought as an extension to real number system.
2 . $\infty - \infty$ is not defined , it is an indeterminate form.
- 5 years, 5 months ago
It's = Infinity Because Infinity is a concept not a number. Thus it cannot be treated as a number. And also two infinitys are never equal.
- 5 years, 5 months ago
infinity - infinity is not 0
- 5 years, 5 months ago
Check into the Hilbert Hotel.
- 5 years, 5 months ago
Infinity isn't a number,its a thing you may say,so you can't treat it as a number
- 5 years, 5 months ago
It is not true. I don't understand what is it .
- 5 years, 5 months ago
Infinity is a concept not a number. Thus it cannot be treated as a number. And also two infinitys are never equal.
- 5 years, 6 months ago
If you go by the standard Set Theory, there are like 5 useful types of infinities. For example set of natural numbers is of the same size as the set of all primes.
- 5 years, 4 months ago
I perfectly agree with you. There may be infinite types of infinities
- 5 years, 5 months ago
You are absolutely true . There are many types of infinities like in various sequences.
- 5 years, 6 months ago
The fallacy is that $\infty$ is not a number and does not follow the laws of algebra.
Otherwise, $\infty = \infty + 1 \implies 1 = 0$, which is obviously not true.
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https://fusion.gat.com/global/theory/weekly/0810 | ### Theory Weekly Highlights for August 2010
##### August 27, 2010
By using an extended energy principle, the response of a tokamak to external magnetic perturbations has been shown to be well approximated by the vacuum response for a force free tokamak when there are no pitch resonances present. The effect of the plasma response introduces a paramagnetic effect that amplifies the vacuum response slightly. The amplification factor is inversely proportional to the aspect ratio and the safety factor. For the resonant components, paramagnetic amplification to relatively large amplitudes is possible inside of the resonant surfaces. However, the ideal MHD constraint of ‘frozen in flux’ at the resonant surfaces imposes a diamagnetic response. The vacuum response is then a poor approximation. The effect of plasma resistivity reduces the effectiveness of the ideal MHD constraint and moves the response towards being less diamagnetic. Similar behavior is expected for a tokamak with finite beta.
##### August 20, 2010
The new version of the GATO mapping which writes input for the NOVA code (see October 9 2009 highlight at Theory Weekly Highlights for October 2009) was simplified to produce only the two 1977-era direct access binary files, bypassing the earlier NetCDF option, since the NOVA code option to read the NetCDF files is not yet publicly available. In addition the changes were merged with the other changes to read inverse equilibria in an arbitrary angle format (see May 14 2010 highlight at Theory Weekly Highlights for May 2010), as well as the traditional equal-arclength formats. The new code, along with compatible changes in the remainder of the GATO code, is now ready to be released publicly. With the new mapping in particular, the NOVA code can now utilize equilibria from all the same sources as the GATO code, including direct equilibria from EFIT and inverse equilibria from a variety of formats, such as that produced by TOQ, CORSICA and CHEASE and JSOLVER, and the PPPL QSOLVER code that is currently embedded in the NOVA code scripts. For use with NOVA, the standard scripts need to be rewritten to bypass the QSOLVER equilibrium iterations and mapping and replaced by running the GATO mapping to produce the required direct access binary files from a selected input equilibrium.
##### August 13, 2010
A more complete analytic calculation of the effects of orbit squeezing on the ion neoclassical transport than the usual heuristic approach has been derived via solution of the hierarchy of drift-kinetic equations ordered in the ion gyroradius relative to the system size ρ*i = ρi/a. For the simple case of a single ion species with uniform temperature and assuming s-alpha geometry, an analytic solution was derived for the first-order (standard local neoclassical) and second-order distribution functions in both the banana and Pfirsch-Schluter collisional regimes and for the third-order distribution function in the latter regime. The third-order solution is necessary to study the non-local transport corrections due to finite-orbit-width effects since the transport coefficients from the second-order solution are zero for up-down symmetric plasmas. The radial dependence of the non-local transport is found to have a coupled dependence on higher-order derivatives of the geometry parameters, the ion density gradient dni/dr, and the radial electric field E = -∇Φ. This dependence is significantly more complicated than the usual orbit-squeezing factor, S ~ (d2 Φ/ dr2)(1/ni)(dni/dr), would imply.
##### August 06, 2010
A new set of IDL-based workflow tools has been developed to quantify statistical uncertainties in turbulent transport calculations and validation studies. These tools use as their starting point ensembles of Monte Carlo trial profile fits generated by the GAPROFILES tools for quantifying profile uncertainties. These profile ensembles are then given as inputs to generate ensembles of power balance calculations using the ONETWO code, and turbulent flux predictions with the TGLF quasilinear transport model. Initial results from application to a typical DIII-D L-mode discharge show that the dominant statistical uncertainty in all transport channels is due to the uncertainty in the ion temperature profile, and that the relative uncertainty generally decreases with radius. These tools were developed and tested in collaboration with P. Namasondhi, who is a summer NUF student, and the results will be presented in a poster at the fall APS meeting.
Disclaimer
These highlights are reports of research work in progress and are accordingly subject to change or modification | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8070369362831116, "perplexity": 1795.9600321497473}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250606872.19/warc/CC-MAIN-20200122071919-20200122100919-00427.warc.gz"} |
http://cpr-mathph.blogspot.com/2012/11/12115676-ricardo-weder.html | ## Universality of Entanglement Creation in Low-Energy Two-Dimensional Scattering [PDF]
Ricardo Weder
We prove that the entanglement created in the low-energy scattering of two particles in two dimensions is given by a universal coefficient that is independent of the interaction potential. This is strikingly different from the three dimensional case, where it is proportional to the total scattering cross section. Before the collision the state is a product of two normalized Gaussians. We take the purity as the measure of the entanglement after the scattering. We give a rigorous computation, with error bound, of the leading order of the purity at low-energy. For a large class of potentials, that are not assumed to be spherically symmetric, we prove that the low-energy behaviour of the purity, $\mathcal P$, is universal. It is given by $\mathcal P= 1- \frac{1}{(\ln (\sigma/\hbar))^2} \mathcal E$, where $\sigma$ is the variance of the Gaussians and the entanglement coefficient, $\mathcal E$, depends only on the masses of the particles and not on the interaction potential. The entanglement depends strongly in the difference of the masses. It takes its minimum when the masses are equal, and it increases rapidly with the difference of the masses.
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http://math.stackexchange.com/questions/260723/volume-of-a-solid | # Volume of a solid.
There's a solid which is a result of revolution of a rhombus by the axis which is paralell to the shorter diagonal and goes through the end-point of a longer diagonal. Longer diagonal is $\frac2{\sqrt{3}}$, sides are 2 and the angle between diagonals and sides is $60^{\circ}$. Calculate its volume.
I thought of using Pappus-Guldin theorem - area of rhombus multiplied by the perimeter of circle with radius made of longer diagonal. Is this a proper or I should divide it to cones?
P.S: Sorry for my english, I've never written anything related to math in this language until now.
-
add comment
## 1 Answer
According to the Pappus-Guldin (second) theorem the volume would be equal to the area of the rhombus times the distance travelled by its centroid which is obviously located at the intersection of the diagonals. Thus half of the larger diagonal must be taken as the radius.
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I was wondering if I should use half of the larger diagonal or it whole. Thanks for clarification on this theorem! – metrampaz Dec 17 '12 at 16:08
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https://motls.blogspot.com/2017/03/erepr-as-schur-orthogonality-relations.html?m=1 | Wednesday, March 29, 2017
ER=EPR as Schur orthogonality relations
The AdS/CFT correspondence relates a murky, effective description with quantum theory in the bulk – in AdS – to a well-defined, microscopic, non-gravitational theory on the boundary – CFT. I think that most people would agree that at least at present, the CFT side is the "more well-defined one", and the relationship therefore helps us to understand what quantum gravity (in this case in AdS) actually is.
I would like to have a more universal definition of quantum gravity that works for any superselection sector, whether the boundary behavior of the spacetime is flat Minkowski, AdS, or otherwise. What is the relationship between the low-energy field and some "detailed microscopic theory" in the most general case?
Witten's monstrous model of pure gravity in $$AdS_3$$ has been one of my favorite toy models that I have employed to check and refine various tools that I proposed for quantum gravity in general. Just to recall, the AdS/CFT dual should describe pure gravity in a 3-dimensional space. In $$D=3$$, the Ricci-tensor $$R_{\mu\nu}$$ and the Riemann tensor $$R_{\kappa\lambda\mu\nu}$$ both have six components. So the Ricci-flatness, i.e. Einstein's vacuum equations, imply the Riemann flatness. The vacuum must be flat. However, sources may create a deficit angle.
In the $$AdS_3$$ space, this statement is deformed by the extra cosmological constant, and the BTZ black hole becomes the most important black hole solution of classical general relativity in $$AdS_3$$. The simplest theory with no extra fields aside from the metric tensor must still allow black hole microstates by consistency.
A decade ago, Witten conjectured that there exists a cool AdS/CFT pair dual to the pure gravity in $$AdS_3$$. Davide Gaiotto later showed that the conjectured duality only holds for the minimum radius, i.e. the $$N=1$$ case. So that's the only case of interest. The dual CFT is a $$CFT_2$$ with $$c=24$$ which is pretty much the same CFT that was used to clarify the monstrous moonshine. It has the discrete monster group symmetry.
So one of the simplest realizations of general relativity – pure gravity in $$AdS_3$$ – seems to secretly carry the most impressive sporadic finite group that relates its microstates. The CFT with the monster group symmetry may be constructed in analogy with the bosonic construction of the heterotic string. But one doesn't use a 16-dimensional even self-dual lattice. Instead, one has to pick the 24-dimensional even self-dual lattice. There are some 24 inequivalent ones.
We must pick one of them, the Leech lattice, which is the unique 24-dimensional even self-dual lattice that only has sites with the length $$\ell^2=0$$, $$\ell^2=4$$, $$\ell^2=6$$, and so on. There are no sites with $$\ell^2=2$$ at all – which is linked to the fact that the dual AdS theory contains no massless fields aside from the metric tensor – which has no allowed vacuum waves, due to the Ricci-Riemann equivalence that I previously mentioned.
Great. So this theory only has the unit operator with $$\ell^2=0$$, at the origin of the lattice, and then various operators at $$\ell^2=4$$ and other operators with similar dimensions that transform as${\bf 196,883}\oplus {\bf 1}$ OK, all these objects must obviously be understood as black hole microstates – there's nothing else in the theory. Their density increases quasi-exponentially with the mass, as you know from CFTs, and they transform as representations of the monster group $$M$$. The GR intuition should be basically right qualitatively but you must be ready to embrace the fact that the corrections to some quantities may be of order 100%.
Now, the monster group has $$K=194$$ conjugacy classes. If you know the basic representation theory of finite groups, you're familiar with the amazing statement$K = R$ saying that the number of conjugacy classes is equal to the number of irreducible representations of the group, too. The dimensions of these 194 irreps may be found e.g. on this Subwiki page. The smallest ones have dimensions $$1$$ and $$196,833$$, of course, while the largest one has the dimension$258823477531055064045234375.$ Many of the large irreps are rather close in size to this one. $$146$$ of these irreps are real, $$48$$ of them are complex, coming as $$24$$ pairs of mutually complex conjugate irreps. (These 24 pairs contain all the DNA chromosome information for chimps, which are included in the monster, and the 146 non-paired chromosomes are those of a unicorn doll, but I don't want to overwhelm you with advanced monster biology.) Another fact that you remember from the basic representation theory is$|G| = \sum_{i=1}^{194} d_i^2.$ The number of elements of the monster group, about $$8\times 10^{53}$$, is the sum of the squared dimensions of all the irreps. This numerical fact, along with $$K=R$$ I mentioned before, may be understood as trivial consequences of stronger Schur orthogonality relations – which hold nicely for finite as well as compact Lie groups.
Wikipedia tells us: The space of complex-valued class functions of a finite group G has a natural inner product:$\left \langle \alpha, \beta\right \rangle := \frac{1}{ \left | G \right | }\sum_{g \in G} \alpha(g) \overline{\beta(g)}$ Just to be sure, a class function is a function mapping the group to the complex numbers that is constant all over each conjugacy class, i.e. one that obeys$\forall g,h\in M:\quad \alpha(hgh^{-1}) = \alpha(g).$ Can you find some really apt physical realization for these things? I think you can. Consider an Einstein-Rosen bridge – a non-traversable wormhole as discussed in the ER-EPR correspondence – and twist the throat by an element $$g\in M$$. So instead of connecting the two black holes, in our case the $$AdS_3$$ black holes, using the most trivial entanglement$\ket\psi = \sum_{i} \ket i \otimes \ket{i'}$ where the sum goes over some interval of masses or something like that, you replace $$\ket{i'}$$ above by the $$g$$-transformed element $$g\ket{i'}$$ for some $$g\in M$$. If there were a single wormhole microstate for every $$g$$, they would form a Hilbert space and the $$\left \langle \alpha, \beta\right \rangle$$ inner product could simply be the inner product on their Hilbert space. I think it's appropriate that we demand "class functions". Why? Because I believe that all symmetries in a theory of quantum gravity (including discrete symmetries) are gauge symmetries. So the physical states must be invariant under all these symmetries. In particular$\forall g\in M: \quad g \ket\psi = \ket\psi$ for all localized objects that may be isolated. Now, a wormhole with a twist given by $$h\in M$$ may also be transformed by the action of $$g$$, but if you do so, I believe that the twist $$h$$ gets conjugated. So the action of $$g$$ on the $$h$$-twisted wormhole is the $$ghg^{-1}$$-twisted wormhole. Does it make sense so far? So the condition that the wave functions on the space of the twisted wormholes are class functions is just a mathematical translation of the gauge invariance for objects with some monodromy – such as cosmic strings or wormholes.
If you continue reading the Wikipedia article, you will also learn that under the very same "sum over $$M$$" inner product above, the characters are orthogonal to each other:$\left \langle \chi_i, \chi_j \right \rangle = \begin{cases} 0 & \mbox{ if } i \ne j, \\ 1 & \mbox{ if } i = j. \end{cases}$ Here, $$\chi_i$$ where $$i=1,2,\dots,194$$ is the character i.e. the trace over the $$i$$-th irrep framed as a function of $$g$$$\chi_i(g) = {\rm Tr}_i (g).$ The orthogonality also works in the opposite direction. If you sum over all $$194$$ irreps (instead of summing over elements of the group),$\sum_{\chi_i} \chi_i(g) \overline{\chi_i(h)} = \begin{cases} \left | C_M(g) \right |, & \mbox{ if } g, h \mbox{ are conjugate } \\ 0 & \mbox{ otherwise.}\end{cases}$ you may determine whether $$g,h$$ belong to the same conjugacy class or not. The normalization $$|C_M(g)|$$ is the number of elements in the centralizer of $$g$$ within $$M$$. The centralizer of $$g$$ is the subgroup of all $$h\in M$$ that commute with $$g$$, i.e. $$gh=hg$$.
Many of these identities are deeply suggestive of the ER-EPR correspondence. In the Schur orthogonality relations, we're basically switching between two bases – one given by elements $$g\in M$$ and their conjugacy classes on one side; and one given by components of irreps and these irreps (and their characters $$\chi_i$$) on the other side. Within the ER-EPR correspondence, I propose to identify the former – the elements with the monster group – with the ER "bulk" description of operators within the wormhole; while the latter – the representations – should be identified with the EPR side of ER-EPR, i.e. with the representations of the monster group and their entanglement.
So I propose to study a "metaduality":
$$ER=EPR$$ is dual to the Schur orthogonality changes of bases.
Something like that should work, I think, but one should get much further. In the construction above, one only discussed the exact symmetry of the vacuum – which are completely unbroken – namely the monster group. This group should be considered a toy model for all gauge and global symmetries (and isometries of compactification etc.) that one encountered in a generic string/M-theoretical vacuum. (I hope that it's not too terrifying for you to call a monster your toy. As a kid, you should have played with monsters, too.)
A monster and a baby monster apparently eat tomatoes and kiwis, respectively.
But I would like to get much more from similar considerations than just some new duality that allows you to pick two different bases in a Hilbert space. I would like to generalize these constructions from the "twist of a wormhole" to all conceivable localized – and then local – operators you may think of, including the low-energy quantum fields. Those aren't generators of exact symmetries but if their energy is low, they may be close to it – think of all low-energy fields as some counterparts of the Nambu-Goldstone bosons (with the same idea that explains why these bosons are massless or light). So all these operators (especially when located in the new region of the spacetime, inside the wormhole) reshuffle some nearby microstates of the two black holes.
If there were some low-energy (and even not so low-energy) quantum fields inside the wormhole, they should be able to act on the Hilbert space of the two black holes – basically the tensor square of the state of all black hole microstates – in a certain way that generalizes the Schur orthogonality relations above. At the end, you should be able to see that the two black holes can't be an "exact" representation of an algebra of low-energy quantum fields. Instead, you should collide with some limits or restrictions of the Raju-Papadodimas type: the number of insertions can't be arbitrarily high etc.
Above, the idea was to start with the known spectrum of black hole microstates, use two such black holes, and study interesting operators acting on that space. In this way, we should get access to the "new region of the spacetime", namely inside the wormhole. The wormhole spacetime has a classically new, non-trivial topology. So the general lesson is that you could construct more complicated spacetimes and states with them from simpler ones.
It's possible that some complex enough spacetime, e.g. one we inhabit, could be constructed from many simpler ones, perhaps even $$AdS_3$$ spacetimes, by similar Schur-like transitions to completely new observables etc. The monster group example shows that quantum gravity demands huge and almost absolute constraints on the spectrum of the "pieces" – the two individual black holes' microstates – that you may entangle and where you may study the interesting operators that may be embedded. In fact, the pieces' microstates are constrained by the new operators in between and vice versa.
There should exist a well-posed definition of this problem and all perturbative string theory vacua – associated with a conformal field theory on the world sheet – should be a subclass of solutions to this problem. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8214274644851685, "perplexity": 365.6987237465364}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886110573.77/warc/CC-MAIN-20170822085147-20170822105147-00072.warc.gz"} |
http://math.stackexchange.com/questions/126625/pointwise-convergent-and-total-variation?answertab=votes | # Pointwise convergent and total variation
I'm preparing for a test for real analysis and I came across this problem in Royden's book:
Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges pointwisely on $[a,b]$ to the real valued function $f$. Show that $TV(f) \leq \liminf ~TV(f_n)?$
This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help?
(TV is short for total variation)
-
Fix $a=t_0<t_1<\ldots<t_N=b$ a subdivision of $[a,b]$. We have $$\sum_{j=1}^N\left|f(t_j)-f(t_{j-1})\right |=\lim_{n\to +\infty}\sum_{j=1}^N\left|f_n(t_j)-f_n(t_{j-1})\right|,$$ and let $u_n:=\sum_{j=1}^N\left|f_n(t_j)-f_n(t_{j-1})\right|$, $v_n:=\mathrm{ TV} (f_n)$. Since $u_n\leqslant v_n$ for each $n$, we have $\liminf_{n\to+\infty}u_n=\lim_{n\to+\infty}u_n\leqslant \liminf_{n\to \infty}v_n$. As a consequence, we have $$\sum_{j=1}^N|f(t_j)-f(t_{j-1})| \leqslant \liminf_{n\to \infty} \mathrm{ TV} (f_n).$$ Since the considered subdivision is arbitrary, it follows that $$\mathrm{ TV} (f)\leqslant \liminf_{n\to \infty} \mathrm{ TV} (f_n)$$ (because if $I$ is a set and $c_i\leqslant M$ for each $i$, then $\sup_{i\in I}c_i\leqslant M$).
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Davide and @Nana: Thanks so much, I see how it works now. You guys are awesome! – Vokram Apr 1 '12 at 4:07
@Davide Graudo, sorry I don't understand here how you took $TV(f)$ into account, since $TV(f)$ is the sup over a partition. You just did the regular sum form, how did you get the conclusion that it's also true when you take the sup of $\sum_{j=1}^N|f(t_j)-f(t_{j-1})|$ – Topoguy Jul 21 at 21:24
@Topoguy I have edited and added some details. – Davide Giraudo Jul 22 at 7:52
@DavideGiraudo Thanks. It makes sense for me. – Topoguy Jul 22 at 15:44
When do you think the equality is true? – Topoguy Jul 22 at 22:46
Hint:
Let $a=x_0\lt \ldots \lt x_n=b$ be a subdivision of $[a,b]$. Let $\varepsilon \gt 0$. Then there is an $M$ such that $$|f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2,$$ whenever $M<n$. Then consider
$$\sum_{k=1}^N |f(x_k) - f(x_{k-1})|\leq \sum_{k=1}^N |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^N |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^N |f_n(x_k) - f_n(x_{k-1})|.$$
Can you continue?
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9868113994598389, "perplexity": 211.0281274840328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065910.17/warc/CC-MAIN-20150827025425-00218-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://getrevising.co.uk/revision-tests/family-and-household-diversity | # Family and household diversity
HideShow resource information
S X P D Y T E R P Q X U B D R U H V G T M V E R G S Y D U A L W O R K E R U A G C T L F A S R H A I D K Y K Y Y E L O T C C G M T A X D Y E S O E V P C D J M U N M E C O H A B I T I N G C D D G P R D N S S X M Q T F G T K P W R M O X T I P E C T W X P I I A I A I T B Q F S V J X F D E P L D C L K K V I C Q K N D R M Q G L A C O V H S A W D C X C O Y Q N M A R Y M M M S P F E H K W S C R G U X U W B E U E F X F M P X N L E K K C T N N B N L F A O M R K A L U R M K L N R N E D H I W H H N J S A U B A U E E P V L Q C W N N D J J N F X P C S A P F I T F J P S V X U H U V L C R U R Q H Q I I F M J R A T X X G I N I J P C C V Q J G R M K Y U Q V A F A M I J S E V G S S V X J Y G R T P W Q T G S F J J Y K M B J N U C Y
### Clues
• Couples that have different religions or are from different countries, ______ diversity. (8)
• Families where both parents are the same gender (4, 3)
• The traditional type of family (7)
• The type of extended family where family members live far away from each other but remain in contact (8)
• When a couple get married but have children from previous relationships (13)
• When both the parents work (4, 6)
• When couples live in separate households (4)
• When you don't have children (9)
• When you live together but aren't married (10) | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9272270798683167, "perplexity": 1448.839768571555}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560285001.96/warc/CC-MAIN-20170116095125-00160-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2522611/how-does-this-numerical-method-of-root-approximation-work | # How does this numerical method of root approximation work?
We can often use iteration to find an approximation for the root of the equation $f(x) = 0$.
For instance, consider $x^2 - 4x + 1 = 0$. This is equivalent to $x = 4 - \frac 1x$. Now we use the iteration formula $$x_{n+1} = 4 - \frac{1}{x_n}$$ starting with, say, $x_0 = 3$.
We iterate over and over until we get sufficiently close to the root.
Now, why does this work? What is the justification for turning the relation into a recurrence relation? Why do successive iterations get us closer to the root of the equation? I understand the path that we take in getting the approximate root:
But I don't understand why this path exists. How do we know that in general, we won't just go back and forth? Or that we won't skip over the root entirely?
• Note that this technique does not work for the other root. – copper.hat Nov 16 '17 at 5:40
## 2 Answers
By writing the equation as $x=4-\frac1x$, you wrote it in the form $$x=f(x).$$ So you are looking for a fixed point for the function $f$. A sufficient condition for a fixed point $x_0$ to be an attractor (that is, that the iteration $x_{n+1}=f(x_n)$ converges to $x_0$) is that $|f'(x_0)|<1$ and $f$ continuously differentiable.
In your example, since $f'(x)=-1/x^2$, a fixed point $x_0>1$ will be an attractor, which is why the sequence converges to the desired root.
The theoretical reason this works is the Banach fixed point theorem.
We can rewrite the iteration as $$x_{n+1}=f(x_{n}) \quad \text{where} \quad f(x)=4-1/x.$$ The Banach fixed point theorem implies that if we can find a closed subset $X$ of $\mathbb{R}$ for which $f(X)\subset X$ and $|f^{\prime}|<1$ on $X$, the iteration started at any point $x_{0}\in X$ converges to a point $x$ satisfying $x=f(x)$. Moreover, no other point $y$ in $X$ satisfies $f(y)=y$.
Can you figure out an appropriate choice of $X$? (Hint: try $X=[1+\epsilon, \infty)$ for any $\epsilon>0$) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9622059464454651, "perplexity": 92.66332617212736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998509.15/warc/CC-MAIN-20190617143050-20190617164711-00070.warc.gz"} |
http://www.maplesoft.com/support/help/Maple/view.aspx?path=SolveTools/Inequality/LinearUnivariateSystem | SolveTools[Inequality] - Maple Programming Help
Home : Support : Online Help : Mathematics : Factorization and Solving Equations : SolveTools : Inequality : SolveTools/Inequality/LinearUnivariateSystem
SolveTools[Inequality]
LinearUnivariateSystem
solve a system of linear inequalities with respect to one variable
Calling Sequence LinearUnivariateSystem(sys, var)
Parameters
sys - system of inequalities var - variable name
Description
• The LinearUnivariateSystem command solves a system of linear inequalities with respect to one variable.
• The LinearUnivariateSystem command returns a set describing the interval of possible values of the variable or a piecewise function of such sets depending on parameters
Examples
> $\mathrm{with}\left(\mathrm{SolveTools}[\mathrm{Inequality}]\right):$
> $\mathrm{LinearUnivariateSystem}\left(\left\{0
$\left\{\frac{{1}}{{2}}{<}{x}\right\}$ (1)
> $\mathrm{LinearUnivariateSystem}\left(\left\{0
$\left\{{-}{1}{<}{x}{,}{x}{<}\frac{{1}}{{2}}\right\}$ (2)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+1<0,2x-1<0\right\},x\right)$
$\left\{{x}{<}{-}{1}\right\}$ (3)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+1<0,0<2x-1\right\},x\right)$
$\left\{{}\right\}$ (4)
> $\mathrm{LinearUnivariateSystem}\left(\left\{x+a<0,0<2x-1\right\},x\right)$
${{}\begin{array}{cc}\left\{\frac{{1}}{{2}}{<}{x}{,}{x}{<}{-}{a}\right\}& {a}{<}{-}\frac{{1}}{{2}}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (5)
> $\mathrm{LinearUnivariateSystem}\left(\left\{2x+a\le 0,0\le 2x-b\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{1}}{{2}}{}{a}{,}\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {a}{\le }{-}{b}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (6)
> $\mathrm{LinearUnivariateSystem}\left(\left\{ax+2\le 0,0\le 2x-b\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{2}}{{a}}{,}\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}\frac{{1}}{{2}}{}{b}{\le }{-}\frac{{2}}{{a}}\right)\\ \left\{{-}\frac{{2}}{{a}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}\frac{{1}}{{2}}{}{b}{<}{-}\frac{{2}}{{a}}\right)\\ \left\{\frac{{1}}{{2}}{}{b}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{-}\frac{{2}}{{a}}{\le }\frac{{1}}{{2}}{}{b}\right)\\ \left\{{}\right\}& {a}{=}{0}\\ \left\{{}\right\}& {\mathrm{otherwise}}\end{array}$ (7)
> $\mathrm{LinearUnivariateSystem}\left(\left\{ax+b<0,0\le cx+d\right\},x\right)$
${{}\begin{array}{cc}\left\{{x}{\le }{-}\frac{{d}}{{c}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{<}{0}{,}{-}\frac{{d}}{{c}}{<}{-}\frac{{b}}{{a}}\right)\\ \left\{{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{<}{0}{,}{-}\frac{{b}}{{a}}{\le }{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}{,}{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{0}{<}{c}{,}{-}\frac{{d}}{{c}}{<}{-}\frac{{b}}{{a}}\right)\\ \left\{{x}{<}{-}\frac{{b}}{{a}}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({0}{<}{a}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{x}{\le }{-}\frac{{d}}{{c}}{,}{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{<}{0}{,}{-}\frac{{b}}{{a}}{<}{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{0}{<}{c}{,}{-}\frac{{b}}{{a}}{<}{-}\frac{{d}}{{c}}\right)\\ \left\{{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{0}{<}{c}{,}{-}\frac{{d}}{{c}}{\le }{-}\frac{{b}}{{a}}\right)\\ \left\{{-}\frac{{b}}{{a}}{<}{x}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{<}{0}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{x}{\le }{-}\frac{{d}}{{c}}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{<}{0}\right)\\ \left\{{-}\frac{{d}}{{c}}{\le }{x}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{0}{<}{c}\right)\\ \left\{{x}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{=}{0}{,}{0}{\le }{d}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{b}{<}{0}{,}{c}{=}{0}{,}{d}{<}{0}\right)\\ \left\{{}\right\}& {\mathrm{And}}{}\left({a}{=}{0}{,}{0}{\le }{b}\right)\end{array}$ (8) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 17, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9602511525154114, "perplexity": 1531.8457232052428}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-44/segments/1476988719754.86/warc/CC-MAIN-20161020183839-00460-ip-10-171-6-4.ec2.internal.warc.gz"} |
https://www.akt.tu-berlin.de/menue/akt_talks/winter_term_20152016/talk_19112015/ | Sie sind hier
# On the One-Dimensional Euclidean Domain
Jiehua Chen (TU Berlin)
Consider a preference profile where each voter has a ranking over a number of alternatives. We say that this profile is one-dimensional Euclidean if there is an embedding of the alternatives and the voters into the real numbers such that every voter ranks alternatives that are embedded closer to him higher than alternatives which are embedded further away.
One-dimensional Euclidean profiles are necessarily single-peaked and single-crossing; single-peakedness describes a natural order of the alternatives, and single-crossingness describes a natural order of the voters.
While it is known that both the single-peaked domain and the single-crossing domain can be characterized by finitely many obstructions, we show that this is not the case for the one-dimensional Euclidean domain. In this talk, we demonstrate that for any number $k\ge 4$, we can construct a preference profile with $2k$ voters and $4k$ alternatives which is not one-dimensional Euclidean, but the deletion of an arbitrary voter from it yields a one-dimensional Euclidean profile.
Date
Speaker
Location
Language
19.11.2015
16:15
Jiehua Chen
TEL 512
English
Back to the research colloquium site. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9030078649520874, "perplexity": 783.6758672695139}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232256281.35/warc/CC-MAIN-20190521062300-20190521084300-00026.warc.gz"} |
http://www.physicsforums.com/showthread.php?t=49998 | # How to explain uncertainty principle to a freshman
P: 124 Hi, What is a good way to explain the uncertainty principle to a new freshman who has never had any quantum mechanics? That is, in plain English and without mentioning momentum space or anything.
Sci Advisor P: 1,474 First make the assertion that particles can be described using wave functions. Consider then a sine function, it has one distinct frequency, however it stretches to infinity in both directions, i.e. it is completely non-localised. Frequency is related to momentum, so one can make the assertion that, when the momentum is precisely known (one frequency), the uncertainity in position is infinite (completely non-localised wavefunction). Add a second frequency component. The wavefunction 'beats', making the particle more localised within the antinodes of the beats. In order to construct the wavefunction such that it only has one 'beat', there must be a frequency range, dF. The frequency range can be identified as the uncertainty in momentum, the 'beat width' can be identified as the uncertainty in space. It is then quite easy to show that the two are complementry, if one gets bigger, the other gets smaller. Include more frequencies, the beat width is reduced and vice-versa. Claude.
HW Helper
P: 11,915
Quote by yxgao Hi, What is a good way to explain the uncertainty principle to a new freshman who has never had any quantum mechanics? That is, in plain English and without mentioning momentum space or anything.
There is one famous description of the "quantum behavior".It's the one made by R.P.Feynman in his famous "Lectures on physics".Everybody trying to explain the Uncertainty Principle without quantum mechanics formalism inspired from this great book.Let it be known that i strongly doubt that Feyman hadn't inspired himself from other book.But at least he got a Nobel Prize and is considered alongside Dirac as the (theoretical) physicist of all time.
So my warm reccomandation is the first chapter from the third volume from any edition of:Richard P.Feynman "Lectures on Physics".
P.S.Hopefully you won't mind my assertion that:"Even a dumbass would understand physics if he was to learn it apud Feynman".
P: 42 How to explain uncertainty principle to a freshman A very short explanation for absolute non-scientific people (like part of my family) is the following. Imagine an electron. If you want to know where it is, you'll have to "see" it, so it must shoot away at least a single photon. When you see the photon, you can know where it originated from (ea location = fixed). By shooting off a photon however, it's speed has changed, and hence it's momentum. So you know where it was, but you don't have a clue where it will be heading. It's a very rough explanation, which rambles on all sides for people knowledgeable, but it gives a hint of explanation. Greetz, Leo
Math Emeritus Sci Advisor Thanks PF Gold P: 39,491 And you can only measure the position of that electron to within one WAVELENGTH of the light you are using. The more precise you want that position, the smaller wavelength light you will need to use- and the shorter the wavelength the more energy the photon has and will "kick" your electron harder.
P: 161
Quote by Leo32 Imagine an electron. If you want to know where it is, you'll have to "see" it, so it must shoot away at least a single photon. When you see the photon, you can know where it originated from (ea location = fixed). By shooting off a photon however, it's speed has changed, and hence it's momentum. So you know where it was, but you don't have a clue where it will be heading.
Hi Leo.
Thanks for your effort to make the HUP (Heisenberg's uncertainty principle) more understandable to us real people. Your argument may be very possible under the conditions you have postulated. However, although it is true that Heisenberg's electrons were singular, their location was limited to those that were very near atomic kernels; i.e., the unpaired valence electrons of the elements. In the interest of simplification I invite you to open the simplist of highschool Chemistry texts where you will see those probability balloons protruding from an atom; such balloons epitomize the proximity of HUP electrons to the quantum field radiating from the element's nucleus.
Now about the conditions whereby an electron can be responsible for the emission of a photon. Normally, when an electron collides with matter, its momentum is transferred to that matter; however, when it is considered that that electron is stopped abruptly by the momentum well of the quantum field, its momentum transfer does effect the production of a single photon.
Once stopped in a radial way, the electron is swept, in an annular/circular way into a paired-electron orbital whose momentum is in the complete control of the quantum field; if the electron is stopped at the position of an empty orbital, then because it is still single, it remains a target of HUP.
Thanks for your audience and patience; Cheers, Jim
Emeritus Sci Advisor PF Gold P: 29,238 The only qualm I have about explaining the HUP this way is that it makes it appears as if the HUP is merely the result of our inability to make a measurement. We need smaller and smaller wavelength to pinpoint the position, but by doing that, we destroy the momentum of the particle we want to measure. While this is true, it makes it appears as if the HUP is really a measurement limitation. This certainly isn't the case since the HUP is a more intrinsic property of QM. I would suggest using the diffraction from a single slit. This is the clearest, in-your-face manifestation of the HUP. I think I've described this at least a couple of times, so I won't do it here and bore a bunch of people. It clearly shows that the HUP isn't the inability to make a measurement of the position and momentum, but rather the inability to make accurate predictions of a momentum as the knowledge of the position improves. This is an extremely profound implication of the HUP that needs to be understood. Zz. Edit: I was responding to the responses BEFORE NeoClassic posting. I am not going to touch what he wrote even with a 20 mile "momentum well".
P: 119 Certain notions and physical quantities routinely measured and analyzed in classical physics aren't even defined for quantum systems and so you cannot talk about micro-systems in their terms. Even then if you try to do that, you can never get a level of accuracy higher than a certain minimum amount in those measurements. That is HUP puts a limit on classical analysis of quantum systems. spacetime www.geocities.com/physics_all/index.html
P: 277 I liked the phrase "when an electron collides with matter".
P: 2
Quote by Leo32 A very short explanation for absolute non-scientific people (like part of my family) is the following. Imagine an electron. If you want to know where it is, you'll have to "see" it, so it must shoot away at least a single photon. When you see the photon, you can know where it originated from (ea location = fixed). By shooting off a photon however, it's speed has changed, and hence it's momentum. So you know where it was, but you don't have a clue where it will be heading. It's a very rough explanation, which rambles on all sides for people knowledgeable, but it gives a hint of explanation. Greetz, Leo
That's the observer effect, not the uncertainty principle. The two aren't connected.
Mentor P: 16,317 That's also 7 years old. The freshman is in grad school by now.
P: 2 I have this medical condition where I can't sleep at night if I don't point out the difference between the observer effect and the uncertainty principle when needed. Really, can you blame me?
Related Discussions Advanced Physics Homework 3 Quantum Physics 1 Quantum Physics 8 Quantum Physics 2 Quantum Physics 3 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8541800379753113, "perplexity": 649.6857293719934}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500814701.13/warc/CC-MAIN-20140820021334-00231-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://www.maa.org/publications/periodicals/convergence/the-rule-of-false-position-and-geometric-problems?device=mobile | # The Rule of False Position and Geometric Problems
Author(s):
Vicente Meavilla Segui and Alfinio Flores
Historical examples in college mathematics courses can help students understand the process of creation in mathematics and how mathematicians have grappled with problems over the ages. This can enliven mathematics and humanize it [1]. Swetz proposes to have students solve some of the problems that were of interest for early mathematicians as a direct approach to enrich mathematics teaching and learning through history [7]. Historical remarks can help students understand the material better and help them see how it fits into the wider domain of mathematics [5]. Historical examples of problems can also provide an interesting background to tell the history of mathematical ideas [4]. Future teachers can benefit especially from the historical perspective. They will learn solution methods alternative to the ones usually taught in schools. Let us look at a problem Simon Stevin published in 1583 [6].
Problem: Construct a square knowing the difference PQ between its diagonal and its side.
Solution: Let BCDE be any square. On the diagonal EC take the point F so that EF = ED.
If FC = PQ, then the square BCDE is the solution to the problem. If not, the side of the solution square (say y) will be the fourth proportional with respect to the segments FC, PQ and ED. That is, FC:PQ = ED:y.
The method used by Stevin is called the rule of false position. The solution is especially interesting because he used this method in a geometrical problem; the method was mainly used in problems of algebraic nature. It is also interesting because our students do not use geometrical proportionality when they face problems about constructing figures that have some restrictions. To solve the problem of the square and its diagonal, students in high school (ages 16-18) frequently use methods involving trigonometry and algebra to compute the side of the square. For them the real challenge is to construct the figure. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8022356033325195, "perplexity": 648.4343724368188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500815861.64/warc/CC-MAIN-20140820021335-00459-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://admin.clutchprep.com/physics/practice-problems/44286/a-sinusoidal-electromagnetic-wave-propogates-in-the-y-direction-in-a-medium-with | # Problem: A sinusoidal electromagnetic wave propogates in the -y direction in a medium with a refractive index of 1.5, with a magnetic field amplitude of 0.001 T and a wavelength of 1.55 nm. (a) What is the angular frequency of the wave? (b) What is the electric field amplitude of the wave? (c) What is the intensity of the wave?
###### Problem Details
A sinusoidal electromagnetic wave propogates in the -y direction in a medium with a refractive index of 1.5, with a magnetic field amplitude of 0.001 T and a wavelength of 1.55 nm.
(a) What is the angular frequency of the wave?
(b) What is the electric field amplitude of the wave?
(c) What is the intensity of the wave? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9668938517570496, "perplexity": 286.8820744287701}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737168.97/warc/CC-MAIN-20200810175614-20200810205614-00207.warc.gz"} |
http://mathhelpforum.com/number-theory/192192-equations.html | 1. ## Equations
Hello,
for years i have accepted equations and knew how to use them. On the other day, i started thinking too much about equations....
if i have a quantity x and i know that x plus 2 equals 5, then i know that x equals 3. But why are equations always valid? How does saying that two things are equal, makes me able to extract the values that would satisfy them?
2. ## Re: Equations
You can't say al the equations are valid, take for example:
$\cos(x)=2$
This equation is not valid, why?
The purpose of an equation is that you're going to search a value of the variable wherefore the equation is true.
(Also I don't think this topic belongs in this section).
3. ## Re: Equations
Equations *similar to* the example you gave are always true for rather advanced (yet fundamental) reasons.
Let's say you're working with the real numbers (R). They form a FIELD. I won't go into all the definitions and properties, but since R is a group under addition, you can always use "additive inverses" (i.e. subtraction) and preserve the equality.
Likewise, having multiplicative inverses allows us to divide.
4. ## Re: Equations
That still does not explain why the statement of an equation allows me to find solutions =S
5. ## Re: Equations
Originally Posted by DarkFalz
That still does not explain why the statement of an equation allows me to find solutions =S
Not all equations have solutions.
x + 1 = 0 has no solution in the naturals.
2x - 3 = 0 has no solution in the integers.
x^2 - 2 = 0 has no solution in the rationals.
x^2 + 2 = 0 has no solution in the reals.
x = x + 1 has no solution for almost every mathematical idea you've ever heard of for "x".
So what's the point of all this? We can constructively approach certain equations. In the case of a linear equation, we can solve for reasons in my first post. If you haven't taken the time to even search "ring/field/group", then it might not be best to call an explanation lacking...
6. ## Re: Equations
Originally Posted by DarkFalz
Hello,
for years i have accepted equations and knew how to use them. On the other day, i started thinking too much about equations....
if i have a quantity x and i know that x plus 2 equals 5, then i know that x equals 3. But why are equations always valid? How does saying that two things are equal, makes me able to extract the values that would satisfy them?
the principle here was giving the name of "muqubalah" or "balancing" by the great algebraist al-Khwārizmī, or in plain english: equals to equals are equal.
x+2 = 5
x+2-2 = 5-2 (by the principle of balancing, stated above)
x+0 = 3 (calculating 2-2 and 5-3, elementary arithmetic operations)
x = 3 (the law of 0: adding nothing changes nothing).
now, if we look closely at what we just did, we find that the whole process depends on being able to subtract something from 2 that gives us 0. one way of saying this, is that -2 is an additive inverse for 2. so that is one important thing we need for "solving" equations, being able to "undo" an operation (in this case, addition).
the other important thing, is that we used a very special property of 0: x+0 = x, no matter which number we use for x. in this case, we say 0 is an identity for addition.
number systems need not possess inverses OR identities. for example, if we are limited to just positive (counting numbers), we CANNOT solve the equation:
x + 5 = 3
because 3 - 5 isn't any positive number.
another common operation used in equations is multiplication. so a similar problem might be:
3x = 5
here, we need a multiplicative inverse, to solve for x. if we are limited to just integers, there is no solution. but if we allow "fractions" (rational numbers), we have:
3x = 5
(1/3)(3x) = (1/3)5
(1/3)(3x/1) = (1/3)(5/1) (because anything divided by 1 is itself: 1 serves as a multiplicative identity, which means 1/1 = 1, since (1)(1) = 1).
(3x/3) = 5/3 (we multiply fractions by multiplying tops and bottoms)
(3/3)(x/1) = 5/3
1x = 5/3
x = 5/3
it is very possible to write equations in number systems without any solutions, for example:
x = x+1 which leads to:
0 = 1, which does not seem very reasonable.
in general, we say that a number system has to have a group structure, if we have only one operation, in order to solve equations. if we have two operations, we usually require what is called a ring structure (this is like a group-and-a-half, with some rules to make sure the two operations are compatible).
the best of all possible worlds, is what we call a field structure, where we have 2 group structures in one set of numbers (with some other rules that make fields "nice" to do algebra in). these important rules are:
for all a,b,c:
(a+b)+c = a+(b+c), called associativity of addition
a+b = b+a, called commutativity of addition
a+0 = 0+a = a, the identity law for 0
a+(-a) = -a+a = 0, the existence of inverses for addition
(ab)c = a(bc), the associative law of multiplication
ab = ba, the law of commutativity of multiplication
a1 = 1a = 1, the identity law of 1 for multiplication
a(1/a) = (1/a)a = 1, the existence of multiplicative inverses for all non-zero a (0 can't have an inverse, for special reasons)
a(b+c) = ab + ac, the law of distributivity of multiplication over addition (this is the "compatability" requirement).
these rules are the usual ones we need to have to start solving equations of most kinds. as you can see, they are very reasonable rules, but some things don't obey them. for example, functions don't always obey them. the "multiplication" in functions is called "composition", and functions don't always have "inverses". this means, in particular, that equations involving functions, can be very hard to solve, and sometimes impossible. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8326261639595032, "perplexity": 611.8483164101675}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280310.48/warc/CC-MAIN-20170116095120-00208-ip-10-171-10-70.ec2.internal.warc.gz"} |
https://forum.allaboutcircuits.com/threads/graphing-decaying-sinusoids-by-hand.50626/ | # Graphing decaying sinusoids by hand?
Discussion in 'Math' started by dewasiuk, Feb 26, 2011.
1. ### dewasiuk Thread Starter New Member
Feb 14, 2011
24
0
How could this be roughly accomplished without making an extensive table of values?
For example if I use the laplace method of analyzing an RLC circuit with a pulse excitation, I know that I will arrive at a decaying sinusoidal current represented by the following equation:
I would like to be able to quickly graph the equation incase I didn't have access to software tools at that moment.
2. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
783
1. Draw the exponential envelope bounded by the curves
$f1(t)=e^{-\alpha t}$
and
$f2(t)=-e^{-\alpha t}$
2. Draw a sinusoid constrained in amplitude by the bounded region with damped period
$T=\frac{2\pi}{\omega}$
Since it is a sine function in this particular case the value at t=0 will be zero. If it were a simple cosine function the initial value would be 1. A phase displaced sine or cosine function would have an initial value determined by the value of the sine or cosine term at t=0 sec.
Last edited: Feb 27, 2011
3. ### t_n_k AAC Fanatic!
Mar 6, 2009
5,448
783
Here's an example sketch of a phase shifted exponentially decaying cosine function.
File size:
144 KB
Views:
37
4. ### dewasiuk Thread Starter New Member
Feb 14, 2011
24
0
Oh I already know how to roughly sketch the actual shapes, but it's mostly about determining the amplitude of the first peak(for a sine wave) so I can draw it more accurately. I should have included that in my first post sorry.
5. ### Georacer Moderator
Nov 25, 2009
5,151
1,266
Peaks and zero-crossings still occur in the multiples of pi/2, so these are points in time you look for.
For example, your first peak will be at the intersection of the envelope with the line x=pi/2 for a sinusoidal curve.
That is, of course, for zero time-shift. Shift your times accordingly for θ<>0.
6. ### dewasiuk Thread Starter New Member
Feb 14, 2011
24
0
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1,536 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 3, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9049324989318848, "perplexity": 2457.664985845302}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501171416.74/warc/CC-MAIN-20170219104611-00502-ip-10-171-10-108.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/why-mole-and-kelvin-are-basic-units.656453/ | # Why mole and kelvin are basic units?
1. Dec 3, 2012
### jd12345
Mole is just a number. It doesn't really measure anything so why is it a fundamental unit?
And with kelvin - it represents the average energy of the atoms/molecules in a compound. Why is it a fundamental unit? Temperature can easily be represented in joules.
2. Dec 3, 2012
### Staff: Mentor
So,moe of the mole usage, history units of measure controversy is described in the wikipedia article:
http://en.wikipedia.org/wiki/Mole_(unit [Broken])
Last edited by a moderator: May 6, 2017
3. Dec 3, 2012
### jd12345
I have basically the same point as the wikipedia article states. So why is it still called a fundamental unit?
Also for kelvin - why is it a fundamental unit as it represents energy
4. Dec 3, 2012
### the_emi_guy
As you said, it is just a word that represents a count, much like the word "dozen" or "gross". Do you understand why we need such as number in chemistry? In the chemistry lab you are measuring out quantities in grams, but you need to keep track of how many elementary entities you have, not in absolute terms but relative to other substance that you are also measuring in grams. Have you studied stoichiometry problems yet?
The kelvin temperature scale is unique in that 0 degrees represents the lowest possible temperature, where all thermal motion ceases.
5. Dec 3, 2012
### jd12345
Well I do understand why we need mole. But why is it a fundamental unit.?
You can just define mole to be 6.022 * 10^23 and use the word.
6. Dec 3, 2012
### f95toli
Fundamental units are called "fundamental" because there is no practical way to express them in terms of other units.
The fundamental units in the SI are there because 1) They are usedful and 2) They can be realized, i.e. it is something that can be measured and used to calibrate instruments.
It is important to understand that the SI is a practical system of units, ultimately it is is system designed in such a way that we can calibrate instruments in a self-consistent way.
This is why the mole and the Kelvin are there, there is no way to express in in terms of other fundamental units in a way that can be used for calibration/comparissons; and both the mole and the Kelvin are (obviously) very important units so they have to be fundamental.
7. Dec 3, 2012
### jd12345
Oh I had the wrong idea of what a fundamental unit is. I thought it should represent something physical which we can measure.
Okay now mole and kelvin makes sense. Thank You!
8. Dec 3, 2012
### Staff: Mentor
In fact, there is a proposal to do exactly this (with a slightly different constant), which may be considered for adoption by the "authorities" in 2014:
http://en.wikipedia.org/wiki/New_SI_definitions#Mole
9. Dec 3, 2012
### D H
Staff Emeritus
No, it can't. For one thing, temperature is an intensive property while energy is an extensive property. Temperature in some simple cases can be represented as energy per mole, perhaps, but not energy.
Even in those simple cases (i.e., ideal gases), using energy/mole in lieu of temperature doesn't quite cut it. Consider a vessel that contains two gases separated by an impermeable wall that transmits heat. Put some quantity of an ideal gas in one half, some quantity of another ideal gas in the other half. Heat will be transferred across the wall from one gas to the other if the two gases are at different temperatures. Heat may or may not be transferred if the two gases have different specific energies. For example, one gas is monatomic, the other diatomic.
Real gases aren't ideal, making the relationship between temperature and specific energy a non-linear one. Things get even worse when you consider the fact that gases can condense, liquids can freeze, chemicals can combine. The concept of temperature is very useful and is measurable. Specific energy is less useful, plus how do you measure it?
10. Dec 3, 2012
### Staff: Mentor
In statistical mechanics, we can define temperature via
$$\frac{1}{T} = {\left( \frac{\partial S}{\partial U} \right)}_{N,V}$$
where $S = k \ln \Omega$ ($\Omega$ being the multiplicity of the system).
If we wanted to be really fundamental about units, we would make entropy a fundamental quantity, and use the numerical value of k to define the unit of entropy which we might call the "boltzmann" (B).
Then the kelvin would be a derived unit: 1 K = 1 J/B.
11. Dec 3, 2012
### f95toli
kB will -unless I am misstaken- actually be defined in 2014. The general "philosophy" of the new SI (which is slowly being introduced) is to have one fundamental constant per unit and then realize the unit by e.g. counting (similar to what we do with the meter and the speed of light).
Hence, the Boltzmann, Avogadros constant, e etc will all eventually be defined to have definite values.
12. Dec 3, 2012
### _Abstraction_
A mole is just a number like "dozen", it's not a unit of measurement since it doesn't measure anything. A meter is a measurement of length, a second is a measurement of time, a gram is a measurement of mass, and a kelvin is a measurement of temperature, but a mole isn't a measurement of anything.
Last edited: Dec 3, 2012
13. Dec 3, 2012
### f95toli
It is a unit of measurement in the SI. This is if you want a political decision more than a a scientific one (but again, the reason is that is practical and useful to let it be a base unit).
Since Avogadro's constant is not a defined number you can't -at the moment- use that to realise the Mole. Hence, there are other methods for realising the mole, but neither of them directly involves counting anything.
14. Dec 3, 2012
### Khashishi
Basically, Boltzmann's constant is just a historical artifact. It should be regarded as a unit conversion factor between energy and temperature units. In plasma physics, and several other fields, temperature is measured in units of energy (typically electron volts), dispensing with Boltzmann's constant.
D H: Temperature and energy don't have to mean the same thing to use the same units.
jtbell: In statistical mechanics, (fundamental) entropy is unitless, being nothing more than the natural logarithm of a number of states. Temperature has the units of energy/entropy, which, therefore, is just the units of energy.
The equipartition theorem states that the average energy in an accessible degree of freedom is 1/2 the temperature times the Boltzmann's constant. If we get rid of the Boltzmann's constant, we can just give temperature in energy units, which is a lot more natural and simple.
Last edited: Dec 3, 2012
15. Dec 3, 2012
### ThinkerofWhat
A mole is the number of atoms that have a mass of 1 gram of hydrogen (single atoms, not H2). It is not arbitrary. Comparitively, the periodic table has on it the atomic mass of each element. You'll note that the atomic mass of hydrogen is not exactly 1 gram/mole, I think this is due to more accurate measurements, or the inclusion of isotopes. The atomic mass of each element on the periodic table is a per mole measurement, all relating to hydrogen's mass. As such, you can determine how much more massive each element is (per quantity) than hydrogen (per same quantity).
Since you can determine the quantity of molecules or atoms of a substance by it's molecular mass/mole, you can figure out how many grams of a substance to add to another substance to predict a chemical reaction.(because molecules combine in predictable quantities with other molecules)
The 'gram' measurement is related to the mass of one cubic centimetre of H20. A meter is the length of 100 cm, and is otherwise recorded somewhere as the number of wavelengths of a certain frequency of light, or the distance light travels in a certain time. I think that length is a truly arbitrary value (check an historical reference), and water is an arbitrarity chosen element (for historically obvious reasons).
Last edited: Dec 3, 2012
16. Dec 3, 2012
### DrDu
The problem with statistical entropy is that to use it as a fundamental quantity we would have to be able to count and identify all the relevant microstates. I don't see that we have even an idea to do so for a real system.
17. Dec 3, 2012
### jbriggs444
While this was the original definition, the current definition is the number of atoms in a mass of 12 grams of carbon-12.
http://en.wikipedia.org/wiki/Mole_(unit)
Last edited by a moderator: Dec 3, 2012
18. Dec 3, 2012
### the_emi_guy
Interesting idea, but is it practical? Let's say I am in a biology lab and I need to measure the temperature of a blood sample. To have the thermometer read out in units of energy, it would have to be calibrated to the specific heat of this particular blood sample.
19. Dec 3, 2012
### Khashishi
No specific knowledge of the specific heat of the sample is needed. The only difference is the unit of measurement. 1 K = 1.3806488*10^-23 J.
However, it isn't practical for everyday usage simply because of the huge difference in magnitude. Room temperature is on the order of 10^-20 joules.
Both Boltzmann's constant and Avogadro's number are used to connect the microscopic scale to human scale, and therefore have similar values. We could get rid of kB and just express temperatures in terms of J/mol, and the scales would be reasonable.
20. Dec 3, 2012
### the_emi_guy
Khashishi,
Forgive me, I am still unclear on this. I thought that heat capacity was the conversion from energy to temperature.
As you mentioned earlier, the constant of proportionality between E and T is 1/2(K) per degree of freedom.
For simple substances (monatomic gasses, perhaps plasmas) we know how many degrees of freedom are involved so we can convert freely between E and T.
But how many degrees of freedom should we assign to a blood sample?
Similar Discussions: Why mole and kelvin are basic units? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9183688163757324, "perplexity": 792.0377918936822}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103910.54/warc/CC-MAIN-20170817185948-20170817205948-00588.warc.gz"} |
http://papers.nips.cc/paper/4156-learning-bounds-for-importance-weighting | # NIPS Proceedingsβ
## Learning Bounds for Importance Weighting
[PDF] [BibTeX] [Supplemental]
### Abstract
This paper presents an analysis of importance weighting for learning from finite samples and gives a series of theoretical and algorithmic results. We point out simple cases where importance weighting can fail, which suggests the need for an analysis of the properties of this technique. We then give both upper and lower bounds for generalization with bounded importance weights and, more significantly, give learning guarantees for the more common case of unbounded importance weights under the weak assumption that the second moment is bounded, a condition related to the Renyi divergence of the training and test distributions. These results are based on a series of novel and general bounds we derive for unbounded loss functions, which are of independent interest. We use these bounds to guide the definition of an alternative reweighting algorithm and report the results of experiments demonstrating its benefits. Finally, we analyze the properties of normalized importance weights which are also commonly used. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9631352424621582, "perplexity": 321.71444119347404}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917124299.47/warc/CC-MAIN-20170423031204-00424-ip-10-145-167-34.ec2.internal.warc.gz"} |
http://ichimokustrategy.com/tag/ichimoku-strategies-2/ | ## Ichimoku Technique Signs
The Ichimoku Kinko Hyo technique offers the chance to obtain many different types of operating signals through the use of a single graph. As for all trading techniques, when we are using the Ichimoku Kinko Hyo for our operations, it is always advisable to take into account other elements (volumes, sentiment, seasonality, oscillators, etc …), but certainly the immediate graphic understanding of the trend and its evolution represent the strongest point of the Ichimoku technique.
As this technique aims to find the best point of entry within a trend, the answers we get from the signals will obviously be probabilistic and will have to conceive a margin of error. A wise money management will curb this phenomenon.
Here below, we will study in deep the 5 trading techniques that can be used by any trader. As we will see below, all the examples have been classified according to the emerged signal strength: strong, neutral and weak. The element that distinguishes the strength of the signal is where the signal is placed respect the Kumo, but in some cases there can be other indicators that can help the trader in the interpretation.
TENKAN SEN /KIJUN SEN CROSS
This trading signal is applied when the Tenkan Sen (black line) cuts the Kijun Sen (red line)upward or downward.
The generated signal is bullish when the Tenkan Sen cuts the Kijun Sen from the bottom to the top. However, these signals, have a different degree of strength according to the position in which this signal occurs with respect to Kumo.
A weak bullish signal is recorded when the cutting of the two lines occurs below the Kumo.
A neutral bullish signal is recorded when the cutting of the two lines occurs inside the Kumo.
A strong bullish signal is recorded when the cutting of the two lines occurs above the Kumo.
The bearish signal is generated when the Tenkan Sen (black line) cuts the Kijun Sen (red line) from the top to the bottom.
A weak bearish signal is recorded when the cutting of the two lines occurs above the Kumo.
A neutral bearish signal is recorded when the cutting of the two lines occurs inside the Kumo.
A strong bearish signal is recorded when the cutting of the two lines occurs below the Kumo.
KIJUN SEN CROSS
When the price crosses the Kijun Sen, then we get this sort of signal.
The generated signal is bullish when the price cuts the Kijun Sen (red line) from the bottom to the top.
A weak bullish signal is recorded when the Kijun is cut from the price below the Kumo.
A neutral bullish signal is recorded when the Kijun is cut from the price inside the Kumo.
A strong bullish signal is recorded when the Kijun is cut from the price above the Kumo.
The generated signal is bearish when the price cuts the Kijun Sen (red line) from the top to the bottom.
A weak bearish signal is recorded when the Kijun is cut from the price above the Kumo.
A neutral bearish signal is recorded when the Kijun is cut from the price within the Kumo.
A strong bearish signal is recorded when the Kijun is cut from the price below the Kumo.
KUMO BREAKOUT
The so-called Kumo Breakout signal occurs when the price cuts one of the two extremes of the Kumo.
A bullish signal occurs when the price enters the Kumo and breaks its upper wall upward.
A bearish signal occurs when the price enters the Kumo and breaks its lower wall downward.
SENKOU SPAN CROSS
The Senkou Span Cross occurs when the Senkou Span A cuts the Senkou Span B.
Since the two Senkou lines are projected forward by 26 periods respect the current price, the element to be checked when the signal is realized is where the price is positioned compared to the Kumo.
The generated signal is bullish when the Senkou Span A (blue line) cuts the Senkou Span B (purple line) from the bottom to the top.
A weak bullish signal is recorded when at the cutting time between the two Senkou, prices are positioned below the Kumo.
A neutral bullish signal is recorded when at the cutting time between the two Senkou, prices are positioned inside the Kumo.
A strong bullish signal is recorded when at the cutting time between the two Senkou, prices are positioned above the Kumo.
The generated signal is bearish when the Senkou Span A (blue line) cuts the Senkou Span B (purple line) from the top to the bottom.
A weak bearish signal is recorded when at the cutting time between the two Senkou, prices are positioned above the Kumo.
A neutral bearish signal is recorded when at the cutting time between the two Senkou, prices are positioned inside the Kumo.
A strong bearish signal is recorded when at the cutting time between the two Senkou, prices are positioned below the Kumo.
CHIKOU SPAN CROSS
When the Chikou Span rises above or falls below the corresponding price then this signal is generated.
An important element is also represented by the inclination of the Chikou Span at the time of the observation. In fact, the Chikou positioning above or below the corresponding price is not enough; it also must have an upward inclination for bullish signals and a downward one for those bearish.
Since the Chikou is nothing more than the current closing price moved back by 26 periods, the necessary elements are: for the long trade, a Chikou above the corresponding price and, for the short trade, a Chikou below the corresponding price. The positioning of the current price compared to the Kumo then contributes to make the signal more or less powerful.
The generated signal is called bullish when the Chikou Span (yellow line) cuts the price from the bottom to the top.
A weak bullish signal is recorded when at the time of the upward cutting of the Chikou, the current price is positioned below the Kumo.
A neutral bullish signal is recorded when at the time of the upward cutting of the Chikou, the current price is positioned inside the Kumo.
A strong bullish signal is recorded when at the time of the upward cutting of the Chikou, the current price is positioned above the Kumo.
The generated signal is called bearish when the Chikou Span (yellow line) cuts the price from the top to the bottom.
A weak bearish signal is recorded when at the downward cutting time of the Chikou, the current price is positioned above the Kumo.
A neutral bearish signal is recorded when at the downward cutting time of the Chikou, the current price is positioned inside the Kumo.
A strong bearish signal is recorded when at the downward cutting time of the Chikou, the current price is positioned below the Kumo. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8230826258659363, "perplexity": 2817.7599899392235}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744368.74/warc/CC-MAIN-20181118114534-20181118140534-00553.warc.gz"} |
http://math.stackexchange.com/questions/59541/do-the-terms-in-these-sequences-tend-to-square-roots | Do the terms in these sequences tend to square roots?
By defining these 3 matrices:
$\displaystyle M_1 = \begin{bmatrix} 1&0&0&0&0 \\ 1&1&0&0&0 \\ 1&1+x&1&0&0 \\ 1&1+x(1+x)&1+2x&1&0 \\ 1&1+x(1+x(1+x))&1+x(1+x)+x(1+2x)&1+3x&1 \end{bmatrix}$
$\displaystyle M_2 = \begin{bmatrix} 1&0&0&0&0 \\ x&1&0&0&0 \\ 1&x&1&0&0 \\ 0&1&x&1&0 \\ 0&0&1&x&1 \end{bmatrix}$
$\displaystyle M_3 = \begin{bmatrix} 1&1&1&1&1 \\ \frac{1}{2}(-1+x)x&1&1&1&1 \\ \frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1&1 \\ 1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1&1 \\ 1&1&\frac{1}{2}(-1+x)x&\frac{1}{2}(-1+x)x&1 \end{bmatrix}$
and multiplying and dividing elementwise $M_1\cdot M_2 / M_3$ , we get the following matrix $A$:
$\displaystyle A = \begin{bmatrix} 1&0&0&0&0 \\ \frac{2}{-1+x}&1&0&0&0 \\ \frac{2}{(-1+x)x}&\frac{2(1+x)}{-1+x}&1&0&0 \\ 0&\frac{2(1+x(1+x))}{(-1+x)x}&\frac{2(1+2x)}{-1+x}&1&0 \\ 0&0&\frac{(2(1+x(1+x)+x(1+2x)))}{((-1+x)x)}&\frac{(2(1+3x))}{(-1 + x)}&1 \end{bmatrix}$
Then the matrix inverse of $A$ is a new matrix $B$ starting:
$\displaystyle B = \begin{bmatrix} 1&0&0 \\ \frac{-2}{-1 + x}&1&0 \\ (\frac{4}{(-1+x)^2} - \frac{2}{(-1+x)x} + \frac{4x}{(-1 + x)^2})&(\frac{-2}{-1+x} - \frac{2x}{-1+x})&1 \end{bmatrix}$
If we then calculate the ratios of the first column in matrix $B$ we get a sequence $c\;$:
$\displaystyle c = \frac {B(n,1)}{B(n+1,1)}, \; n=1,2,3\;...$
and if we define a sequence $b$:
$\displaystyle b = 1,\; 1+x,\; 1+2x,\; 1+3x \;...$
and multiply element-wise $c$ with $b$ and add $x$, we get a sequence $d\;$:
$\displaystyle d = x + b*c$
$\displaystyle d = \frac{1+x}{2}, \frac{x(2+x+x^2)}{1+x+2x^2}, \frac{1+x(6+x(7+2x(4+x)))}{2(1+x)(2+x+3x^2)}\;...$
My question is: Do the terms in this sequence $d$ tend to $\displaystyle \sqrt x\;$?
Example: $x=2\;$
$\displaystyle \frac{3}{2}, \frac{16}{11}, \frac{137}{96}, \frac{1642}{1157}, \frac{8429}{5950}, \frac{67952}{48001}, \frac{1509939}{1066976}, \frac{38701726}{27353183}, \frac{1124000429}{794502270}, \frac{36478904464}{25787223797} \;...$
$\displaystyle \frac{36478904464}{25787223797} = 1.41461$ which is close to $\displaystyle \sqrt 2 = 1.41421$
As a Mathematica program this is:
(*x is the value to calculate the square root of*)
Clear[x, t, tt, ttt, M1, M2, M3, A, B, b, c, n, k, i, j,
squarerootofx];
(*x=2; uncomment this to calculate sqrt of 2 or any other number*)
(*nn is size of matrix*)
nn = 5;
(*Variant of the Pascal triangle*)
t[n_, 1] = 1;
t[n_, k_] := t[n, k] = If[n > 1, t[n - 1, k - 1] + x*t[n - 1, k], 0];
M1 = Table[Table[t[i, j], {j, 1, nn}], {i, 1, nn}];
MatrixForm[M1]
tt[n_, k_] :=
tt[n, k] = If[Or[n == k, n == k + 2], 1, If[n == k + 1, x, 0]];
M2 = Table[Table[tt[i, j], {j, 1, nn}], {i, 1, nn}];
MatrixForm[M2]
ttt[n_, k_] :=
ttt[n, k] =
If[n == k, 1, If[Or[n == k + 1, n == k + 2], (x - 1)*x/2, 1]];
M3 = Table[Table[ttt[i, j], {j, 1, nn}], {i, 1, nn}];
MatrixForm[M3]
(*Elementwise multiplication and division of the 3 matrices*)
A = M1*M2/M3;
MatrixForm[A]
B = Inverse[A];
MatrixForm[B]
b = Range[1, x*(nn - 2) + 1, x]
c = Ratios[B[[All, 1]]]^-1;
d = squarerootofx = x + b*c
FullSimplify[squarerootofx]
N[squarerootofx];
-
@t-laarhoven Your assumption about the notation is correct. I am used to the notation $T(n,k)$ that is often seen for matrices in Neil Sloane's OEIS, therefore I used $n$ as index. The Mathematica program should be correct however. Regarding an expression for the entries in matrix $B$, no I have not thought about that. And yes, I agree that there are easier ways to calculate $\sqrt x$. – Mats Granvik Aug 24 '11 at 21:25
I am not quite sure I understand what the rule is for generating $M_1$. It looks like each row may be able to be generated recursively from the previous one(s); can you write down the relation? – Willie Wong Aug 25 '11 at 1:55
@Willie Wong: Yes, $M_1$ is generated recursively. The recurrence is: $M_1 (n,1)=1\;$,$n>1:\;M_1 (n,k)=M_1(n-1,k-1)+x \cdot M_1(n-1,k)$ – Mats Granvik Aug 25 '11 at 8:47
1 Answer
This reminds me a lot of Do these series converge to the Mangoldt function? :-). I'm sure you have an explanation for how you managed to come up with this hypothesis in the first place, as you did in that case...
The sequence $d$ does indeed converge to $\sqrt x$. The solution of the recurrence
$$M_1 (n,k)=\begin{cases}1&k=1\\M_1(n-1,k-1)+x \cdot M_1(n-1,k)&k>1\\\end{cases}$$
is
$$M_1(n,k)=\sum_{j=0}^{n-k}\binom{k+j-2}jx^j\;,$$
where the $k=1$ case works out if the binomial coefficient is generalized to $\binom{j-1}j$. But most of the matrix isn't used because of the zeros in $M_2$; so we will only need the cases $0\le n-k\le2$:
$$\begin{eqnarray} M_1(n+2,n+2) &=& 1\;, \\ M_1(n+2,n+1) &=& 1+nx\;, \\ M_1(n+2,n+0) &=& 1+(n-1)x+\frac{n(n-1)}2x^2\;. \end{eqnarray}$$
Here and in the following $n\ge1$. Then the corresponding entries of $A$ are
$$\begin{eqnarray} A(n+2,n+2) &=& 1\;, \\ A(n+2,n+1) &=& \frac2{x-1}(1+nx)\;, \\ A(n+2,n+0) &=& \frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)\;. \end{eqnarray}$$
The first column of the inverse $B$ is determined by the condition
$$\sum_{l=0}^2A(n+2,n+l)B(n+l,1)=0\;.$$
Since $A(n+2,n+2)=1$, we can directly solve for $B(n+2,1)$ to get a recurrence for $B$:
$$\begin{eqnarray} B(n+2,1) &=& -A(n+2,n+1)B(n+1,1)-A(n+2,n)B(n,1) \\ &=& -\frac2{x-1}(1+nx)B(n+1,1) \\ && -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)B(n,1) \;. \end{eqnarray}$$
Dividing through by $B(n+1,1)$ yields
$$\begin{eqnarray} \frac1{c(n+1)} &=& -A(n+2,n+1)-A(n+2,n)c(n) \\ &=& -\frac2{x-1}(1+nx) -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)c(n) \;, \end{eqnarray}$$
which is a one-term recurrence for $c$. With $d(n)=x+(1+(n-1)x)c(n)$, and thus
$$c(n)=\frac{d(n)-x}{1+(n-1)x}\;,$$
this yields a one-term recurrence for $d$:
$$\frac{1+nx}{d(n+1)-x} = -\frac2{x-1}(1+nx) -\frac2{x(x-1)}\left(1+(n-1)x+\frac{n(n-1)}2x^2\right)\frac{d(n)-x}{1+(n-1)x} \;,$$
$$\begin{eqnarray} \frac1{d(n+1)-x} &=& -\frac2{x-1}\left(1+\frac{1+(n-1)x+\frac12n(n-1)x^2}{(1+nx)(1+(n-1)x)}\frac{d(n)-x}x\right) \\ &=& -\frac2{x-1}\left(1+\gamma(n)\frac{d(n)-x}x\right) \;, \end{eqnarray}$$
where $\gamma(n)$ contains the dependence on $n$.
Now to get our bearings let's first see without full rigour what happens for large $n$. Then $\gamma(n)$ goes to $1/2$, and we get
$$\begin{eqnarray} \frac1{d(n+1)-x} &=& -\frac2{x-1}\left(1+\frac12\frac{d(n)-x}x\right) \\ &=& \frac{d(n)+x}{x(1-x)} \;, \\ d(n+1) &=& x+\frac{x(1-x)}{d(n)+x} \\ &=& x\frac{d(n)+1}{d(n)+x}\;. \end{eqnarray}$$
Now we can look at what happens to the difference from $\sqrt x$:
$$\begin{eqnarray} d(n+1)-\sqrt x &=& x\frac{d(n)+1}{d(n)+x}-\sqrt x \\ &=& \frac{x-\sqrt x}{d(n)+x}(d(n)-\sqrt x)\;. \end{eqnarray}$$
So if we could show that $d(n)$ is always positive, then we might be able to show that the sequence of the differences is dominated by a convergent geometric sequence. To make the approximation with $\gamma(n)=1/2$ rigorous, we will also need to bound $d(n)$ from above.
So let's go back to the full recurrence for $d$:
$$d(n+1) = x-\frac{x-1}2\left(1+\gamma(n)\frac{d(n)-x}x\right)^{-1}\;.$$
Let's see how $\gamma(n)$ differs from its limit:
$$\begin{eqnarray} \gamma(n)-\frac12 &=& \frac{1+(n-1)x+\frac12n(n-1)x^2}{(1+nx)(1+(n-1)x)}-\frac12 \\ &=& \frac12\frac{1-x}{(1+nx)(1+(n-1)x)} \end{eqnarray}$$
Thus we have $0<\gamma(n)\lt1/2$ for $x\gt1$ and $1/2\lt\gamma(n)\lt1$ for $x\lt1$. Also note that for $d(n)>0$ we have
$$1+\gamma(n)\frac{d(n)-x}x\gt1+\gamma(n)\frac{-x}x=1-\gamma(n)\gt0\;.$$
Now first assume $x\gt1$. Then for $d(n)\gt0$,
$$\begin{eqnarray} d(n+1) &\gt& x-\frac{x-1}2\left(1+\gamma(n)\frac{0-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\left(1-\gamma(n)\right)^{-1} \\ &\gt& x-\frac{x-1}2\left(\frac12\right)^{-1} \\ &=& x-(x-1) \\ &=& 1\;. \end{eqnarray}$$
On the other hand, for $d(n)\lt x$,
$$\begin{eqnarray} d(n+1) &\lt& x-\frac{x-1}2\left(1+\gamma(n)\frac{x-x}x\right)^{-1} \\ &=& x-\frac{x-1}2 \\ &=& \frac{x+1}2\;. \end{eqnarray}$$
Since $1\lt d(1)=(x+1)/2\lt x$, it follows that $1\lt d(n)\le(x+1)/2\lt x$ for all $n$.
Now assume $x<1$. Then $d(n+1)\gt x$. On the other hand, for $d(n)\gt x$,
$$\begin{eqnarray} d(n+1) &\lt& x-\frac{x-1}2\left(1+\frac12\frac{d-x}x\right)^{-1} \\ &\lt& x-\frac{x-1}2\left(1+\frac12\frac{-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\left(\frac12\right)^{-1} \\ &=& x-(x-1) \\ &=& 1\;. \end{eqnarray}$$
Since $x\lt d(1)=(x+1)/2\lt1$, it follows that $x\lt d(n)\lt1$ for all $n$.
Now we have all the ingredients for the proof of convergence. Let's put $\gamma(n)=\frac12(1+\epsilon(n))$, with $\epsilon(n)\to0$ as $n\to\infty$, so
$$\begin{eqnarray} d(n+1) &=& x-\frac{x-1}2\left(1+\frac12(1+\epsilon(n))\frac{d(n)-x}x\right)^{-1} \\ &=& x-\frac{x-1}2\frac{x}{x+\frac12(1+\epsilon(n))(d(n)-x)} \\ &=& \frac{x(d(n)+x+\epsilon(n)(d(n)-x))-(x-1)x}{d(n)+x+\epsilon(n)(d(n)-x)} \\ &=& \frac{x(d(n)+1)+\epsilon(n)x(d(n)-x)}{d(n)+x+\epsilon(n)(d(n)-x)}\;, \end{eqnarray}$$
and consider the difference from $\sqrt x$ again:
$$\begin{eqnarray} d(n+1)-\sqrt x &=& \frac{x(d(n)+1)+\epsilon(n)x(d(n)-x)}{d(n)+x+\epsilon(n)(d(n)-x)}-\sqrt x \\ &=& \frac{(x-\sqrt x)(d(n)-\sqrt x+\epsilon(n)x(d(n)-x))}{d(n)+x+\epsilon(n)(d(n)-x)} \\ &=& \frac{(x-\sqrt x)(1+\epsilon(n)x(d(n)+\sqrt x))}{d(n)+x+\epsilon(n)(d(n)-x)}(d(n)-\sqrt x) \;. \end{eqnarray}$$
Since $d(n)$ is bounded and positive (and $x$ is constant and positive), we have
$$\limsup_{n\to\infty}\left|\frac{(x-\sqrt x)(1+\epsilon(n)x(d(n)+\sqrt x))}{d(n)+x+\epsilon(n)(d(n)-x)}\right| =\limsup_{n\to\infty}\left|\frac{x-\sqrt x}{d(n)+x}\right| \le\left|\frac{x-\sqrt x}x\right|\lt1\;,$$
and so the sequence of the differences is dominated by a convergent geometric sequence. That completes the proof.
- | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9509173631668091, "perplexity": 304.0900438313625}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783395620.56/warc/CC-MAIN-20160624154955-00075-ip-10-164-35-72.ec2.internal.warc.gz"} |
http://math.stackexchange.com/users/24190/scotty?tab=activity&sort=comments | Scotty
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Jun6 comment I'm looking a compound interest formula. The future value formula from that link is what I was looking for. Jun6 comment I'm looking a compound interest formula. @Amzoti Yeah, I suppose that's the correct terminology. Feb12 comment Finding derivatives of Trigonometric Functions These are starting to click now. Feb12 comment Finding derivatives of Trigonometric Functions That makes perfect sense now for that problem, but I'm having trouble applying that same approach to this problem $$p = \frac{6+secq}{6-secq}$$ I tried moving the bottom -sec(q) to the top and making it -cos(q) but am not able to get even close to the correct derivative of that function. Feb10 comment Using the alternative formula to find the derivative of a function? Good point. I should have included the lim z->x on every step. Feb10 comment Using the alternative formula to find the derivative of a function? @MJD Somebody fixed it for me. There was a red /[ before and after each equation. And when I tried to delete those, it would break my formatting. Feb1 comment How can I calculate the time two people will meet if they are paddling towards each other on a lake? Yeah, it just clicked. thanks Feb1 comment How can I calculate the time two people will meet if they are paddling towards each other on a lake? @HenningMakholm Ken is 4miles closer and kara is 7 miles closer. Therefor the gap between them is now 19 miles. After 2 hrs the gap would be 8miles. Got it: 4x + 7x = 30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.886932909488678, "perplexity": 724.8098604345074}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042988598.68/warc/CC-MAIN-20150728002308-00206-ip-10-236-191-2.ec2.internal.warc.gz"} |
https://byjus.com/ncert-solutions-class-12-physics/chapter-10-wave-optics/ | NCERT Solutions For Class 12 Physics Chapter 10
NCERT Solutions Class 12 Physics Wave Optics
NCERT solutions for class 9 maths chapter 10 is an essential study material. Class 12 is an important phase of a student's life because after class 12 most of the students opt of higher studies and come go for government and private jobs. These competitive exams have lots of questions which are asked from class 12th physics. So one must solve the questions provided at the end of each chapter. Solving these questions will help you to understand the chapter.
NCERT Solutions for class 12th maths chapter 10 wave optics is provided here so that students can have look at these for better understanding and clarification.
Question 1:
Monochromatic light having a wavelength of 589nm from the air is incident on a water surface. Find the frequency, wavelength and speed of (i) reflected and (ii) refracted light? [1.33 is the Refractive index of water]
Monochromatic light incident having wavelength, $$\lambda$$ = 589 nm = 589 x 10-9 m
Speed of light in air, c = 3 x 108 m s-1
Refractive index of water, $$\mu$$ = 1.33
(i) In the same medium through which incident ray passed the ray will be reflected back.
Therefore the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.
Frequency of light can be found from the relation:
$$v=\frac{c}{\lambda}$$ = $$\frac{3\times10^{8}}{589\times10^{-9}}$$ = 5.09 × 1014 Hz
Hence, the speed, frequency, and wavelength of the reflected light are:
c = 3 x 108 m s-1, 5.09 × 1014 Hz, and 589 nm respectively.
(b) The frequency of light which is travelling never depends upon the property of the medium. Therefore, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.
Refracted frequency, v = 5.09 x 1014 Hz
Speed of light in water Is related to the refractive Index of water as:
$$v=\frac{c}{\lambda}$$ = $$v=\frac{3\times10^{8}}{1.33}$$ = 2.26 × 108 m s‑1
Wavelength of light in water can be found by the relation:
$$\lambda=\frac{v}{V}$$ = $$\frac{2.26\times10^{8}}{5.09\times10^{14}}$$ = 444.007 × 10-9 m = 444.01nm
Hence the speed, frequency and wavelength of refracted light are: 444.007 × 10-9 m, 444.01nm, and 5.09 × 1014 Hz respectively.
Question 2:
Find the shape of the wave front in each of the following cases:
(i) Light diverging from a point source.
(ii) Light emerging out of a convex lens when a point source is placed at its focus.
(iii) The portion of the wave front of light from a distant star intercepted by the Earth.
(i) The shape of a wave front is spherical in the case of a light diverging from a point source. The wave front is shown in the figure
(ii) The shape of a wave front is a parallel odd in the case of a light emerging out of a convex lens when a point source is placed at its focus.
(iii) The shape of the wave front is a plane when a portion of the wave front of light from a distant star intercepted by the earth.
Question 3:
(i) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in vacuum is ( 3.0 x 108 m s-1 )
(ii) Is the speed of light in glass Independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
(i) Refractive Index of glass, $$\mu$$ = 1.5
Speed of light, c = 3 × 108 ms-1
Speed of light in glass is given by the relation; $$v=\frac{c}{\mu}$$
= $$\frac{3\times10^{8}}{1.5}$$ = $$2\times 10^{8}m/s$$
Hence, the speed of light in glass is 2 × 108 m s-1
(ii) The speed of light in glass is not independent of the colour of light. The refractive Index of a violet component of white light is greater than the refractive Index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.
Question 4:
In Young’s double-slit experiment, 0.28mm separation between the slits and the screen is placed 1.4m away. 1.2cm is the distance between the central bright fringe and the fourth bright fringe. Determine the wavelength of light used in the experiment.
Distance between the slits and the screen, D = 1.4 m
and the distance between the slits, d = 0.28 mm = 0.28 x 10-3 m
Distance between the central fringe and the fourth (n = 4) fringe,
u = 1.2cm = 1.2 × 10-2 m
The relation for the distance between the two fringes as in case of a constructive interference:
$$u=n\;\lambda\;\frac{D}{d}$$
Where, n = order of fringes
= 4$$\lambda$$ = Wavelength of light used
$$u=n\;\lambda\;\frac{D}{d}$$
= $$\frac{1.2\times10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}$$
= 60 × 10-7 = 600nm
Therefore the wavelength of the light is 600 nm.
Question 5:
In Young’s double-slit experiment using the monochromatic light of wavelength $$\lambda$$, the intensity of light at a point on the screen where path difference is $$\lambda$$, is K units. What is the intensity of light at a point where path difference is $$\frac{\lambda}{3}$$?
Let $$I_{1}$$ and $$I_{2}$$ be the intensity of the two light waves. Their resultant intensities can be obtained as:
$$I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi$$
Where,
$$\phi$$ = Phase difference between the two waves
For monochromatic light waves:
$$I_{1}$$ = $$I_{2}$$
Therefore $$I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi$$
= $$2I_{1}+2I_{1}\;cos\phi$$
Phase difference = $$\frac{2\pi}{\lambda}\times\;Path\;difference$$
Since path difference = $$\lambda$$, Phase difference, $$\phi=2\pi$$ and I’ = K [Given]
Therefore $$I_{1}=\frac{K}{4}$$ . . . . . . . . . . . . . . . (i)
When path difference= $$\frac{\lambda}{3}$$
Phase difference, $$\phi=\frac{2\pi}{3}$$
Hence, resultant intensity:
$$I’_{g}=I_{1}+I_{1}+2\sqrt{I_{1}\;I_{1}}\; cos\frac{2\pi}{3}$$
= $$\\2I_{1}+2I_{1}(-\frac{1}{2})$$
Using equation (i), we can write:
$$I_{g}=I_{1}=\frac{K}{4}$$
Hence, the intensity of light at a point where the path difference is $$\frac{\lambda}{3}$$ is $$\frac{K}{4}$$ units.
Question 6: 650 nm and 520 nm are two wavelengths of a beam of light which is used to obtain interference fringes in Young’s double slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Wavelength of the light beam, $$\lambda_{1}$$ = 650 nm
Wavelength of another light beam, $$\lambda_{2}$$ = 520 nm
Distance of the slits from the screen = D
Distance between the two slits = d
(i) Distance of the $$n^{th}$$ bright fringe on the screen from the central maximum is given by the relation,
x =$$n\;\lambda_{1}\;(\frac{D}{d})$$
For third bright fringe, n=3
Therefore x = $$3\times650\frac{D}{d}= 1950\frac{D}{d}nm$$
(b) Let, the $$n^{th}$$ bright fringe due to wavelength $$\lambda_{2}$$ and $$(n – 1)^{th}$$ bright fringe due to wavelength $$\lambda_{2}$$ coincide on the screen. We can equate the conditions for bright fringes as:
$$n\lambda_{2}=(n-1)\lambda_{1}$$
520n = 650n – 650
650=130n
Therefore n = 5
Hence, the least distance from the central maximum can be obtained by the relation:
x = $$n\;\lambda_{2}\;\frac{D}{d}$$ = $$5\times 520\frac{D}{d}=2600\frac{D}{d}$$ nm
Note: The value of d and D are not given in the question.
Question 7:
In a double-slit experiment, 0.2° is found to be the angular width of a fringe on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be $$\frac{4}{3}$$.
Distance of the screen from the slits, D=1m
Wavelength of light used, $$\lambda_{1}$$ = 600 nm
Angular width of the fringe in air $$\theta_{1}$$ = 0.2°
Angular width of the fringe in water=$$\theta_{2}$$
Refractive index of water, $$\mu=\frac{4}{3}$$
Refractive index is related to angular width as:
$$\mu=\frac{\theta_{1}}{\theta_{2}}$$ $$\theta_{2}=\frac{3}{4}\theta_{1}$$ $$\frac{3}{4}\times 0.2=0.15$$
Therefore, the angular width of the fringe in water will reduce to 0.15°
Question 8: What is the Brewster angle for air to glass transition? (Refractive index of glass=1.5.)
Refractive index of glass, $$\mu=1.5$$
Let, Brewster angle = $$\theta$$
Brewster angle is related to refractive index as:
$$tan\theta=\mu$$
$$\theta=tan^{-1}(1.5)$$ = 56.31°
Therefore, the Brewster angle for air to glass transition is 56.31°
Question 9: Light of wavelength 5000 Armstrong falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?
Wavelength of incident light, $$[\lambda]$$ = 5000 Armstrong = 5000 x 10-10 m
Speed of light, c =3 x 108 m
Frequency of incident light is given by the relation,
v = $$\frac{c}{\lambda}$$ = $$\frac{3\;\times \;10^{8}}{5000\;\times \;10^{-10}}$$ = 6 × 1014
The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 Armstrong and its frequency is $$6 \times 10^{14}$$Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, $$\angle i$$ and angle of reflection, $$\angle r$$ is 90°
According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:
$$\angle i+\angle r$$ = 90°
i.e. $$\angle i+\angle i$$ = 90°
Hence, $$\angle i=\frac{90}{2}$$ = 45°
Therefore, the angle of incidence for the given condition is 45°
Question 10:
Estimate the distance for which ray optics is a good approximation for an aperture of 4 mm and wavelength 400 nm.
Fresnel’s distance ($$Z_{F}$$) is the distance for which the ray optics is a good approximation. It is given by the relation,
$$Z_{F}=\frac{a^{2}}{\lambda}$$
Where,
Aperture width, a = 4 mm = 4 × 10-3 m
Wavelength of light, $$\lambda$$ = 400 nm = 400 × 10-9 m
$$Z_{F}=\frac{(4\times10^{-3})^{2}}{400\times10^{-9}}$$ = 40m
Therefore, the distance for which the ray optics is a good approximation is 40 m. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8172391057014465, "perplexity": 489.2312622314714}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267867364.94/warc/CC-MAIN-20180625014226-20180625034226-00269.warc.gz"} |
http://mathhelpforum.com/calculus/107196-areas-integration.html | # Math Help - Areas with integration
1. ## Areas with integration
A region is bounded by y=1/2x - 3, by x=14 and the x-axis. Find the area of the region by
a) using the formula for the area of the triangle
2. Originally Posted by creatively12
A region is bounded by y=1/2x - 3, by x=14 and the x-axis. Find the area of the region by
a) using the formula for the area of the triangle
making a sketch of the given graphs will make finding a solution rather simple.
can you do that?
3. actually i was unable to post the whole question, the graph is given in the question, but m still unable to find the base and height for the area of the triangle
4. ok thanks I finally sorted what I neede to do, thanks for the help
5. Originally Posted by creatively12
actually i was unable to post the whole question, the graph is given in the question, but m still unable to find the base and height for the area of the triangle
$y = \frac{1}{2}x - 3$ crosses the x-axis at $x = 6$
triangle base is $14 - 6 = 8$
triangle height is $y(14) = 4$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8538066148757935, "perplexity": 524.9226719233376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-22/segments/1432207927863.72/warc/CC-MAIN-20150521113207-00221-ip-10-180-206-219.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/question-about-evolution-of-a-wave-function.569427/ | # Question about evolution of a wave function
1. Jan 21, 2012
### andrewkirk
As I understand it, where a system’s Hamiltonian is not time-dependent, the wave function of a system that is in state psi(0) at time t=0 evolves as:
psi(t) = sum, over all eigenvalues E of operator H, of exp(-i*E*t / hbar) * <E|psi(0)> * | E>
If the eigenvalues are continuous it is an integral rather than a sum but let’s assume they are discrete.
In Shankar’s ‘Principles of Quantum Mechanics’, he claims on p118 (2nd edition) that multiplying a ket by a complex number of modulus one does not change the physical state. The first factor in the above formula has modulus one so, we can re-write the equation as:
psi(t) = sum, over all eigenvalues E of operator H, of <E|psi(0)> * f(t,|E>)
where f(t,|E>)=exp(-i*E*t / hbar) * | E> is a ket representing a state that is physically identical to |E>.
Since f(t,|E>) is physically identical to |E> and the first postulate of QM tells us that the ket represents the physical state, it should make no physical difference if we replace f(t,|E>) by |E> in the formula.
That then gives us:
psi(t) = sum, over all eigenvalues E of operator H, of <E|psi(0)> * |E>
But now we have a formula from which t has disappeared, which implies that the state does not change over time.
Where did I go wrong?
2. Jan 21, 2012
### tom.stoer
very hard to read; please try LaTeX
$$|\psi,t\rangle = e^{-iHt}\,|\psi,0\rangle$$
3. Jan 21, 2012
### andrewkirk
OK here is a Tex version of the post. I hope somebody can help explain this puzzle.
As I understand it, where a system’s Hamiltonian is time-independent and has discrete eigenvalues, the wave function of a system that is in state $|\psi(0)\rangle$ at time $t=0$ evolves as:
$|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ |E\rangle\ \ e^{-iEt/h}\ \ \ \ \ \$ Formula 1
where $J$ is the set of eigenvalues of the Hamiltonian operator.
In Shankar’s ‘Principles of Quantum Mechanics’, he claims on p118 (2nd edition) that multiplying a ket by a complex number of unit modulus does not change the physical state. The last factor $e^{-iEt/h}$ in the above formula has unit modulus, because E is an eigenvalue of a Hermitian operator and hence real. So so we can re-write the equation as:
$|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ f(t,|E\rangle)\ \ \ \ \ \$ Formula 2
where $f(t,|E\rangle) = \ |E\rangle\ e^{-iEt/h}$ is a ket representing a state that is, according to Shankar, physically identical to $|E\rangle$.
Since $f(t,|E\rangle)$ is physically identical to $|E\rangle$ and the first postulate of QM tells us that the ket represents the physical state, it should make no physical difference if we replace $f(t,|E\rangle)$ by $|E\rangle$ in formula 2.
That then gives us:
$|\psi(t)\rangle = {\LARGE{\Sigma}}_{\small{E\in J}}\ \langle E\ |\ \psi(0)\rangle\ |E\rangle\ \ \ \ \ \$ Formula 3
But now we have a formula from which $t$ has disappeared, which implies that the state does not change over time, which is clearly not the case.
Where did I go wrong?
4. Jan 21, 2012
### akhmeteli
No, it isn't. You did not multiply the ket by a complex number, you multiplied it by a complex function (unlike a complex number, it depends on t). And I don't think this is the only problem.
5. Jan 21, 2012
### The_Duck
If you multiply the overall state vector by a phase, the physical state does not change. The *overall* phase is not physically significant. But time evolution does not do this, it multiplies different components of the overall state vector by different phases. If you write the state as a superposition of energy eigenstates, the *relative* phases between these eigenstates *are* physically significant.
I suggest taking an explicit example, like
|1> + exp(i alpha) |2>
where |1> and |2> are the first two energy eigenstates of the infinite square well and alpha is a number between 0 and 2*pi. Try computing the expectation value of the position operator (this is clearly physically meaningful) in this state and you will see that it depends on alpha.
If you want you could do the same thing for the state
exp(i alpha) |1> + exp(i beta) |2>
and you will find that physically meaningful numbers depend only on the relative phase (alpha - beta).
6. Jan 21, 2012
### andrewkirk
Why do you think that observation would change things? If it were a function of position or momentum, I can see this would be an issue, but it is not - it's only a function of t. At any given future time t', $e^{-iEt'/h}$ is a simple complex number, not a function of position or momentum, so $f(t',|E\rangle) = \ |E\rangle\ e^{-iEt'/h}$ is just $|E\rangle$ multiplied by a complex number of unit modulus.
In fact, Shankar even uses the term "stationary state". This is what he says:
"The normal modes $|E(t)\rangle\ =\ |E\rangle\ \ e^{-iEt/h}$ are also called stationary states for the following reason: the probability distribution P(ω) for any variable Ω is time-independent in such a state."
This makes me think the time-dependence must somehow be introduced via the summation over different eigenvalues, but I can't see how that would work.
7. Jan 22, 2012
### akhmeteli
With all due respect, we are talking about physics and mathematics, not about abstract painting, so we have to deal with some precise statements and take them seriously. What's happening, actually? You quote Shankar, who discusses multiplying by a complex number, then you multiply by a time-dependent function and get some nonsense as a result. I would say this is your problem, not Shankar's, not mine. This is pretty much the same as if you read in a book that the sum of angles of a planar triangle always equals 180 degrees and then complained that the sum of angles of a triangle on a sphere is less than 180 degrees. I don't know how to explain that this is wrong. As I said, this is not the only problematic thing about your reasoning, but until you remove this problem, further discussion of your reasoning does not make much sense.
8. Jan 22, 2012
### tom.stoer
As said physical states are rays in an Hilbert space; as such physics doesn't change if all state vectors $|\psi\rangle$ of a given system are multiplied by a constant number z with modulus one.
But b/c the dynamics (time dependence) of a stationary state is encoded in it's phase it's nonsense to use a time dependent phase instead.
9. Jan 22, 2012
### andrewkirk
Thanks Duck and Tom, your comments really helped me to get a clearer view of this. That example you suggested Duck clarified things a lot.
What follows is how I understand it now, including where the problem was, really written just for my own benefit to help solidify my thoughts, and in case anybody else with this query comes across this discussion:
A ket in the Hilbert space completely describes a state. Any ket that is a complex scalar multiple of that ket (ie in the same ray) describes the same state. However that does not mean they are the same ket. They are not. It is a many-to-one relationship from kets to states.
When we express the evolved state of a system in terms of the energy basis, we write the system's representative ket as a sum of the basis kets. Note this is a sum of kets, not a sum of states. We can multiply the whole sum by a phase factor exp(iBt) without changing the system state, as that is just multiplying the state's ket by a complex scalar. That is why a state corresponding to an energy eigenket is called stationary, because it physically does not change as time goes by.
For a state with nonzero projection on more than one of the energy eigenkets, its expansion in the energy basis is a sum of more than one basis ket. These kets evolve over time by changing phase with different frequencies. While the state to which each ket in the sum can be mapped does not change as the phase changes, that is irrelevant, as the overall state is specified via a ket that is a sum of kets, not states. That ket sum changes over time because of the differing phases of the components, and that change involves a change of state.
I think the source of the confusion is that people sometimes talk of a quantum state as being a superposition of states which, in my mind at least, conjures up images of some kind of a sum of states, which is an error. Indeed I wonder whether the concept of superposition is actually intrinsically misleading and best avoided.
10. Jan 22, 2012
### Ken G
I think the problem is actually that you are associating the "superposition" with the coefficients of the kets. The problem is not that the superposition is an invalid concept because states not unambiguously associated with kets, the problem is that the superposition is not the list of ket coefficients. The kets are acting like a basis here, and the coefficients of the kets are acting like coordinates, and like any coordinates, the same state can be associated with many different coordinates if the basis is different. All the same, there is a physical connection between the superposition state and the physical states that make up that superposition. There is just some ambiguity in the coordinates of that superposition.
Similar Discussions: Question about evolution of a wave function | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9150401949882507, "perplexity": 431.78961421175484}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818694719.89/warc/CC-MAIN-20170926032806-20170926052806-00099.warc.gz"} |
https://www.gradesaver.com/textbooks/math/applied-mathematics/elementary-technical-mathematics/chapter-2-section-2-6-scientific-notation-exercises-page-127/35 | Elementary Technical Mathematics
When the exponent is positive, move the decimal that many places to the right to change from scientific notation to decimal form. $5\times10^{10}=5\,\underrightarrow{0000000000}.$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9091727137565613, "perplexity": 687.9019654345325}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583850393.61/warc/CC-MAIN-20190122120040-20190122142040-00520.warc.gz"} |
https://www.groundai.com/project/domain-dependent-stability-analysis-and-parameter-classification-of-a-reaction-diffusion-model-on-spherical-geometries/ | [
# [
## Abstract
In this work an activator-depleted reaction-diffusion system is investigated on polar coordinates with the aim of exploring the relationship and the corresponding influence of domain size on the types of possible diffusion-driven instabilities. Quantitative relationships are found in the form of necessary conditions on the area of a disk-shape domain with respect to the diffusion and reaction rates for certain types of diffusion-driven instabilities to occur. Robust analytical methods are applied to find explicit expressions for the eigenvalues and eigenfunctions of the diffusion operator on a disk-shape domain with homogenous Neumann boundary conditions in polar coordinates. Spectral methods are applied using chebyshev non-periodic grid for the radial variable and Fourier periodic grid on the angular variable to verify the nodal lines and eigen-surfaces subject to the proposed analytical findings. The full classification of the parameter space in light of the bifurcation analysis is obtained and numerically verified by finding the solutions of the partitioning curves inducing such a classification. Furthermore, analytical results are found relating the area of (disk-shape) domain with reaction-diffusion rates in the form of necessary conditions for the different types of bifurcations. These results are on one hand presented in the form of mathematical theorems with rigorous proofs, and, on the other hand using finite element method, each claim of the corresponding theorems are verified by obtaining the theoretically predicted behaviour of the dynamics in the numerical simulations. Spatio-temporal periodic behaviour is demonstrated in the numerical solutions of the system for a proposed choice of parameters and a rigorous proof of the existence of infinitely many such points in the parameter plane is presented under a restriction on the area of the domain, with a lower bound in terms of reaction-diffusion rates.
R
\checkfont
eurm10 \checkfontmsam10 \newdefinitiondefinition[theorem]Definition European Journal of Applied Mathematics]Domain-dependent stability analysis and parameter classification of a reaction-diffusion model on spherical geometries W. Sarfaraz et al.]W.\nsS\lsA\lsR\lsF\lsA\lsR\lsA\lsZ,\ns and A.\nsM\lsA\lsD\lsZ\lsV\lsA\lsM\lsU\lsS\lsE 2000 \volume000 \pagerange[References
eaction-diffusion systems, Dynamical systems, Bifurcation analysis, Stability analysis, Turing diffusion-driven instability , Hopf Bifurcation, Transcritical bifurcation, Parameter spaces, Polar coordinates, Curved boundary
## 1 Introduction
Analysis of reaction-diffusion systems (RDSs) in the context of pattern formation is a widely studied topic [50, 56, 1, 2, 23, 30, 31, 32] in many branches of scientific research. Scholars of mathematical and computational biology [3, 4, 5, 6, 7, 11, 13, 17] devoted a great deal of attention to fully explore the theory of the dynamics governed by reaction-diffusion systems for a variety of reaction kinetics. Few of the routinely used and popular reaction kinetics in the theory of reaction-diffusion systems are activator-depleted [29, 28, 26, 27, 22, 20, 63], Gierer-Meinhardt [24, 23] and Thomas reaction kinetics [25, 37]. It is crucial to realise that the ultimate and complete knowledge encapsulating all aspects of reaction-diffusion systems for all types of possible reaction-kinetics is beyond the scope and feasibility of a single research paper or even a single book. Therefore, literature to date [37]-[50] on the analysis of reaction-diffusion systems falls under a natural classification of approaches based on the application of methods corresponding to the speciality of a particular scholar. It is in this spirit that the current work is a natural extension of [38], in which a full classification of the admissible parameter space associated to the dynamical behavior of activator-depleted reaction-diffusion system was explored on stationary rectangular domains. It is a common hypothesis in the theory of biological pattern formation [39, 13], that the emergence of spatially periodic pattern in the computational simulations of reaction-diffusion systems is connected to the properties of the associated eigenfunctions of the diffusion operator in the corresponding domain. The associated drawback with computational approaches [10, 12, 15, 16, 19, 21, 37, 53, 54, 55] for reaction-diffusion systems is that it lacks to provide rigorous insight of the role of the eigenfunctions in the emergence of spatial pattern, thus forming a natural platform to explore the evolution of spatial patterns from a perspective of dynamical systems. Numerous researchers have contributed to exploring the computational aspect of reaction-diffusion systems [4, 12, 14, 15, 26, 27], in which a restricted choice of parameter values are used with a partial reliance on trial and error to obtain the emergence of an evolving pattern either in space or in time. A large variety of scholars have also employed the analytical approach [1, 9, 13, 17, 18, 20, 28], to explore reaction-diffusion systems from a perspective of dynamical systems. Majority of cases with analytical approaches concluded with certain conditions [9, 17, 18, 20, 28, 31] on the characteristics of the stability matrix that theoretically predicts reaction-diffusion systems to exhibit spatial patterns, lacking to extend the analysis to computational consequences of these conditions on the admissible parameter spaces in terms of diffusion-driven instability. Furthermore, from the literature to date [28, 30, 39, 45, 46, 47, 48], it is evident that insufficient attention is given to rigorously explore the influence of reaction-diffusion rates on the dynamical behaviour of such systems in the context of domain size. The motivation of the current work originates from the hypothesis of the spatial dependence of the eigenfunctions of the diffusion operator with the domain size itself and extending this idea to further investigate how this spatial dependence induces an influence on the linearised stability-matrix of a reaction-diffusion system through the properties of the eigenfunctions of the diffusion operator. With an attempt to further contribute to the current knowledge of reaction-diffusion systems, the present paper employs a set of rigorous findings obtained from bifurcation analysis of the activator-depleted reaction-diffusion model to explore the corresponding influence induced on the admissible parameter spaces and diffusion-driven instability. Despite the restriction of the current work to a particular type of reaction kinetics namely activator-depleted, the methodology can however serve as a framework for developing a standard independent method by combining the known aspects of exploring the topic namely bifurcation analysis, parameter spaces and computational methods, that could be utilised for general reaction-diffusion systems. The lack of a self-contained complete methodology that combines all the known distinct aspects of exploring this topic creates a credible argument for the importance of this work. The contents of the current paper offer a natural extension of the work presented in [38], where the domain of solution was restricted to rectangular geometries. Equivalent results to those presented in [38] are found in the present work, except that the domain of solution in this work is a two dimensional disk-shape geometry bounded by a circle. The bulk of the current work consists of rigorously proven statements that quantitatively relate the radius of a disk-shape solution domain to reaction-diffusion rates in light of bifurcation analysis, which are computationally verified using the finite element method.
The contents of this work are structured such that in Section 2 the model equations for activator-depleted reaction-diffusion system are stated in cartesian coordinates and transformed to polar coordinates in its non-dimensional form with the corresponding initial and boundary conditions. Section 3 consists of a detailed analytical method to explicitly derive eigenfunctions and the corresponding eigenvalues satisfying the boundary conditions prescribed for activator-depleted reaction-diffusion system in Section 2. Furthermore, in Section 3, an application of spectral approach on polar coordinates using chebyshev and Fourier grid method is presented [57] to demonstrate and verify the analytical derivation of the eigenfunctions as well as the stability matrix and the corresponding characteristic polynomial for a linearised approximation of the original system. Section 4 presents analytical results on the relationship of domain-size (radius of disk) with different types of bifurcations in the dynamics. Section 5 is devoted to the parameter space classification in light of bifurcation analysis of the uniform steady-state of the system. A numerical method for computing the partitioning curves for such classification is presented in detail. The shift of parameter spaces as a consequence of changing reaction-diffusion rates is investigated and theorems are proven that rigorously establish the relation of domain size (radius of disk) with the corresponding types of diffusion-driven instabilities. Furthermore, a full classification of the admissible parameter space in terms of stability and types of the uniform steady state of the system is also presented in Section 4, where it is shown that if the radius of a disk-shape domain satisfies certain inequalities in terms of reaction-diffusion rates, then the admissible parameter space allows or forbids certain types of bifurcations in the dynamics of the reaction-diffusion system. Section 6 presents numerical simulations of an activator-depleted reaction-diffusion system using finite element method to verify the proposed classification of parameter spaces and the theoretically predicted behaviour in the dynamics. Due to the curved boundary of the domain a non-standard technique called distmesh [40, 41, 44] is used to obtain discretisation for simulating the finite element method. The technicality of the algorithm for distmesh is briefly explained. Section 7 presents the conclusion and possible extensions of the current work.
## 2 Model equations
Two chemical species and are modelled by the well-known activator-depleted reaction-diffusion system, where both species are coupled through nonlinear reaction terms. The system assumes independent diffusion rates for both of the species. The RDS is considered on a two dimensional circular domain denoted by , where is defined by
Ω={(x,y)∈R2:x2+y2<ρ2}.
This forms a two dimensional disk-shape domain with a circle forming its boundary denoted by , which contains the points given by
∂Ω={(x,y)∈R2:x2+y2=ρ2}.
The RDS with activator-depleted reaction kinetics in its non-dimensional form on cartesian coordinates has the form
⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩{∂u∂t=△u+γ(α−u+u2v),(x,y)∈Ω,t>0,∂v∂t=d△v+γ(β−u2v),∂u∂n=∂v∂n=0,on(x,y)∈∂Ω,t≥0,u(x,y,0)=u0(x,y),v(x,y,0)=v0(x,y),(x,y)∈Ω,t=0, (1)
where , , and are strictly positive real constants. In system (1) the positive parameter denotes the non-dimensional ratio of diffusion rates given by , where and are the independent diffusion rates of and respectively. The non-dimensional parameter is known as the scaling parameter of (1), which quantifies the reaction rate. The boundary of under the current study assumes homogeneous Neumann boundary conditions (also known as zero flux boundary conditions), which means that the chemical species and can neither escape nor enter through . denotes the normal to in the outward direction. Initial conditions for (1) are prescribed as positive bounded continuous functions and . System (1) can be spatially transformed to polar coordinates using and to obtain
⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩{∂u∂t=△pu+γf(u,v),∂v∂t=d△pv+γg(u,v),∂u∂r∣∣r=ρ=∂v∂r∣∣r=ρ=0,(r,θ)∈∂Ω,t≥0,u(r,θ,0)=u0(r,θ),v(r,θ,0)=v0(r,θ),(r,θ)∈Ω,t=0, (2)
where denotes the Laplace operator in polar coordinates written as
△pu(r,θ)=1r∂∂r(r∂u∂r)+1r2∂2u∂θ2. (3)
In (2), and depend on coordinates and the functions and are defined by and . Initial and boundary conditions are transformed in a similar fashion. In (2) the strictly positive constants namely , , and remain to satisfy exactly the same definitions as in (1).
## 3 Stability analysis of the reaction-diffusion system
### 3.1 Stability analysis of the reaction-diffusion system in the presence of diffusion
Let and denote the uniform steady state solution satisfying system (1) (and equivalently system (2)) with activator-depleted reaction kinetics in the absence of diffusion and these are given by [50, 56, 38, 17]. For linear stability analysis, system (2) is perturbed in the neighbourhood of the uniform steady state and the dynamics of the perturbed system are explored i.e. , where and are assumed small. In system (2) the variables and are substituted by the expression in terms of and and expanded using Taylor expansion for functions of two variables up to and including the linear terms with the higher order terms discarded. This leads to the linearised version of system (2) written in matrix form as
∂∂t[¯u¯v]=[100d][△p¯u△p¯v]+⎡⎢⎣∂f∂u(us,vs)∂f∂v(us,vs)∂g∂u(us,vs)∂g∂u(us,vs)⎤⎥⎦[¯u¯v]. (4)
The next step to complete the linearisation of system (2) is to compute the eigenfunctions of the diffusion operator namely , which will require to find the solution to an elliptic 2 dimensional eigenvalue problem on a disk that satisfies homogeneous Neumann boundary conditions prescribed for system (2).
The solution of eigenvalue problems on spherical domains is a well studied area [58, 33, 59], with the majority of research focused on problems with boundary-free manifolds such as circle, torus and/or sphere. Considering the restriction imposed from the boundary conditions prescribed for the current problem, entails that the case requires explicit detailed treatment to rigorously find the eigenfunctions satisfying these boundary conditions. Therefore, it is important to present a step-by-step demonstration of the process, starting with writing out the relevant eigenvalue problem all the way through to finding the closed form solution in the form of an infinite set of eigenfunctions satisfying such an eigenvalue problem. The eigenvalue problem we want to solve is of the form
{△pw=−η2w,η∈R,∂w∂r∣∣r=ρ=0,ρ∈R+∖{0}, (5)
where is the diffusion operator in polar coordinates defined by (3), on a disk with radius . We use separation of variables to obtain the solutions of (5). Application of separation of variables to problem (5) requires a solution of the form , which is substituted into problem (5) to obtain
d2Rdr2Θ+1rdRdrΘ+1r2Rd2Θdθ=−η2RΘ. (6)
Dividing both sides of (6) by , multiplying throughout by and rearranging, yields
r21Rd2Rdr2+r1RdRdr+η2r2=−1Θd2Θdθ2. (7)
Using anzats of the form and a change of variable is applied to (7). Taking the usual steps of Frobenius method it can be shown that the general solution satisfying (7) takes the form , where and are respectively given by
R1(x)=xn∞∑j=0(−1)jC0x2j4j×j!×(n+j)×(n+j−1)×⋅⋅⋅×(n+1) (8)
and
R2(x)=x−n∞∑j=0(−1)jC0x2j4j×j!×(−n+j)×(−n+j−1)×⋅⋅⋅×(−n+1). (9)
Series solutions (8) and (9) are referred to as the Bessel functions of the first kind [60, 59]. Before writing the general solution to problem (5), it must be noted that we require to be a function of and not of , bearing in mind that is related to under the linear transformation given by . Therefore, the general solution to problem (5) can be written in the form
w(r,θ)=R(x(r))Θ(θ), (10)
where and , with and defined by (8) and (9) respectively. The homogeneous Neumann boundary conditions are imposed on the set of eigenfunctions (10), and noting that the flux is independent of the variable , we have which implies that . A straightforward application of chain rule yields .
Let and denote the coefficients corresponding the term in the infinite series for and respectively, then for every the expressions for and take the forms
aj=(−1)jC04j×j!×(n+j)×(n+j−1)×⋅⋅⋅×(n+1), (11)
bj=(−1)jC04j×j!×(−n+j)×(−n+j−1)×⋅⋅⋅×(−n+1). (12)
This entails that can be written in the form . Differentiating with respect to and equating it to zero, on substituting and using the chain rule we obtain the equation
0=dRdr∣∣r=ρ=η∞∑j=0[aj(n+2j)xn+2j−1+bj(−n+2j)x−n+2j−1]∣∣x=ηρ=η∞∑j=0[aj(n+2j)(ηρ)n+2j−1+bj(−n+2j)(ηρ)−n+2j−1]=η[(ηρ)n∞∑j=0aj(n+2j)(ηρ)2j−1+(ηρ)−n∞∑j=0bj(−n+2j)(ηρ)2j−1], (13)
which holds true if and only if both of the summations in (13) are independently zero. Investigating the first summation in (13) and expanding it for a few successive terms, it can be shown that the successive terms carry alternating signs (due to the expression for ), therefore, the only way the infinite series can become zero is if each of the successive terms cancel one another. Let and denote the terms of the first and second summations in (13) respectively, then for (13) to hold true, and must be true for all . The full expressions for and can be written as
Fj=(ηρ)n(−1)jC0(n+2j)(ηρ)2j−14j×j!×(n+j)×(n+j−1)×⋅⋅⋅×(n+1) (14)
and
Fj+1=(ηρ)n(−1)j+1C0(n+2j+2)(ηρ)2j+14j+1×(j+1)!×(n+j+1)×(n+j)×⋅⋅⋅×(n+1). (15)
For the first , second and third successive pairs, independently equating the sum of the expressions (14) and (15) to zero yields that can be written as
η2n,k=4(2k+1)(n+2k+1)(n+4k)ρ2(n+4k+2). (16)
#### Remark
The restriction on the order of the corresponding Bessel’s equation namely in Theorem 3.1 can be relaxed by employing Bessel’s function of the second kind [58, 59, 60] and imposing on it the homogeneous Neumann boundary conditions. Therefore, for the purpose of the current study the order of the corresponding Bessel’s equation is precluded from becoming a full or half integer. Bear in mind that the proposed choice of makes the set of eigenvalues a semi discrete infinite set. The spectrum is discrete with respect to positive integers and continuous (uncountable) with respect to .
###### Theorem 3.1
Let satisfy problem (5) with homogeneous Neumann boundary conditions. Given that the order of the associated Bessel’s function belongs to the set , [see Remark 3.1.1] then for a fixed there exists an infinite set of eigenfunctions of the diffusion operator as defined in (3), which is given by
wn,k=[R1n,k(r)+R2n,k(r)]Θn(θ) (17)
with the explicit expressions for , and as
R1n,k(r)=∞∑j=0(−1)jC0(ηkr)2j+n4jj!(n+j)(n+j−1)⋅⋅⋅(n+1), (18)
R2n,k(r)=∞∑j=0(−1)jC0(ηkr)2j−n4jj!(−n+j)(−n+j−1)⋅⋅⋅(−n+1) (19)
and
Θn(θ)=exp(inθ), (20)
where for the successive pair satisfies (16), when ever . {proof} The proof consists of all the steps from (6) to (16).
It is worth noting that for a set containing terms from the series solutions defining the eigenfunctions (17), one can only obtain eigenvalues that satisfy problem (5). Therefore, it is important to realise that unlike the eigenfunctions of a diffusion operator [38, 63] on a rectangular geometry, here only pairwise terms can qualify in the series solutions to be claimed as the eigenfunction satisfying problem (5) and not individual terms associated discretely to the summation index .
### 3.2 Numerical experiments using the spectral method
Let denote the expression , then the full set of eigenfunctions to problem (5), can be written as . For numerical demonstration of Theorem 3.1, the spectral method [57] is employed on a unit disk centred at the origin of the coordinates. A spectral mesh in polar coordinates is constructed on a unit disk , where a periodic Fourier grid is applied to the angular axis and a non-periodic chebyshev grid is applied to the radial axis [57]. In order to tackle the singularity at , chebyshev discretisation is applied to the whole of the diameter of , which means that is used instead of . The interval is discretised in the form , for , where , is a positive odd integer. The odd number of points on the chebyshev grid serves to preclude complications that could rise from the singular point by locating it radially in the middle of two successive chebyshev grid points. The second coordinate is discretised using , for , where , is a positive even integer. For the full details on the implementation of the spectral method in polar coordinates the reader is referred to [57]. Figure 3 (a) shows a coarse mesh structure on a unit disk using and , with angular step-size of . Similarly, Figure 3 (b) shows the refined mesh using and with angular step-size of , on which all the simulations for eigenfunctions are performed to visualise the corresponding nodal lines and surfaces. Colour encoded plots, corresponding to nine different modes are simulated on the mesh given in Figure 3 (b) using a technique presented in [35, 36]. Due to the fact that eigenfunctions are complex valued functions, therefore, trivial methods of visualising a real valued function of two variables do not suffice to give a meaningful representation to a complex valued function. Note that in the expression for the eigenfunctions , only the function contains the non-zero imaginary part, therefore, the variable is encoded by a polar colour scheme (Hue, Saturation Value) and displayed directly on . For full details on the implementation of this process the reader is referred to [35, 36]. Figure 13 shows a colour encoded representation of nine different modes, each of which corresponds to one of the nine nodal line depictions of in Figure 14.
Note that in Figure 14 different modes namely modes 2 and 3 result in the same eigenvalues that correspond to a different orientation of the nodal line depiction for and . Similarly, one may find that there are a lot of such pairs of positive integers and , that correspond to different orientations of the same nodal line depiction. This occurs, when the multiplicity of an eigenvalue is 2, and in fact there are numerous such pairs of and , that correspond to the same , but for different orientation of eigenmodes. However applying the colour encoded representation using the (HSV) colour scheme indicates a distinction between the plots corresponding to each integer value for as shown in Figure 13.
### 3.3 Stability matrix and the characteristic polynomial
The solution to system (2) using separation of variables, can be written (with bars omitted from and ) as the product of the eigenfunctions of the diffusion operator and in the form
u(r,θ,t)=∞∑k=0Un,kexp(σn,kt)Rn,k(r)Θn(θ),v(r,θ,t)=∞∑k=0Vn,kexp(σn,kt)Rn,k(r)Θn(θ),
where and are the coefficients associated with the mode of the eigenfunctions in the infinite expansion. Substituting this form of solution in (4) and the steady state values in terms of the parameters and for , one obtains the fully linearised form of (2) as a system of two algebraic equations of the form
Unknown environment '%' (21)
where . In turn, this can be written as a two dimensional discrete eigenvalue problem of the form
⎡⎢⎣γβ−αβ+α−η2n,kγ(β+α)2−γ2ββ+α−γ(β+α)2−dη2n,k⎤⎥⎦[uv]=σ[uv]. (22)
The left-hand matrix in (22) is referred to as the stability matrix [50, 56, 12] for system (2), with given by (16). In order to investigate the stability of the uniform steady state , it is required to analyse the eigenvalues satisfying (22), for which the characteristic polynomial takes the form of a quadratic equation in , written as
∣∣ ∣ ∣∣γβ−αβ+α−η2n,k−σγ(β+α)2−γ2ββ+α−γ(β+α)2−dη2n,k−σ∣∣ ∣ ∣∣=0. (23)
Let and respectively denote the trace and determinant of the stability matrix given by (22), then the quadratic polynomial (23) can be written in terms of and as
σ2−T(α,β)σ+D(α,β)=0, (24)
with and expressed by
⎧⎪⎨⎪⎩T(α,β)=γβ−α−(β+α)3β+α−(d+1)η2n,k,D(α,β)=(γβ−αβ+α−η2n,k)(−γ(β+α)2−(d+1)η2n,k)+2γ2β(β+α). (25)
The roots of equation (24) in terms of and are . Stability of the uniform steady state is determined by the signs of the two roots namely if they are real. However, if are a complex conjugate pair (with non-zero imaginary parts), then the sign of the real part is sufficient to predict the stability of the uniform steady state . Since there are always exactly two roots on a complex plane for a quadratic equation, therefore, it is impossible with the same choice of parameters , , , and , for to be real and to be complex or vice versa. Therefore, a reasonable approach to encapsulate all the possibilities for the stability and types of the uniform steady state in light of parameters and is to consider the cases when and . In each case the parameter space is rigorously analysed and the classification of the parameter plane in relation to the diffusion parameter is studied. In light of such classification, the analysis is further extended to explore the effects of domain size on the existence of regions in parameter space that correspond to spatial and/or temporal bifurcations.
## 4 Parameter spaces and bifurcation analysis
Bifurcation analysis of system (2) is better conducted when the parameter plane is appropriately partitioned for both of the cases when as well as when . To obtain such a partition on the parameter plane, it is required to find the equations of the partitioning curves and these can be found through a detailed analysis of the expression for , which in turn requires to explore the domains of and .
### 4.1 Equations of the partitioning curves
Starting with the curve on the parameter plane that forms a boundary for the region that corresponds to eigenvalues containing non-zero imaginary part. It must be noted that the only possibility through which can have a non-zero imaginary part is if the inequality is true. It means that those parameter values and satisfying the equation must be lying on a partitioning curve that determines the boundary between the region on the parameter plane that corresponds to eigenvalues with non-zero imaginary part and that which corresponds to a pair of real eigenvalues. We therefore, state that the set of points on the parameter plane satisfying the implicit equation
Missing or unrecognized delimiter for \Big (26)
forms the partitioning curve between the region that corresponds to a real pair of and that corresponding to a complex conjugate pair of . For the solution of (26) refer to Section 5, where a numerical method is employed to find combinations of on the plane satisfying (26). It is worth noting that a combination of and satisfying (26), entails that the expression for possesses a repeated real root of the form , whose sign will require to be explored in Section 5 for real eigenvalues. Another important point to be made about the curve satisfying (26) is that it also partitions the parameter space for regions of spatial and temporal bifurcations. Because on one side of curve (26), both of the eigenvalues are always real, which can never excite temporal instability of the uniform steady state , whereas on the other side, the eigenvalues are always a complex conjugate pair, which can never excite spatial instability. Therefore, on the side where are a pair of real values, any instability that occurs in the dynamics of system (2) will be strictly relevant to spatial variation, hence any pattern that system (2) can evolve to will be strictly spatial with stable and invariant evolution in time. However, on the side where have non-zero imaginary part, every possible instability in the dynamics of system (2) will be strictly concerned with temporal periodicity, therefore any pattern that emerges from the dynamics of system (2) is expected to be periodic along the time axis. This analysis raises the interesting question whether it is possible for the dynamics of (2) to cause a spatially periodic pattern to undergo an unstable temporal periodicity as well? To answer this question, on the first sight it sounds absurd to claim on one hand that a specific choice of admissible can either cause spatial or temporal instability, and not both at the same time, which is an intuitive claim to make. Because, any admissible choice of either yields a pair of real or a complex conjugate pair of , therefore, a fixed choice of is not expected to yield a real and a complex or vice versa. Therefore, it is intuitive to presume that system (2) should not admit the evolution of such dynamics, in which a combination of spatial and temporal instabilities can occur. However, the current study finds that this counter-intuitive behaviour is possible for system (2) to exhibit, which is related to the existence and shift in the location of a partitioning curve on which the real part of becomes zero during the course of the temporal evolution. Such a curve is referred to as the transcritical curve, for which the implicit equation in terms of admissible and is of the form
γβ−α−(β+α)3β+α=(d+1)η2n,k, (27)
under the assumption that . It must be noted, that equations (26) and (27) are the only two equations that fully partition the parameter plane . However, one may expect the number of partitions subject to the types and stability of the uniform steady state , on the parameter plane to be four regions separated by three curves. These regions are obtained by sub-partitioning the region corresponding to a pair of real , into two sub-regions where is a pair of real negative values and another where at least or is positive. The region on the parameter plane , that corresponds to complex eigenvalues can also be sub-divided into two regions, one where is a complex conjugate pair with negative real part, and the other where is a complex conjugate pair, but with a positive real part. The numerical solution of equation (26) will reveal that the region corresponding to complex eigenvalues is in fact bounded by two curves in the parameter plane , each of which satisfies (26). Therefore, (26) is implicitly the equation of two partitioning curves instead of one, which including (27) makes a total of three curves dividing the admissible parameter space into four different regions. It is worth bearing in mind, that subject to the types and stability of the uniform steady state, one obtains at most four regions separated by three curves, with the possibility that parameters such as , and may induce equations (26) and (27) in such a way that the number of partitioning curves may become less than three in total. Consequently this entails that a region corresponding to a certain type of bifurcation may completely disappear from the admissible parameter space. This kind of influence on the location and existence of the partitioning curves is quantitatively investigated, in particular, the effect of embedded in the expression for in (16) is explored to analyse its influence on the bifurcation of the uniform steady state.
#### Analysis for the case of complex eigenvalues
Before determining the region with complex eigenvalues on the admissible parameter plane using a numerical treatment of (26), the real part of is investigated analytically, when it is a complex conjugate pair. It can be noted that can only become a pair of complex roots, if satisfies the inequality
T2(α,β)−4D(α,β)<0. (28)
Given that (28) is satisfied, then the stability of the uniform steady steady state is decided purely by the sign of the real part of , which is the expression
Re(σ1,2)=12(γβ−α−(β+α)3β+α−(d+1)η2n,k). (29)
If the sign of the expression given by (29) is negative, simultaneously with assumption (28) satisfied, then it can be predicted that no choice of parameters can cause temporal instability in the dynamics of system (2). Therefore, under the assumption (28), if the dynamics of system (2) do exhibit diffusion-driven instability, it will be restricted to spatially periodic behaviour only, which uniformly converges to a temporal steady state, consequently one obtains spatial pattern that is invariant in time. The sign of the expression given in (29) is further investigated to derive from it, relations between the parameter controlling the domain size and reaction-diffusion rates denoted by and respectively. Given that assumption (28) is satisfied then the sign of expression (29) is negative if parameters , , and satisfy the inequality
β−α−(β+α)3β+α<(d+1)η2n,kγ, (30)
with defined by (16). Note that the expression on the left hand-side of (30) is a bounded quantity by the constant value of 1 [38], for all the admissible choices of , therefore, substituting for in expression (16) and rearranging, it can be obtained that for inequality (30) to remain true, it induces a restriction on the value of , which is of the form
ρ2<4(d+1)(2k+1)(n+2k+1)(n+4k)γ(n+4k+2). (31)
Inequality (31) conversely implies that so long as the radius of the disk shape domain satisfies (31), then the dynamics of system (2) is guaranteed to exhibit global temporal stability, which also means that any possible instability in the dynamics must be restricted to spatial periodicity or spatial pattern. The formal proof of this claim is presented in Theorem 33. This type of instability concerning space and not time is referred to as Turing instability [50, 56]. On the other hand assuming that (28) is satisfied and using the upper bound of the quantity on the left hand-side of (30) with as defined in (16), it can be shown that a necessary condition for the sign of expression (29) to become positive is for to satisfy the inequality
ρ2≥4(d+1)(2k+1)(n+2k+1)(n+4k)γ(n+4k+2). (32)
Conditions (31) and (32) both have quantitative influence on the location and topology of the partitioning curves obtained from the numerical solutions of (26) and (27) in the admissible parameter plane, namely . System (2) is restricted from any type of temporal bifurcation if the radius of the disk shape domain is related to the reaction-diffusion parameters and , through inequality (31). It also means that on a disk shape domain with radius satisfying (31), the dynamics of system (2), either exhibit spatially periodic pattern or no pattern at all, both of which are globally stable in time. If the dynamics of a system become unstable along the time axis and exhibits temporal periodicity, then the system is said to undergo Hopf bifurcation [61, 50, 39, 2]. If the real part of a pair of complex eigenvalues become zero, then the system is expected to exhibit oscillations with orbital periodicity. This behaviour is known as transcritical bifurcation [50, 2, 39]. The consequences of the restriction on in the sense of bifurcation analysis means that, whenever the radius of a disk-shape domain is bounded by (31) in terms of and , then the dynamics of system (2) is guaranteed to forbid Hopf and transcritical bifurcations, only allowing for Turing instability to occur. If system (2) allows only Turing type instability to occur under condition (31), it indicates that the eigenvalues only become positive, when they are a pair of real values with zero imaginary parts. Therefore, it is also an indication that diffusion-driven instability is still possible, but it just becomes strictly spatial with (31) satisfied. If the values of parameters and are chosen such that inequality (32) is satisfied, then the possibility of all three types of diffusion-driven instabilities exist on the admissible parameter plane , namely Turing, Hopf and transcritical types of bifurcations. The influence from the area of through the relationship (32) of with the reaction-diffusion rates namely and is summarised in Theorem 33 with a detailed sketch of the proof.
###### Theorem 4.1 (Hopf or transcritical bifurcation)
Let and satisfy the non-dimensional reaction-diffusion system with activator-depleted reaction kinetics (2) on a disk-shape domain with radius and positive real parameters , , and . For the system to exhibit Hopf or transcritical bifurcation in the neighbourhood of the unique steady state , the necessary condition on the radius of the disk-shape domain is that it must be sufficiently large satisfying
ρ≥2√(d+1)(2k+1)(n+2k+1)(n+4k)γ(n+4k+2), (33)
where is the associated order of the Bessel’s equations and is any positive integer.
{proof}
[Proof:] For system (2) to exhibit Hopf or transcritical bifurcations the eigenvalues of the stability matrix (22) must have non-zero imaginary part with non-negative real part. Consider the real part of , which is precisely given by under the assumption that the admissible choice of parameters satisfies (28). When , then the stability of the uniform steady state is precisely determined by the sign of , which is given by
T(α,β)=γβ−α−(β+α)3β+α−(d+1)η2n,k. (34)
System (2) undergoes Hopf or transcritical bifurcation if , given that the strict inequality (28) is satisfied, which can only hold true if . In (34) is given by
η2n,k=4(2k+1)(n+2k+1)(n+4k)ρ2(n+4k+2), (35)
where is the order of the associated Bessel’s equation and is any positive integer. To show the condition on for Hopf or transcritical bifurcation, one may substitute (35) into (34) and requiring the resulting quantity to be non-negative, which yields the inequality
γβ−α−(β+α)3β+α≥4(d+1)(2k+1)(n+2k+1)(n+4k)ρ2(n+4k+2). (36)
Noting that the left hand-side of (36) can be written as the difference between two non-negative functions and in the form , where and are given by
f1(α,β)=βα+β,f2(α,β)=α+(α+β)3α+β. (37)
Note also that resides in the denominator of the right hand-side of (36) and parameter is multiplied by the expression on the left hand-side. In order to find what this inequality induces on the relationship between parameters , and , it is essential to analyse the supremum and infemum of and within their respective domains which is . The range for and are independently analysed to find the supremum of the expression on the left of (36). Starting with , which is bounded below and above in the domain , we have , and the for all Similarly considering the expression for , we have and the , for all Since the ranges of both and are non-negative within their respective domains, therefore the supremum of their difference is determined by the supremum of the function with positive sign, which is . Therefore, inequality (36) takes the form
4(d+1)(2k+1)(n+2k+1)(n+4k)ρ2(n+4k+2)≤γβ−α−(β+α)3β+α≤γsupα,β∈R+(f1(α,β)−f2(α,β))=γsupα,β∈R+f1(α,β)=γ,
which by rearranging and writing the inequality for in terms of everything else, yields the desired statement of Theorem 33, which is condition (33). The claim of Theorem 33, is also numerically verified by showing that a region in the admissible parameter plane that corresponds to Hopf or transcritical bifurcations emerges only if radius of a disk shape domain is sufficiently large satisfying inequality (33). Otherwise, no choice of parameters exist in the admissible parameter plane , allowing the dynamics of (2) to exhibit Hopf or transcritical bifurcation.
#### Analysis for the case of real eigenvalues
The eigenvalues are both real if the discriminant of the roots is either zero or positive, which in turn means that the relationship between and is such that
T2(α,β)≥4D(α,β). (38)
The equal case of (38) is looked at first, where we have
T2(α,β)=4D(α,β), (39)
which implies that the discriminant is zero, hence the roots are repeated real values of the form , given by
σ1=σ2=12(γβ−α−(β+α)3β+α−4(d+1)(2k+1)(n+2k+1)(n+4k)ρ2(n+4k+2)). (40)
When and satisfy condition (39), the stability of the steady state is determined by the sign of the root itself. The expression given by (40) can be easily shown to be negative if the radius of the disk-shape domain satisfies the inequality
ρ<2√(α+β)(d+1)(2k+1)(n+2k+1)(n+4k)γ(β−α−(α+β)3)(n+4k+2). (41)
Otherwise, the repeated root is positive provided that satisfies
ρ>2√(α+β)(d+1)(2k+1)(n+2k+1)(n+4k)γ(β−α−(α+β)3)(n+4k+2). (42)
Analysing (41) and (42) carefully, it can be observed that the only terms that can possibly invalidate the inequalities are in the denominator of the right hand-side, namely the expression . Therefore, a restriction is required to be stated on this term to ensure that the radius of is not compared against an imaginary number, such a restriction is
β>α+(β+α)3. (43)
It must be noted that (43) is the same restriction on the parameter choice obtained for the case of repeated real eigenvalues in the absence of diffusion [38]. By further comparing with (42), it can be noted that it is very similar to condition (33) of Theorem 33, except that (33) is free from any dependence of the parameters and . This makes (33) a sharper version of (42) in the sense, that the curve satisfying (39) subject to condition (42) must be the one forming the boundary of the region on the admissible parameter plane, that corresponds to complex eigenvalues with positive real parts, which is the region for Hopf bifurcation. Therefore, the region of the admissible parameter plane that corresponds to Hopf bifurcation is on one side bounded by curve (39) and on the other side it is bounded by the curve satisfying (27) under their respective assumptions. In Section 5 it is verified to be the case by the numerical computation of the partitioning curve (26), which is the same curve (39). This analysis motivates to explore the possibility of similar comparison between the conditions (41) and (31). A reasonable intuition behind this comparison is that the sub-region on the admissible parameter plane that corresponds to complex eigenvalues with negative real parts must be bounded by curve (39) subject to condition (41), outside of which every possible choice of parameters and will guarantee the eigenvalues to be a pair of distinct real values, which promotes the necessity to state and prove Theorem 44.
###### Theorem 4.2 (Turing type diffusion-driven instability)
Let and satisfy the non-dimensional reaction-diffusion system with activator-depleted reaction kinetics (2) on a disk-shape domain with radius and positive real parameters , , and . Given that the radius of domain satisfies the inequality
ρ<2√(d+1)(2k+1)(n+2k+1)(n+4k)γ(n+4k+2), (44)
where is the associated order of the Bessel’s equations and is any positive integer, then for all | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9440467953681946, "perplexity": 305.5444757964299}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267165261.94/warc/CC-MAIN-20180926140948-20180926161348-00074.warc.gz"} |
http://platonicrealms.com/encyclopedia/binary-operation | PRIME
Platonic
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# binary operation
A binary operation is a function that maps ordered pairs of elements of a set to elements of the set. For example, addition of natural numbers maps every pair of natural numbers to their sum, so addition is a binary operation on natural numbers.
Apart from the common operations such as addition, multiplication, dot-product, etc., a binary operation is commonly denoted by placing an asterisk between the elements: $$a*b$$.
If a binary operation has the property that $$a*b=b*a$$ for every $$a$$ and $$b$$ in the set, then the operation is said to be commutative. If a binary operation has the property that $$(a*b)*c=a*(b*c)$$ for all $$a$$, $$b$$, and $$c$$ in the set, then the operation is said to be associative.
Citation Info
• [MLA] “binary operation.” Platonic Realms Interactive Mathematics Encyclopedia. Platonic Realms, 19 Mar 2013. Web. 19 Mar 2013. <http://platonicrealms.com/>
• [APA] binary operation (19 Mar 2013). Retrieved 19 Mar 2013 from the Platonic Realms Interactive Mathematics Encyclopedia: http://platonicrealms.com/encyclopedia/binary-operation/ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8560842275619507, "perplexity": 476.80056021363424}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891814300.52/warc/CC-MAIN-20180222235935-20180223015935-00683.warc.gz"} |
https://stats.stackexchange.com/questions/365349/standard-error-of-circular-standard-deviation | # Standard error of circular standard deviation
How to calculate the standard error of circular standard deviation estimated with equation
1. $s = (2(1 − R))^{-\frac{1}{2}}$
or
1. $s_{0} = (−2 \ln{R})^{-\frac{1}{2}}$
where R is the length of the mean resultant vector (from here).
It should first be noted that the problem is not fully defined currently, because one would need to know the (assumed) distribution of the data before getting to a result that we can actually calculate. I will assume you are using the von Mises distribution.
Most of the theory devoted to this problem is in terms of the distribution of $R,$ $\bar{R} = R/n,$ or in the specific case of the von Mises distribution $\kappa$.
From Mardia & Jupp (2000), 5.3.17, we can get the standard error for $\hat\kappa$ following
$$n \text{var}(\hat\kappa) \approx \frac{1}{1 - A(\kappa) - A(\kappa)/\kappa},$$
so the standard error is
$$\text{sd}(\hat\kappa) \approx \frac{1}{\sqrt{n}} \frac{1}{\sqrt{1 - A(\kappa) - A(\kappa)/\kappa}},$$
where $A(\kappa) = \frac{I_1(\kappa)}{I_0(\kappa)},$ and $I_j(\kappa)$ the modified Bessel function of the first kind and order $j$.
From there, note that $A(\hat\kappa) = \bar{R},$ so $s = (2(1 - R))^{-1/2} = g(\hat\kappa) = (2(1 - nA(\hat\kappa)))^{-1/2}.$
You can see that I use a function $g(\hat\kappa)$ to show clearly that $s$ can be calculated directly from $\hat\kappa.$
We can then use the delta method to get
$\text{var}(s) \approx \left(\frac{d g(\hat\kappa)}{d\kappa}\right)^2 \text{var}(\hat\kappa) = (-2n A'(\kappa))^2 \text{var}(\hat\kappa)$
and finally the approximation to the standard error
$\text{sd}(s) \approx (-2n A'(\kappa)) \text{sd}(\hat\kappa) = (-2n A'(\kappa)) \frac{1}{\sqrt{n}} \frac{1}{\sqrt{1 - A(\kappa) - A(\kappa)/\kappa}}$
Whether this approximation is any good remains to be seen, but at least it's a start! A problem is that estimates of $\hat\kappa$ are usually biased, which will likely make this approximation a bit worse. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9716286063194275, "perplexity": 200.5240042399452}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590348521325.84/warc/CC-MAIN-20200606222233-20200607012233-00551.warc.gz"} |
http://clay6.com/qa/20157/integrate-int-limits-0-min-m-dx | # Integrate : $\int \limits_0^{\large\frac{\pi}{2}} min ^m \{ \tan x , \cot x \} dx$
$\begin {array} {1 1} (a)\;\log \sqrt 2 \\ (b)\;\log 2 \sqrt 2 \\ (c)\;\log(\frac{1}{2}) \\ (d)\;\log 2 \end {array}$
To solve this type are throw graph. to $min ^m$, take lower portion. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9792818427085876, "perplexity": 4264.272207350353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267860776.63/warc/CC-MAIN-20180618183714-20180618203714-00134.warc.gz"} |
https://www.askmehelpdesk.com/gardening-plants/red-sunset-maple-tree-root-system-475468.html | I am trying to find out as much information as I can about Red Sunset maple root systems. Is the root system mostly shallow or is it a combination of shallow & deep roots? I have allot of shallow roots within the first 3-4 inches of significant size (thickness) and was hoping to cut them off. If I can cut them I prevent further damage to my lawn. Thanks for any information you can provide.
Bob M. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8451495170593262, "perplexity": 736.9517750433225}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738661953.95/warc/CC-MAIN-20160924173741-00235-ip-10-143-35-109.ec2.internal.warc.gz"} |
http://garden.irmacs.sfu.ca/comment/reply/60002 | Importance: Medium ✭✭
Author(s): Morrison, Natasha Noel, Jonathan A. Scott, Alex
Subject: Combinatorics
Keywords: cycles hypercube minimum saturation saturation
Recomm. for undergrads: no
Posted by: Jon Noel on: September 20th, 2015
Question What is the saturation number of cycles of length in the -dimensional hypercube?
Let and be graphs. Say that a spanning subgraph of is -saturated if contains no copy of but contains a copy of for every edge . Let denote the minimum number of edges in a -saturated graph. Saturation was introduced by Erdős, Hajnal and Moon [EHM] who proved the following:
Theorem (Erdős, Hajnal and Moon) For we have .
Let denote the -dimensional hypercube. Saturation of -cycles in the hypercube has been studied by Choi and Guan [CG] who proved that . This was drastically improved by Johnson and Pinto [JP] to . The saturation number for longer cycles in the hypercube is not known, though. The question above addresses this.
Another open problem is to determine the saturation number of sub-hypercubes in . This was first considered by Johnson and Pinto [JP] who proved that for fixed and . This upper bound was improved to by Morrison, Noel and Scott [MNS]. The best known lower bound on for fixed and large , also due to [MNS], is .
Problem Improve the upper and lower bounds on for fixed and large .
The results of [MNS] show that for fixed . Howver, the precise asymptotic behaviour of this quantity is unknown.
Question (Morrison, Noel and Scott) For fixed , is it true that converges as ?
## Bibliography
[CG] S. Choi and P. Guan, Minimum critical squarefree subgraph of a hypercube, Proceedings of the Thirty-Ninth Southeastern International Conference on Combinatorics, Graph Theory and Computing, vol. 189, 2008, pp. 57–64.
[EHM] P. Erdős, A. Hajnal, and J. W. Moon, A problem in graph theory, Amer. Math. Monthly 71 (1964), 1107–1110.
[JP] J. R. Johnson and T. Pinto, Saturated subgraphs of the hypercube, arXiv:1406.1766v1, preprint, June 2014.
[MNS] N. Morrison, J. A. Noel and A. Scott, Saturation in the Hypercube and Bootstrap Percolation, arXiv:1408.5488v2, June 2015.
* indicates original appearance(s) of problem. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9215623736381531, "perplexity": 2715.056204358273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-04/segments/1547583660529.12/warc/CC-MAIN-20190118193139-20190118215139-00055.warc.gz"} |
https://socratic.org/questions/595ccb6eb72cff185cc0da0e | Chemistry
Topics
# An 8*g mass of methane, and a 2*g mass of dihydrogen gas, were enclosed in a piston. What volume with the gases occupy under standard conditions?
## $\text{1. 24.5 L;}$ $\text{2. 39.0 L;}$ $\text{3. 33.6 L;}$ $\text{4. 49.0 L?}$
Jul 5, 2017
$\text{Option 3.........}$
#### Explanation:
$\text{Dalton's Law of partial pressures.........}$ states that in a gaseous mixture, the partial pressure exerted by a component gas is the same as the pressure it would exert if it ALONE occupied the container. The total pressure is the sum of the individual partial pressures........
And so we works out.....
${n}_{C {H}_{4}} = \text{Mass"/"Molar Mass} = \frac{8 \cdot g}{16 \cdot g \cdot m o {l}^{-} 1} = \frac{1}{2} \cdot m o l$
${n}_{{H}_{2}} = \text{Mass"/"Molar Mass} = \frac{2 \cdot g}{2.0 \cdot g \cdot m o {l}^{-} 1} = 1 \cdot m o l$
And thus we have $\frac{3}{2}$ moles of gas, THAT WE ASSUME to BEHAVE IDEALLY. Since we know (or should know) that 1 mole of Ideal Gas occupies $22.4 \cdot L$ under standard conditions.....(and this should be given as supplementary material on any exam paper).
${V}_{\text{gaseous mixture}} = \frac{3}{2} \cdot m o l \times 22.4 \cdot L \cdot m o {l}^{-} 1 = 33.6 \cdot L$
Oh, and by the way, $1 \cdot L \equiv 1 \cdot {\mathrm{dm}}^{-} 3$. Just to expand a bit on this, I thought it might be useful to explain the equivalence of $1 \cdot L$ and $1 \cdot {\mathrm{dm}}^{3}$. Now chemists commonly use $\text{litres}$ as they are convenient units, and of course they also use $\text{cubic centimetres}$ which are equivalent to $\text{millilitres}$; $1 \cdot c {m}^{3} \equiv 1 \cdot m L$.
Now $1 \cdot {m}^{3}$ is an impossibly large volume, and this is equal to $1000 \cdot L$, which is also equal to $1000 \cdot {\mathrm{dm}}^{3}$. If you ever shift $4 - 5 \cdot {m}^{3}$ of concrete, this is a hot and hard day's work.
Now $1 \cdot {\mathrm{dm}}^{3} = 1 \times {\left({10}^{-} 1 \cdot m\right)}^{3}$ (i.e. the $d$ means $\text{deci}$, i.e. $\times {10}^{-} 1$), and so $1 \cdot {\mathrm{dm}}^{3} = 1 \times {\left({10}^{-} 1 \cdot m\right)}^{3} = 1 \times {10}^{-} 3 \cdot {m}^{3}$ $\equiv 1 \cdot L$ as required.
##### Impact of this question
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https://www.physicsforums.com/threads/is-this-definition-of-reynolds-transport-theorem-correct.905034/ | # Is this definition of Reynolds' Transport Theorem correct?
Tags:
1. Feb 22, 2017
### FluidStu
One thing I find frustrating when trying to get a handle on this theorem is the number of different forms presented in the literature. I understand this to be due to it being very general theorem applicable to many different contexts.
Not that the world needs a new, slightly different looking definition of RTT, but I have tried to look at it from a 'fluid parcel' (a.k.a. material volume) and 'control volume' point of view. Based on general reading from multiple sources, this is a definition I've come up with which makes sense to me because it's very similar to the idea of the "Material Derivative". But, is it correct?
2. Feb 27, 2017
### PF_Help_Bot
Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
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http://mathhelpforum.com/calculus/15151-help-integration-problem-trigonometric-substitution.html | # Thread: Help with a integration problem (trigonometric substitution)
1. ## Help with a integration problem (trigonometric substitution)
I'm having a test on integration on Monday and I'm having some major problems with solving integrations involving trigonometric substitution. There's a sample of a problem I'm having issues with:
Code:
I = Integration symbol
u = dummy variable for substitution
I tan^2(6x) * dx
My work:
sin^2(6x)
I ---------- * dx u = cos(6x)
cos^2(6x) du = -6sin(6x) * dx
du
dx = ------------
-6sin(6x)
sin^2(6x) du
I ---------- * --------
u -6sin(6x)
-1 sin(6x)
--- I -------- * du
6 u
This is where I get stuck. I remember my teacher saying that I have to get everything in terms of u before I go substituting u back into the equation, however, I don't see a feasible way to do this. If someone could step my through this one so I can see how it's done, I'd really appreciate it.
If you all need any more information about the problem or something I'll do my best to provide. Regards.
2. Originally Posted by Gamerdude
I'm having a test on integration on Monday and I'm having some major problems with solving integrations involving trigonometric substitution. There's a sample of a problem I'm having issues with:
Code:
I = Integration symbol
u = dummy variable for substitution
I tan^2(6x) * dx
My work:
sin^2(6x)
I ---------- * dx u = cos(6x)
cos^2(6x) du = -6sin(6x) * dx
du
dx = ------------
-6sin(6x)
sin^2(6x) du
I ---------- * --------
u -6sin(6x)
-1 sin(6x)
--- I -------- * du
6 u
This is where I get stuck. I remember my teacher saying that I have to get everything in terms of u before I go substituting u back into the equation, however, I don't see a feasible way to do this. If someone could step my through this one so I can see how it's done, I'd really appreciate it.
If you all need any more information about the problem or something I'll do my best to provide. Regards.
You need to cancel the common sin(6x) first.
Then
u = cos(6x)
So
u^2 = cos^2(6x) = 1 - sin^2(6x)
sin^2(6x) = 1 - u^2
and
sin(6x) = (+/-)sqrt{1 - u^2}
You'll need to look at the limits of integration to decide where you use the + and - signs.
-Dan
3. Thanks a ton. I'm going to do some more practice problems in a few hours and see how I do. If I run into trouble again I'll post back here. Thanks for such a quick response.
4. Hello, Gamerdude!
tan²(6x) dx
We have: .
[sec²(6x) - 1] dx .= .(1/6)tan(6x) - x + C
5. Thanks a lot, I've been able to solve most of the problems I've been having trouble with over the past few days. My headache is now basically gone . I'm only stuck on two problems (which implement the same concept), here they are:
Code:
∫csc^2(3x) * cot(3x) dx
and
Code:
∫tan(2x) * sec^2(2x) dx
I know that these two problems are essentially the same as far as the answering them goes. I found a integration calculator online which shows me the answer to the problems, but unfortunately it doesn't show the steps. Could someone walk me through one of these? If I can do one I should be able to do the other. Thanks again all.
6. Hello, Gamerdude!
You seem to be making hard work out of these.
. . Just pick an appropriate substitution.
csc²(3x)·cot(3x) dx
Let u = cot(3x) . . . . du = -3·csc²(3x) dx
The integral becomes: .(-1/3)
u du
tan(2x)·sec²(2x) dx
Let u = tan(2x) . . . . du = 2·sec²(2x) dx
The integral becomes: .(1/2)
u du
7. Soroban, I think where I'm getting lost is that once you have
Code:
(1/2) ∫ u du
for: ∫ tan(2x)·sec²(2x) dx
and you plug u and du back into the equation, you end up with almost the same function as when you started. I know this is where I'm not getting it, but I don't know why.
So, in the end, after I plug u and du back in, this is what I always come up with:
Code:
(1/2)∫ tan(2x)·2sec²(2x)
I know this isn't the answer (The answer should be getting is (1/4) * sec²(2x) from what I've gathered). What am I doing wrong?
8. Originally Posted by Gamerdude
Soroban, I think where I'm getting lost is that once you have
Code:
(1/2) ∫ u du
for: ∫ tan(2x)·sec²(2x) dx
and you plug u and du back into the equation, you end up with almost the same function as when you started. I know this is where I'm not getting it, but I don't know why.
So, in the end, after I plug u and du back in, this is what I always come up with:
Code:
(1/2)∫ tan(2x)·2sec²(2x)
I know this isn't the answer (The answer should be getting is (1/4) * sec²(2x) from what I've gathered). What am I doing wrong?
remember, du = 2sec^2(2x) dx, so plugging in u and du back into the formula we would get exactly the one we started with. however, that is not what you should do. you should integrate with respect to u first, and then plug in the value for u. try it
9. So, if I integrate with respect to u, I will get
Code:
(1/4) * tan²(2x) + c
right? Is this my final answer or do I have to do something else?
10. Originally Posted by Gamerdude
So, if I integrate with respect to u, I will get
Code:
(1/4) * tan²(2x) + c
right? Is this my final answer or do I have to do something else?
that's the final answer, if you were doing the definate integral it wouldn't be
11. Awesome. I think I finally get it now. Just to clarify my understanding, the following integration:
Code:
∫csc²(3x) * cot(3x) dx
will come out to equal:
Code:
(-1/6) * cot²(3x)
Correct?
12. Originally Posted by Gamerdude
Awesome. I think I finally get it now. Just to clarify my understanding, the following integration:
Code:
∫csc²(3x) * cot(3x) dx
will come out to equal:
Code:
(-1/6) * cot²(3x)
Correct?
Correct
13. Awesome! Thank you everyone that helped me out. I'm feeling much more confident about my test tomorrow. If I run into any last minute issues I'll be sure to post back. Positive rep points to all!
14. Originally Posted by Gamerdude
Awesome! Thank you everyone that helped me out. I'm feeling much more confident about my test tomorrow. If I run into any last minute issues I'll be sure to post back. Positive rep points to all!
great! keep practicing. and good luck! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8712795376777649, "perplexity": 877.5244654817998}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948513478.11/warc/CC-MAIN-20171211105804-20171211125804-00116.warc.gz"} |
http://www.ck12.org/book/Algebra-I/r2/section/6.7/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 6.7: Linear Inequalities in Two Variables
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Graph linear inequalities in one variable on the coordinate plane.
• Graph linear inequalities in two variables.
• Solve real-world problems using linear inequalities
## Introduction
A linear inequality in two variables takes the form
\begin{align*} y > mx+b\end{align*} or \begin{align*} y < mx+b\end{align*}
Linear inequalities are closely related to graphs of straight lines. A straight line has the equation \begin{align*} y = mx+b\end{align*}. When we graph a line in the coordinate plane, we can see that it divides the plane in two halves.
The solution to a linear inequality includes all the points in one of the plane halves. We can tell which half of the plane the solution is by looking at the inequality sign.
> The solution is the half plane above the line.
\begin{align*}\geq\end{align*} The solution is the half plane above the line and also all the points on the line.
< The solution is the half plane below the line.
\begin{align*}\leq\end{align*} The solution is the half plane below the line and also all the points on the line.
(Above the line means for a given \begin{align*}x-\end{align*}coordinate, all points with \begin{align*}y-\end{align*}values greater than the \begin{align*}y-\end{align*}value are on the line)
For a strict inequality, we draw a dashed line to show that the points on the line are not part of the solution.
For an inequality that includes the equal sign, we draw a solid line to show that the points on the line are part of the solution.
Here is what you should expect linear inequality graphs to look like.
The solution of \begin{align*} y>mx+b\end{align*} is the half plane above the line. The dashed line shows that the points on the line are not part of the solution.
The solution of \begin{align*} y \geq mx+b\end{align*} is the half plane above the line and all the points on the line.
The solution of \begin{align*} y is the half plane below the line.
The solution of \begin{align*} y\leq mx+b\end{align*} is the half plane below the line and all the points on the line.
## Graph Linear Inequalities in One Variable in the Coordinate Plane
In the last few sections, we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type \begin{align*}x=a\end{align*} we get a vertical line and when we graph an equation of the type \begin{align*}y=b\end{align*} we get a horizontal line.
Example 1
Graph the inequality \begin{align*}x>4\end{align*} on the coordinate plane.
Solution
First, let’s remember what the solution to \begin{align*}x>4\end{align*} looks like on the number line.
The solution to this inequality is the set of all real numbers \begin{align*}x\end{align*} that are bigger than four but not including four. The solution is represented by a line.
In two dimensions we are also concerned with values of \begin{align*} y\end{align*}, and the solution to \begin{align*} x>4\end{align*} consists of all coordinate points for which the value of \begin{align*}x\end{align*} is bigger than four. The solution is represented by the half plane to the right of \begin{align*}x=4\end{align*}.
The line \begin{align*}x=4\end{align*} is dashed because the equal sign is not included in the inequality and therefore points on the line are not included in the solution.
Example 2
Graph the inequality \begin{align*} y \leq 6\end{align*} on the coordinate plane.
Solution
The solution is all coordinate points for which the value of \begin{align*}y\end{align*} is less than or equal than 6. This solution is represented by the half plane below the line \begin{align*}y=6\end{align*}.
The line \begin{align*}y=6\end{align*} is solid because the equal sign is included in the inequality sign and the points on the line are included in the solution.
Example 3
Graph the inequality \begin{align*} \mid 6 \mid <5\end{align*}
Solution
The absolute value inequality \begin{align*} \mid 6 \mid <5\end{align*} can be re-written as \begin{align*}-5 < y <5\end{align*}. This is a compound inequality which means
\begin{align*}y>-5\end{align*} and \begin{align*}y<5\end{align*}
In other words, the solution is all the coordinate points for which the value of \begin{align*}y\end{align*} is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines \begin{align*}y=-5\end{align*} and \begin{align*}y=5\end{align*}.
Both horizontal lines are dashed because points on the line are not included in the solution.
Example 4
Graph the inequality \begin{align*}\mid x \mid \geq 2\end{align*}.
Solution
The absolute value inequality \begin{align*} \mid x \mid \geq 2\end{align*} can be re-written as a compound inequality:
\begin{align*} x \leq -2 \end{align*} or \begin{align*} x \geq 2 \end{align*}
In other words, the solution is all the coordinate points for which the value of \begin{align*}x\end{align*} is smaller than or equal to -2 and greater than or equal to 2. The solution is represented by the plane to the left of the vertical line \begin{align*}x= -2\end{align*} and the plane to the right of line \begin{align*}x=2\end{align*}.
Both vertical lines are solid because points on the line are included in the solution.
## Graph Linear Inequalities in Two Variables
The general procedure for graphing inequalities in two variables is as follows.
Step 1: Re-write the inequality in slope-intercept form \begin{align*}y=mx+b\end{align*}. Writing the inequality in this form lets you know the direction of the inequality
Step 2 Graph the line of equation \begin{align*}y=mx+b\end{align*} using your favorite method. (For example, plotting two points, using slope and \begin{align*}y-\end{align*}intercept, using \begin{align*}y-\end{align*}intercept and another point, etc.). Draw a dashed line if the equal sign in not included and a solid line if the equal sign is included.
Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than.
Example 5
Graph the inequality \begin{align*} y \geq 2x-3\end{align*}.
Solution
Step 1
The inequality is already written in slope-intercept form \begin{align*} y \geq 2x-3\end{align*}.
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-1\end{align*} \begin{align*}2(-1) - 3 = -5\end{align*}
0 \begin{align*}2(0) - 3 = -3\end{align*}
1 \begin{align*}2(1) - 3 = -1\end{align*}
Step 2
Graph the equation \begin{align*}y=2x-3\end{align*} by making a table of values.
Step 3
Graph the inequality. We shade the plane above the line because \begin{align*}y\end{align*} is greater than. The value \begin{align*}2x-3\end{align*} defines the line. The line is solid because the equal sign is included.
Example 6
Graph the inequality \begin{align*} 5x - 2y > 4\end{align*}.
Solution
Step 1
Rewrite the inequality in slope-intercept form.
\begin{align*} -2y & > -5x + 4\\ y & > \frac{5} {2}x - 2\end{align*}
Notice that the inequality sign changed direction due to division of negative sign.
Step 2
Graph the equation \begin{align*} y>\frac{5} {2}x - 2\end{align*} by making a table of values.
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-2\end{align*} \begin{align*}\frac{5} {2}(-2) - 2 = -7\end{align*}
0 \begin{align*}\frac{5} {2}(0) - 2 = -2\end{align*}
2 \begin{align*}\frac{5} {2}(2) - 2 = 3\end{align*}
Step 3
Graph the inequality. We shade the plane below the line because the inequality in slope-intercept form is less than. The line is dashed because the equal sign in not included.
Example 7
Graph the inequality \begin{align*} y+4 \leq -\frac{x} {3}+5\end{align*}.
Solution
Step 1
Rewrite the inequality in slope-intercept form \begin{align*} y \leq -\frac{x} {3}+1\end{align*}
Step 2
Graph the equation \begin{align*} y= -\frac{x} {3}+1\end{align*} by making a table of values.
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
\begin{align*}-3\end{align*} \begin{align*}-\frac{(-3)} {3}+1=2\end{align*}
0 \begin{align*}-\frac{0} {3}(0)+1=1\end{align*}
3 \begin{align*}-\frac{3} {3}+1=0\end{align*}
Step 3
Graph the inequality. We shade the plane below the line. The line is solid because the equal sign in included.
## Solve Real-World Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve real-world applications.
Example 8
A pound of coffee blend is made by mixing two types of coffee beans. One type costs $9 per pound and another type costs$7 per pound. Find all the possible mixtures of weights of the two different coffee beans for which the blend costs 8.50 per pound or less. Solution Let’s apply our problem solving plan to solve this problem. Step 1: Let \begin{align*}x =\end{align*} weight of9 per pound coffee beans in pounds
Let \begin{align*}y =\end{align*} weight of 7 per pound coffee beans in pounds Step 2 The cost of a pound of coffee blend is given by \begin{align*}9x + 7y\end{align*}. We are looking for the mixtures that cost8.50 or less.
We write the inequality \begin{align*}9x + 7y \leq 8.50\end{align*}.
Step 3
To find the solution set, graph the inequality \begin{align*}9x + 7y \leq 8.50\end{align*}.
Rewrite in slope-intercept \begin{align*}y \leq -1.29x + 1.21\end{align*}.
Graph \begin{align*}y=-1.29x+1.21\end{align*} by making a table of values.
\begin{align*}x\end{align*} \begin{align*}y\end{align*}
0 1.21
1 -0.08
2 -1.37
Step 4
Graph the inequality. The line will be solid. We shade below the line.
Notice that we show only the first quadrant of the coordinate plane because the weight values should be positive.
The blue-shaded region tells you all the possibilities of the two bean mixtures that will give a total less than or equal to $8.50. Example 9 Julian has a job as an appliance salesman. He earns a commission of$60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and refrigerators must Julian sell in order to make$1000 or more in commission?
Solution Let’s apply our problem solving plan to solve this problem.
Step 1
Let \begin{align*}x =\end{align*} number of washing machines Julian sells
Let \begin{align*}y =\end{align*} number of refrigerators Julian sells
Step 2
The total commission is given by the expression \begin{align*}60x+130y\end{align*}.
We are looking for total commission of 1000 or more. We write the inequality. \begin{align*}60x + 130y \geq 1000\end{align*}. Step 3 To find the solution set, graph the inequality \begin{align*}60x + 130y \geq 1000\end{align*}. Rewrite it in slope-intercept \begin{align*}y \geq -.46x + 7.7\end{align*}. Graph \begin{align*}y = -.46x + 7.7\end{align*} by making a table of values. \begin{align*}x\end{align*} \begin{align*}y\end{align*} 0 7.7 2 6.78 4 5.86 Step 4 Graph the inequality. The line will be solid. We shade above the line. Notice that we show only the first quadrant of the coordinate plane because dollar amounts should be positive. Also, only the points with integer coordinates are possible solutions. ## Lesson Summary • The general procedure for graphing inequalities in two variables is as follows: Step 1 Rewrite the inequality in slope-intercept form \begin{align*}y=mx+b\end{align*}. Step 2 Graph the line of equation \begin{align*}y=mx+b\end{align*} by building a table of values. Draw a dashed line if the equal sign in not included and a solid line if the it is included. Step 3 Shade the half plane above the line if the inequality is greater than. Shade the half plane under the line if the inequality is less than. ## Review Questions Graph the following inequalities on the coordinate plane. 1. \begin{align*}x<20\end{align*} 2. \begin{align*}y \geq -5\end{align*} 3. \begin{align*} \mid x \mid >10 \end{align*} 4. \begin{align*} \mid y \mid \leq 7 \end{align*} 5. \begin{align*} y \leq 4x+3\end{align*} 6. \begin{align*}y > -\frac{x}{2}-6\end{align*} 7. \begin{align*}3x-4y \geq 12\end{align*} 8. \begin{align*} x+7y <5 \end{align*} 9. \begin{align*}6x + 5y>1\end{align*} 10. \begin{align*}y + 5 \leq -4x + 10\end{align*} 11. \begin{align*}x-\frac{1}{2}y \geq 5\end{align*} 12. \begin{align*}30x + 5y< 100\end{align*} 13. An ounce of gold costs670 and an ounce of silver costs $13. Find all possible weights of silver and gold that makes an alloy that costs less than$600 per ounce.
14. A phone company charges 50 cents cents per minute during the daytime and 10 cents per minute at night. How many daytime minutes and night time minutes would you have to use to pay more than \$20 over a 24 hour period?
## Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9616.
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Feb 22, 2012 | {"extraction_info": {"found_math": true, "script_math_tex": 106, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8690369129180908, "perplexity": 1102.2846006467867}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461862047707.47/warc/CC-MAIN-20160428164727-00152-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria | Skip to main content
# Chemical Equilibria
• Balanced Equations and Equilibrium Constants
In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. The coefficients relate to the equilibrium constants because they are used to calculate them.
• Calculating an Equilibrium Concentration
To calculate an equilibrium concentration from an equilibrium constant, an understanding of the concept of equilibrium and how to write an equilibrium constant is required. Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.
• Calculating An Equilibrium Concentrations
• Calculating an Equilibrium Constant, Kp, with Partial Pressures
• Determining the Equilibrium Constant
• Difference Between K And Q
Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, Qc , and the equilibrium constant, Kc , is essential in solving for the net change. With this relationship, the direction in which a reaction will shift to achieve chemical equilibrium, whether to the left or the right, can be easily calculated.
• Dissociation Constant
Page notifications Off Share Table of contents The dissociation constant specifies the tendency of a substance AxBy to reversibly dissociate (separate) in a solution into smaller components A and B.
• Effect of Pressure on Gas-Phase Equilibria
Le Chatelier's Principle states that a system at equilibrium will adjust to relieve stress when there are changes in the concentration of a reactant or product, the partial pressures of components, the volume of the system, and the temperature of reaction.
• Equilibrium Calculations
• Kc
This page defines the equilibrium constant and introduces the equilibrium constant expressed in terms of concentrations, Kc. It assumes familiarity with the concept of dynamic equilibrium, as well as the terms "homogeneous" and "heterogeneous" as applied to chemical reactions. The two types of dynamic equilibria (homogeneous and heterogeneous) are discussed separately below, because the equilibrium constants are defined differently.
• Kp
This page explains equilibrium constants expressed in terms of partial pressures of gases, Kp. It covers an explanation of the terms mole fraction and partial pressure, and looks at Kp for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about Kc - an equilibrium constant expressed in terms of concentrations
• Law of Mass Action
• Mass Action Law
• The Equilibrium Constant
The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant.
• The Reaction Quotient | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9453492760658264, "perplexity": 871.1841272761178}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00311.warc.gz"} |
https://ai.stackexchange.com/questions/22678/reinforcement-learning-with-action-consisting-of-two-discrete-values/22680#22680 | # Reinforcement learning with action consisting of two discrete values
I'm new to reinforcement learning. I have a problem where an action is composed of an order (rod with a required length) and an item from a warehouse (an existing rod with a certain length, which will be cut to the desired length and the remainder put back to the warehouse).
I imagine my state as two lists of a defined size: orders and warehouse, and my action as an index from the first list and an index from the second list. However, I have only worked with environments where it was only possible to pick single action and I'm not sure how to deal with two indexes. I'm not sure how DQN architecture should look like to give me such action.
Can anyone validate my general idea and help me find a solution? Or maybe just point me to some papers where similar problems are described?
You would still be picking a single action. Your action space is now $$\mathcal{A} = \mathcal{O} \times \mathcal{I}$$ where I've chosen $$\mathcal{O}$$ to be the set of possible orders from your problem and $$\mathcal{I}$$ to be the set of possible items.
Provided both of these sets are finite, then you should still be able to approach this problem with DQN. Theoretically, this should be easy to see, as any element from $$\mathcal{A}$$ is still a single element it just happens that this element is now a tuple.
From a programming point of view, let's consider the simple example of cartpole, where the possible actions are left and right. Your $$Q$$-function obviously won't know the meanings of 'left' and 'right', you just assign it to an element of a vector, i.e. your $$Q$$-function would output a vector in $$\mathbb{R}^2$$ with e.g. the first element corresponding to the score for 'left' and the second element corresponding to the score for 'right'. This is still the case in your problem formulation, you will just have a $$Q$$-function that outputs a vector in $$\mathbb{R}^d$$ where $$d = |\mathcal{A}|$$ - you would just have to make sure you know which element corresponds to which action.
Also, there is the possibility that this approach could leave you with a large dimensional vector output, which I imagine would probably mean you'd need more simulations to properly explore the action space.
Hope this helps.
• Thank you for the answer!
– GUZ
Jul 26 '20 at 21:30 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8104149699211121, "perplexity": 263.8455942379055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588341.58/warc/CC-MAIN-20211028131628-20211028161628-00363.warc.gz"} |
https://planetmath.org/independenceofcharacteristicpolynomialonprimitiveelement | # independence of characteristic polynomial on primitive element
The simple field extension $\mathbb{Q}(\vartheta)/\mathbb{Q}$ where $\vartheta$ is an algebraic number of degree (http://planetmath.org/DegreeOfAnAlgebraicNumber) $n$ may be determined also by using another primitive element $\eta$. Then we have
$\eta\in\mathbb{Q}(\vartheta),$
whence, by the entry degree of algebraic number, the degree of $\eta$ divides the degree of $\vartheta$. But also
$\vartheta\in\mathbb{Q}(\eta),$
whence the degree of $\vartheta$ divides the degree of $\eta$. Therefore any possible primitive element of the field extension has the same degree $n$. This number is the degree of the number field (http://planetmath.org/NumberField), i.e. the degree of the field extension, as comes clear from the entry canonical form of element of number field.
Although the characteristic polynomial
$g(x)\;:=\;\prod_{i=1}^{n}[x-r(\vartheta_{i})]\;=\;\prod_{i=1}^{n}(x-\alpha^{(i% )})$
of an element $\alpha$ of the algebraic number field $\mathbb{Q}(\vartheta)$ is based on the primitive element $\vartheta$, the equation
$\displaystyle g(x)\;=\;(x-\alpha_{1})^{m}(x-\alpha_{2})^{m}\cdots(x-\alpha_{k}% )^{m}$ (1)
in the entry http://planetmath.org/node/12050degree of algebraic number shows that the polynomial is fully determined by the algebraic conjugates of $\alpha$ itself and the number $m$ which equals the degree $n$ divided by the degree $k$ of $\alpha$.
The above stated makes meaningful to define the norm and the trace functions in an algebraic number field as follows.
Definition. If $\alpha$ is an element of the number field $\mathbb{Q}(\vartheta)$, then the norm $\mbox{N}(\alpha)$ and the trace $\mbox{S}(\alpha)$ of $\alpha$ are the product and the sum, respectively, of all http://planetmath.org/node/12046$\mathbb{Q}(\vartheta)$-conjugates $\alpha^{(i)}$ of $\alpha$.
Since the coefficients of the characteristic equation of $\alpha$ are rational, one has
$\mbox{N}\!:\,\mathbb{Q}(\vartheta)\to\mathbb{Q}\quad\mbox{and}\quad\mbox{S}\!:% \,\mathbb{Q}(\vartheta)\to\mathbb{Q}.$
In fact, one can infer from (1) that
$\displaystyle\mbox{N}(\alpha)\;=\;a_{k}^{m},\qquad\mbox{S}(\alpha)\;=\;-ma_{1},$ (2)
where $x^{k}\!+\!a_{1}x^{k-1}\!+\ldots+\!a_{k}$ is the minimal polynomial of $\alpha$.
Title independence of characteristic polynomial on primitive element Canonical name IndependenceOfCharacteristicPolynomialOnPrimitiveElement Date of creation 2014-02-04 8:07:18 Last modified on 2014-02-04 8:07:18 Owner pahio (2872) Last modified by pahio (2872) Numerical id 10 Author pahio (2872) Entry type Topic Classification msc 11R04 Classification msc 12F05 Classification msc 11C08 Classification msc 12E05 Synonym norm and trace functions in number field Related topic Norm Related topic NormAndTraceOfAlgebraicNumber Related topic PropertiesOfMathbbQvarthetaConjugates Related topic DiscriminantInAlgebraicNumberField Defines norm in number field Defines trace in number field Defines norm Defines trace | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 34, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.881881833076477, "perplexity": 541.9073308131128}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703565541.79/warc/CC-MAIN-20210125092143-20210125122143-00095.warc.gz"} |
https://www.physicsforums.com/threads/the-derivative-of-a-derivative.297469/ | # The derivative of a derivative
1. Mar 5, 2009
### Emethyst
1. The problem statement, all variables and given/known data
Using implicit differentation, find d^2y/dx^2 (the second derivative of y with respect to x) of the following in terms of x and y: (a) xy=4 (b)4y^2-3x^2=1
2. Relevant equations
All the simplifying laws for derivatives
3. The attempt at a solution
I found the derivatives for both the starting equations ((a) is -y/x and (b) is 3x/4y), but I cannot seem to find the derivatives of these derivatives. I know that the answer for (a) is 2y/x^2 and the answer for (b) is 3/16y^3, but I don't know where these answers come from. The way I have tried to solve for the second derivative always ends up with a dy/dx somewhere in the solution:
ex. for (a) I wound up with (dy/dx)(-x)+y/x^2 and for (b) 12y-(12x)(dy/dx)/16y^2
If someone could be of assistance for the second part of these questions and show me where i'm going wrong it would be greatly appreciated. Thanks in advance.
2. Mar 5, 2009
### Dick
Take the first one. Start with y'=(-y/x). Differentiate both sides using the quotient rule on the right. You'll get some y' terms on the right side, but you know y'=(-y/x).
3. Mar 5, 2009
### Emethyst
Thanks a lot for the help, can't believe I missed that :tongue: | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9502493143081665, "perplexity": 531.352555381148}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202635.43/warc/CC-MAIN-20190322054710-20190322080710-00489.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/186025-proving-variable-has-poission-distribution.html | # Math Help - Proving a variable has a poission distribution
1. ## Proving a variable has a poission distribution
Hi
I am given that:
P(Y=k)= $\sum$ $\binom{n}{k}$p^k.(1-p)^(n-k).e^-x.(x^n/n!)
where the sum is from n=k to infinity
and I need to show that Y is poission distributued with parameter $\lambda$p
I'm not really sure where to start, should I take out all the variables which are not raised to the power of n out of the sum and change $\binom{n}{k}$ to n!/(n-k)!k!?
Any tips to get me started would be much appreciated
2. ## Re: Proving a variable has a poission distribution
Hello,
Yes you can take out the variables where n is not in the expression.
You can also, in a first time, change the summation from n=0 to infinity (since this is the range of a Poisson distribution). It should simplify quite a lot, since in the $n \choose k$, the n! simplifies, the (n-k)! becomes 0!=1 and k! can be pulled out from the sum.
3. ## Re: Proving a variable has a poission distribution
The question is:
P(Y=k)= $\sum$ $\binom{n}{k}$ $p^k$(1-p)^(n-k) $e^{-x}$ $\frac{x^n}{n!}$
where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)
is this right?
P(Y=k)= $\sum$ x^n/(n-k)!k! $p^k$ (1-p)^(n-k) $e^{-x}$
P(Y=k)= $p^k$ $e^{-x}$ $\sum$ x^n/(n-k)!k! . (1-p)^(n-k)
then change the sum so that it runs from n=0 to infinity
P(Y=k)=(p^k)/k! $e^{-x}$ $\sum$
then I'm not sure what I'll have left inside? x^n...
any hints appreciated
4. ## Re: Proving a variable has a poission distribution
After the change of summation, we have $P(Y=k) = \frac{p^k}{k!}e^{-x}\sum_{j=0}^{+\infty}\frac{x^{j+k}}{j!}(1-p)^j = \frac{p^k}{k!}e^{-x}x^k\sum_{j=0}^{+\infty}\frac{((1-p)x)^j}{j!}$.
We recognize the series of $\exp$, and now you an conclude.
5. ## Re: Proving a variable has a poission distribution
Originally Posted by yellowcarrotz
The question is:
P(Y=k)= $\sum$ $\binom{n}{k}$ $p^k$(1-p)^(n-k) $e^{-x}$ $\frac{x^n}{n!}$
where the sum runs from n=k to infinity (sorry i'm rubbish at using latex)
is this right?
P(Y=k)= $\sum$ x^n/(n-k)!k! $p^k$ (1-p)^(n-k) $e^{-x}$
P(Y=k)= $p^k$ $e^{-x}$ $\sum$ x^n/(n-k)!k! . (1-p)^(n-k)
then change the sum so that it runs from n=0 to infinity
P(Y=k)=(p^k)/k! $e^{-x}$ $\sum$
then I'm not sure what I'll have left inside? x^n...
any hints appreciated
Replace n-k by j, that's a way to change the summation index
And if there's just an n, n=j+k
6. ## Re: Proving a variable has a poission distribution
yep i've done it, thank you so much for all your help | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 33, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.924305260181427, "perplexity": 1763.7160248437433}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011056097/warc/CC-MAIN-20140305091736-00027-ip-10-183-142-35.ec2.internal.warc.gz"} |
http://hal.in2p3.fr/in2p3-01148934 | # Shape coexistence at N=20 and N=28 : study of 0+2 states in $^{34}$Si and $^{44}$S
Abstract : It is well known that the nuclear shell structure changes for the most exotic nuclei. One of the consequences of this phenomenon is the modification of the "classical" magic numbers, as experimentally observed at N = 20 and N = 28. Nevertheless, the mechanisms responsible for such changes are still under discussion and more experimental information is needed to better constrain the theoretical models. In these proceedings, we report on the discovery and the experimental study by precise spectroscopy experiments of the 0+2 state in 34Si and 44S. The 34Si is located between the magic spherical 36S and the deformed 32Mg, member of the so-called island of inversion, whereas 44S is located between the magic spherical 48Ca and the deformed 42Si. Therefore, the structure of these nuclei, and in particular the phenomenon of shape coexistence, is of crucial importance to understand how the intruder configurations progressively dominate the ground state structure of the most exotic nuclei at both N = 20 and N = 28.
Type de document :
Communication dans un congrès
Sivaramakrishnan Muralithar. Frontiers in Gamma Ray Spectroscopy 2012 (FIG-12), Mar 2012, New Delhi, India. AIP Conference Proceedings, 1609, pp.167-172, 2014, 〈10.1063/1.4893271〉
http://hal.in2p3.fr/in2p3-01148934
Contributeur : Michel Lion <>
Soumis le : mardi 5 mai 2015 - 16:52:05
Dernière modification le : jeudi 26 juillet 2018 - 12:08:20
### Citation
S. Grévy, F. Rotaru, F. Negoita, C. Borcea, R. Borcea, et al.. Shape coexistence at N=20 and N=28 : study of 0+2 states in $^{34}$Si and $^{44}$S . Sivaramakrishnan Muralithar. Frontiers in Gamma Ray Spectroscopy 2012 (FIG-12), Mar 2012, New Delhi, India. AIP Conference Proceedings, 1609, pp.167-172, 2014, 〈10.1063/1.4893271〉. 〈in2p3-01148934〉
### Métriques
Consultations de la notice | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8535122871398926, "perplexity": 3168.161120552848}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215843.55/warc/CC-MAIN-20180820042723-20180820062723-00060.warc.gz"} |
https://tex.stackexchange.com/questions/3323/vertical-space-before-and-after-align-environment?noredirect=1 | # Vertical space before and after align environment
I am wondering if it is possible in LaTeX to generally adjust the spacing before and after math in the align environment from amsmath.
From the Mathmode-guide (by Herbert Voss) I gather that \belowdisplayskip and friends should do the job, but manipulating them doesn't yield any visible results.
It has a pretty clear effect to me:
\hrule
\begin{align}
f(x)&=x
\end{align}
\hrule
\vskip1cm
\belowdisplayskip=0pt
\hrule
\begin{align}
f(x)&=x
\end{align}
\hrule
Can you provide a minimal working example where it doesn't?
• It seems that the difference between yours and mine was that I put \belowdisplayskip=0pt in the preamble rather than after \begin{document}. Indeed, putting it after the preamble yields results. This had never occured to me :-/ .. but thanks for the answer :-) – plc Sep 20 '10 at 19:58
• @plc -- in many document classes, \belowdisplayskip is reset every time the text size (e.g., \small, etc.) is reset. although the reset is local, \normalsize is invoked as part of \begin{document}, overruling the setting you put in the preamble. – barbara beeton Oct 16 '12 at 13:10 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9708673357963562, "perplexity": 2271.5683058773475}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739182.35/warc/CC-MAIN-20200814070558-20200814100558-00248.warc.gz"} |
http://math.stackexchange.com/questions/313020/parameters-of-an-elliptic-equation | # Parameters of an elliptic equation?
I know that an ellipse equation described by:
$\frac {x^2} {a^2} + \frac {y^2} {b^2} =1$
My question is in the equation above how many parameters we need to estimate? Two or four? The unknows parameters are only a, b or are also x, y???
-
I think there is no elliptic function here. Please consider retagging it. Thanks. – awllower Feb 24 '13 at 16:31
For every fixed ellipse, $a$ and $b$ are determined, and $x$ and $y$ are variables. The ellipse itself is the collection of points in the plane whose $x$ and $y$-coordinates statisfy the equation. For instance, take the ellipse $$\frac{x^2}{1} + \frac{y^2}{4} = 1$$ then $a=1$ and $b=2$. For any point $(x_0, y_0)$, if it is the case that $\frac{x_0^2}{1} + \frac{y_0^2}{4} = 1$, then that point is on the ellipse. For any point on the ellipse, if you take the coordinates and put them into the equation, the left-hand side and right-hand side will turn out equal. For any other point in the plane (like the origin, or say $(535, -1234)$), the equation is not statisfied, and therefore they are not on the ellipse.
Any other coice of $a$ and $b$ will result in a different ellipse. So, $a$ and $b$ are numbers that are determined beforehand, like $1$ and $2$ above, or any other pair of numbers, really. The reason they are called $a$ and $b$, and not written out specifically is in order to cover any concievable choice of numbers. This is something done often in mathematics (and quite successfully at that, if you can get your head around it).
The weakness of using letters instead of numbers like this is that you lose the immediate and concrete grasp on the numbers involved if you're not used to it. The strengths, however, is that you can talk about them all at once, instead of taking care of every single choice one by one. Imagine if you want to say something about all those ellipses, and having to go through the same calculations for every possible choice of denominators. Or just call them $a$ and $b$, whatever they are, and do the calculations once. It's pretty clear which approach will be more fruitfull in the long run. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8611217737197876, "perplexity": 205.64912066246328}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507443901.32/warc/CC-MAIN-20141017005723-00042-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://chemistry.stackexchange.com/questions/14700/adsorption-on-surface-of-adsorbent | Why is energy released during the process of adsorption? How is this energy change observed and how is it calculated? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9872546792030334, "perplexity": 208.59509152514445}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00144.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2014_AIME_II_Problems/Problem_7&oldid=88748 | # 2014 AIME II Problems/Problem 7
## Problem
Let . Find the sum of all positive integers for which
## Solution 1
First, let's split it into two cases to get rid of the absolute value sign
Now we simplify using product-sum logarithmic identites:
Note that the exponent is either if is odd or if is even.
Writing out the first terms we have
This product clearly telescopes (i.e. most terms cancel) and equals either or . But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where is odd and another where is even.
For odd , it telescopes to where is clearly .
For even , it telescopes to where is the only possible value. Thus the answer is
## Solution 2
Note that is when is odd and when is even. Also note that for all . Therefore Because of this, is a telescoping series of logs, and we have Setting each of the above quantities to and and solving for , we get possible values of and so our desired answer is | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9870743751525879, "perplexity": 765.1207681728849}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00642.warc.gz"} |
http://mathhelpforum.com/pre-calculus/216874-vector-equation-question.html | # Math Help - Vector Equation Question
1. ## Vector Equation Question
Question:
Find the vector equation of the line r = (0,2,-1)+t(1,0,1), t belongs to real numbers, reflected in the plane 2x-y+z-9 = 0.
I am not sure on how to start solving this question and how do you reflect a line in a plane? What method would you use?
2. ## Re: Vector Equation Question
Hey SeerAlpha.
What you are essentially doing is finding the vector that goes from the plane to the point which is normal to the plane, reversing it and adding two times the reverse vector to the point.
To find the perpendicular distance between a point and a plane you calculate n . p where n is the plane normal (nx,ny,nz) [A unit normal vector] and p is the point in question.
You then take your point and calculate p = p - n*2*d where n is the plane normal and d is the distance you calculated above.
If you do this for a particular value of t and leave the final reflected point arbitrary you will get your solution.
If you need a more intuitive explanation is going on I can help in that way as well.
3. ## Re: Vector Equation Question
Perfect I solved it Thanks for your help! | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8788697123527527, "perplexity": 338.9074395011872}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535924131.19/warc/CC-MAIN-20140901014524-00177-ip-10-180-136-8.ec2.internal.warc.gz"} |
https://ai.stackexchange.com/questions/21542/how-to-prevent-deep-q-learning-algorithms-to-overfit | # How to prevent deep Q-learning algorithms to overfit?
I have recently solved the Cartpole problem using double deep Q-learning. When I saw how the agent was doing, it used to go right every time, never left, and it did similar actions all the time.
Did the model overfit the environment? It seems that the agent just memorized the environment.
What are the common techniques to prevent the agent to overfit like that? Is that a common problem?
• The concept of overfitting in reinforcement learning is not well-defined (see e.g. ai.stackexchange.com/q/20127/2444). In supervised learning, overfitting generally means that you learn the training data, but don't perform well on the test data. However, if you train your RL agent in the same exact environment where you want it to behave, you actually want it to "overfit", i.e. learn the environment dynamics. So, why don't you want the RL to learn the environment? Can you elaborate a little more? – nbro Jun 2 at 15:18 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9169468879699707, "perplexity": 837.5519995437948}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655886095.7/warc/CC-MAIN-20200704073244-20200704103244-00540.warc.gz"} |
http://math.eretrandre.org/tetrationforum/showthread.php?tid=876&pid=7083 | • 0 Vote(s) - 0 Average
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Negative, Fractional, and Complex Hyperoperations KingDevyn Junior Fellow Posts: 2 Threads: 2 Joined: May 2014 05/30/2014, 06:58 AM Is there a way to continue the patterns we see within the natural numbers of current hyper-operations (Hyper-1, Hyper-2, Hyper-3, Hyper-4, ect...) or at least prove that we cannot extend the value of the operation to fractional numbers? E.g. Hyper-1/2. Negative numbers? E.g. Hyper-(-2) Or even imaginary numbers? E.g. Hyper-3i. They need not be defined, but are these operations technically there, just without practical use? Or are our names for the hyper-operations strictly for listing and naming purposes, with no way to derive meaning from such a number? Could a fractional, or negative hyper-operation represent an operator we have already defined? E.g. Hyper-(-2)= Division, or Hyper-1/2 = Division? Comments on the controversy of Zeration are also encouraged. Thanks! MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 05/30/2014, 07:57 AM (This post was last modified: 05/30/2014, 07:58 AM by MphLee.) $s$-rank hyperoperations have meaning as long as we can iterate $s$ times a function $\Sigma$ defined in the set of the binary functions over the naturals numbers (or defined over a set of binary functions.) let me explain why. There are many differente Hyperoperations sequences, end they are all defined in a different way: we start with an operation $*$ and we obtain its successor operation $*'$ applying a procedure $\Sigma$ (usually a recursive one). $\Sigma(*)=*'$ So every Hyperoperation sequence is obtained applying that recursive procedure $\Sigma$ to a base operation $*$ (aka the first step of the sequence) $*_0:=*$ $*_1:=\Sigma(*_0)$ $*_2:=\Sigma(\Sigma(*_0))$ and so on or in a formal way $*_0:=*$ $*_{n+1}:=\Sigma(*_n)$ That is the same as $*_{n}:=\Sigma^{\circ n}(*_{0})$ so if we can extend the iteration of $\Sigma^{\circ n}$ from $n \in \mathbb{N}$ to the real-complex numbers the work is done. ---------------------- MathStackExchange account:MphLee MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 05/30/2014, 08:19 AM (This post was last modified: 05/30/2014, 08:22 AM by MphLee.) I'm not sure but I think that bo198214(Henrik Trappmann) had this idea in 2008 http://math.eretrandre.org/tetrationforu...l+function With his idea we can reduce the problem of real-rank hyperoperations to an iteration problem Later this idea was better developed by JmsNxn (2011) with the concept of "meta-superfunctions" http://math.eretrandre.org/tetrationforu...hp?tid=708 I'm still working on his point of view but there is a lot of work to do... MathStackExchange account:MphLee « Next Oldest | Next Newest »
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https://rd.springer.com/article/10.1007/s10587-004-6424-6 | Czechoslovak Mathematical Journal
, Volume 54, Issue 3, pp 767–771
# Some Characterization of Locally Nonconical Convex Sets
• Witold Seredyński
Article
## Abstract
A closed convex set Q in a local convex topological Hausdorff spaces X is called locally nonconical (LNC) if for every x, y ∈ Q there exists an open neighbourhood U of x such that $${\text{(}}U \cap Q{\text{)}} + \frac{{\text{1}}}{{\text{2}}}{\text{(}}y - x{\text{)}} \subset Q$$. A set Q is local cylindric (LC) if for x, y ⊂ Q, x ≠ y, z ⊂ (x, y) there exists an open neighbourhood U of z such that U ∩ Q (equivalently: bd(Q) ∩ U) is a union of open segments parallel to [x, y]. In this paper we prove that these two notions are equivalent. The properties LNC and LC were investigated in [3], where the implication LNC ⋒ LC was proved in general, while the inverse implication was proved in case of Hilbert spaces.
stable convex set
## References
1. [1]
J. Cel: Tietze-type theorem for locally nonconical convex sets. Bull. Soc. Roy. Sci Liège 69 (2000), 13–15.Google Scholar
2. [2]
S. Papadopoulou: On the geometry of stable compact convex sets. Math. Ann. 229 (1977), 193–200.Google Scholar
3. [3]
G. C. Shell: On the geometry of locally nonconical convex sets. Geom. Dedicata 75 (1999), 187–198.Google Scholar | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9828439950942993, "perplexity": 2136.547422854314}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00068.warc.gz"} |
https://www.explodingdots.org/station/I8S8D | 06 285 000
# Station 8DConverting Fractions into Decimals
A fraction is a number that is an answer to a division problem. For example, the fraction $\dfrac {1} {8}$ is the result of dividing $1$ by $8$. And we can compute $1 \div 8$ in a $1 \leftarrow 10$ machine by making use of decimals. The method is exactly the same as for division without decimals.
### Question 1
For $1 \div 8$ we seek groups of eight in the following picture.
None are to be found right away, so let’s unexplode.
We have one group of $8$, leaving two behind.
Two more unexplosions.
This gives two more groups of $8$ leaving four behind.
Unexploding again
reveals five more groups of $8$ leaving no remainders.
We see that, as a decimal, $\dfrac {1}{8}$ turns out to be $0.125$.
And as a check we have $0.125 = \dfrac {125}{1000} = \dfrac {25}{200} = \dfrac {5}{40} = \dfrac {1}{8}$.
Perform the division in a $1 \leftarrow 10$ machine to show that $\dfrac {1}{4}$, as a decimal, is $0.25$.
### Question 2
Perform the division in a $1 \leftarrow 10$ machine to show that $\dfrac {1}{2}$, as a decimal, is $0.5$.
### Question 3
Perform the division in a $1 \leftarrow 10$ machine to show that $\dfrac {3}{5}$, as a decimal, is $0.6$.
### Question 4
Perform the division in a $1 \leftarrow 10$ machine to show that $\dfrac {3}{16}$, as a decimal, is $0.1875$.
### Question 5
In simplest terms, what fraction is represented by each of these decimals?
$0.75$
$0.625$
$0.16$
$0.85$
$0.0625$
### Question 6
Not all fractions lead to simple decimal representations. For example, consider the fraction $\dfrac {1}{3}$. To compute it, we seek groups of three in the following picture.
Let’s unexplode.
We see three groups of $3$ leaving one behind.
Unexploding gives another ten dots to examine.
We find another three groups of ${3}$ leaving one behind.
And so on. We are caught in an infinitely repeating cycle.
This puts us in a philosophically interesting position. As human beings we cannot conduct this, or any, activity for an infinite amount of time. But it seems very tempting to write $\dfrac{1}{3}=0.33333...$ with the ellipsis representing the instruction “keep going with this pattern forever.” In our minds we can almost imagine what this means. But as a practical human being it is beyond our abilities: one cannot actually write down those infinitely many $3$s represented by the ellipses.
Nonetheless, many people choose not to contemplate what an infinite statement like this means and just carry on to say that some decimals are infinitely long and not be worried by it. In which case, the fraction $\dfrac{1}{3}$ is one of those fractions whose decimal expansion goes on forever.
Notation: Many people make use of a vinculum (a horizontal bar) to represent infinitely long repeating decimals. For example, $0.\overline {3}$ means “repeat the $3$ forever”
$0.\overline {3}=0.33333...$
and $0.38\overline {142}$ means “repeat the group $142$ forever” after the beginning “$38$” hiccup:
$0.38\overline {142}=0.38142142142142...$.
As another (complicated) example, here is the work that converts the fraction $\dfrac {6}{7}$ to an infinitely long repeating decimal. Make sure to understand the steps one line to the next.
Do you see, with this $6$ in the final right-most box that we have returned to the very beginning of the problem? This means that we shall simply repeat the work we have done and obtain the same sequence $857142$ of answers, and then again, and then again. We have
$\dfrac{6}{7}=0.857142857142857142857142...$.
Compute $\dfrac{4}{7}$ as an infinitely long repeating decimal.
### Question 7
Compute $\dfrac{1}{11}$ as an infinitely long repeating decimal.
### Question 8
Which of the following fractions give infinitely long decimal expansions?
$\dfrac{1}{2}, \dfrac{1}{3}, \dfrac{1}{4}, \dfrac{1}{5}, \dfrac{1}{6}, \dfrac{1}{7}, \dfrac{1}{8}, \dfrac{1}{9}, \dfrac{1}{10}$
### Question 9
Use a $1 \leftarrow 10$ machine to compute $133 \div 6$, writing the answer as a decimal.
### Question 10
Use a $1 \leftarrow 10$ machine to compute $255 \div 11$, writing the answer as a decimal.
You can either play with some of the optional stations below or go to the next island!
### Let's Go Wild!
Register NOW and unlock all islands! | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 53, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9313495755195618, "perplexity": 739.5122193156659}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500076.87/warc/CC-MAIN-20230203221113-20230204011113-00269.warc.gz"} |
http://swmath.org/software/13892 | # Algorithm 689
Algorithm 689: Discretized collocation and iterated collocation for nonlinear Volterra integral equations of the second kind. This paper describes a FORTRAN code for calculating approximate solutions to systems of nonlinear Volterra integral equations of the second kind. The algorithm is based on polynomial spline collocation, with the possibility of combination with the corresponding iterated collocation. It exploits certain local superconvergence properties for the error estimation and the stepsize strategy.
This software is also peer reviewed by journal TOMS.
## References in zbMATH (referenced in 8 articles , 1 standard article )
Showing results 1 to 8 of 8.
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https://www.physicsforums.com/threads/finding-the-point-of-inflection-for-the-function.295381/ | # Finding the point of inflection for the function
#### meeklobraca
189
0
1. The problem statement, all variables and given/known data
Find any points of inflection for the function y = x - cos x on the interval {0,2pi}
2. Relevant equations
3. The attempt at a solution
Well i know the second derivative is cos x, and I know it equals 0. But im just not sure what to give as a final answer. cos x = 0 at pi/2 and 3pi/2 but is that the final answer?
#### Mark44
Mentor
31,932
3,895
1. The problem statement, all variables and given/known data
Find any points of inflection for the function y = x - cos x on the interval {0,2pi}
2. Relevant equations
3. The attempt at a solution
Well i know the second derivative is cos x, and I know it equals 0. But im just not sure what to give as a final answer. cos x = 0 at pi/2 and 3pi/2 but is that the final answer?
cos(x) is not identically zero. To confirm that there are inflection points for x = pi/2 and x = 3pi/2, check to see whether the concavity changes around these values. IOW, does y'' change sign at pi/2 and 3pi/2? If so, you're almost done, and the only other thing is to find the y values at these points. Those will be your inflection points.
#### meeklobraca
189
0
how do I check around those values, Im not sure how to use something like pi/2 in this case.
#### HallsofIvy
41,626
821
Then use pi/2+ .001 and pi/ - .001! What do you mean "use pi/2"? It is a number! enter it into your calculator. Be sure you calculator is set to "radian" mode of course.
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# Constitutive relation error for FFT-based methods
Abstract : Since their introduction by Moulinec and Suquet (1994, 1998), FFT-based methods have become popular alternatives to finite element methods (FEM) for the numerical homogenization of heterogeneous materials. They have been successfully applied to a variety of constitutive laws. Even geometric non-linearities can now be accounted for (Kabel et al., 2014). Theoretical analysis of this numerical technique has also greatly improved. Convergence of the iterative solvers is now well understood (Michel et al., 2001; Brisard & Dormieux, 2010; Zeman et al., 2010; Monchiet & Bonnet, 2012; Moulinec & Silva, 2014), as well as convergence with respect to the grid-size (Brisard & Dormieux, 2012; Schneider 2014). Alternative discretizations aiming at improving the quality of the solution (at fixed grid-size) have also been proposed (Brisard & Dormieux, 2012; Willot et al., 2014; Vondřec et al., 2015; Willot, 2015). However, this quality improvement is usually assessed in a qualitative way. Some progress has recently been made by Vondřec et al. (2015) towards quantitative analysis of the quality of the solution, but their approach is restricted to discretizations by trigonometric polynomials. In this talk, we present a strategy for the aposteriori error analysis of the numerical solution to the Lippmann–Schwinger equation. We use the framework of the error in constitutive relation, initially introduced by Ladevèze and Leguillon (1983) for the FEM. This requires to produce a kinematically admissible displacement field and a statically admissible stress field. We will show that a kinematically admissible displacement field can be efficiently reconstructed. Our approximation does not use trigonometric polynomials; rather it is local (Q1) in space (thus free of spurious Gibbs-like oscillations). Besides, the reconstructed displacement field has some properties that allow us to reuse standard techniques from the FEM world to produce a statically admissible stress field. A posteriori error estimates are then readily computed. Our strategy applies to any discretization/solver, which allows a quantitative comparison of the various FFT-based strategies proposed in the literature. This will be illustrated on a few examples, both in two and three dimensions.
Type de document :
Documents associés à une manifestation scientifique, support de présentation orale, résumé
https://hal-enpc.archives-ouvertes.fr/hal-01330790
Contributeur : Sébastien Brisard <>
Soumis le : lundi 13 juin 2016 - 09:32:15
Dernière modification le : mardi 8 décembre 2020 - 10:20:37
Archivage à long terme le : : mercredi 14 septembre 2016 - 10:53:24
### Fichier
ECCOMAS2016-BRISARD_Sebastien-...
Fichiers produits par l'(les) auteur(s)
### Identifiants
• HAL Id : hal-01330790, version 1
### Citation
Sébastien Brisard, L Chamoin. Constitutive relation error for FFT-based methods. VII European Congress on Computational Methods in Applied Sciences and Engineering (ECCOMAS Congress 2016), Jun 2016, Crete Island, Greece. 2016. ⟨hal-01330790v1⟩
### Métriques
Consultations de la notice
## 71
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http://math.stackexchange.com/questions/764354/nice-functions | # “nice functions”
I see the statement of "nice functions" in textbooks and the authors usually don't need to give the definition of "nice functions". For example in a book which I read now the authors write
"Morrey spaces is not separable. A version of Morrey space where it is possible to approximate by "nice functions" is vanishing Morrey space."
and don't give the definition of "nice functions" anywhere in the book.
I wonder in here what is the meaning of "nice functions" ?
and
Is there a fixed definition of "nice functions" ?
-
It means a function that has sufficiently many derivatives, and which converges fast enough at infinity, that we don't have to worry about rigour when considering any expression involving derivatives, integrals, or integration by parts. Usually you can replace "nice" by "Schwarz class." But it has no real definition, and the term is intentionally vague. – Stephen Montgomery-Smith Apr 22 '14 at 12:11
I read "nice" or "well-behaved" as "satisfying all requirements for the appropriate theorems I am using." – apnorton Apr 22 '14 at 14:01
@anorton You should make that an answer, so I can vote for it (that's the most correct answer I see right now) – Mario Carneiro Apr 22 '14 at 16:52
@MarioCarneiro Done. – apnorton Apr 22 '14 at 18:16
The terms "nice" and "good" are used in an ad-hoc way throughout mathematics, and so giving them a fixed definition is counterproductive. The idea is to build intuition: we can expect our theory to work when we only consider objects that are not too strange, or we can guarantee the existence of objects satisfying certain properties that are easy to work with. Concrete definitions do not need to be introduced unless a technical discussion is forthcoming, and can inhibit readability otherwise.
-
It might mean continuous, it might mean differentiable, it might mean smooth etc. The only common theme is that the author didn't bother explaining it in more detail.
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https://www.ora.ox.ac.uk/objects/uuid:9b305cb7-1c55-49f8-9ffe-940a67beeeb3 | Internet publication
### The geometry of decoupling fields
Abstract:
We consider 4d field theories obtained by reducing the 6d (1,0) SCFT of N M5-branes probing a C2/Zk singularity on a Riemann surface with fluxes. We follow two different routes. On the one hand, we consider the integration of the anomaly polynomial of the parent 6d SCFT on the Riemann surface. On the other hand, we perform an anomaly inflow analysis directly from eleven dimensions, from a setup with M5-branes probing a resolved C2/Zk singularity fibered over the Riemann surface. By comparing ...
Publication status:
Published
### Access Document
Files:
• (Pre-print, 4.6MB)
Publication website:
https://arxiv.org/abs/2112.07796v1
### Authors
More by this author
Institution:
University of Oxford
Division:
MPLS
Department:
Mathematical Institute
Role:
Author
ORCID:
0000-0003-4469-6890
Publication date:
2021-12-14
Language:
English
Keywords:
Pubs id:
1227595
Local pid:
pubs:1227595
Deposit date:
2022-01-04 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9405577778816223, "perplexity": 4395.595811364418}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335124.77/warc/CC-MAIN-20220928051515-20220928081515-00268.warc.gz"} |
https://www.physicsforums.com/threads/tubular-bells.195180/ | # Tubular bells
1. Oct 31, 2007
### ELESSAR TELKONT
I have the next theoretical-practical problem. I have to build a tubular bell array (like that at symphonic orchrestas) with tubes (not rods) of aluminium or copper. The principal problem I have is I don't know how to state the wave equation for a tube (I have done it for a string). How I do it?
When I have the wave equation stated I must get the frecuencies to get a standing wave. Obviously this is what I want.
Could you help me?
2. Oct 31, 2007
### tyco05
If the bells are open at both ends, the wave equation is the same as for strings.
3. Oct 31, 2007
### ELESSAR TELKONT
Ok but of what quantities the velocity of propagation depends on that's my problem.
4. Oct 31, 2007
### tyco05
The velocity of propagation is the velocity of sound.
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http://mathoverflow.net/feeds/question/58732 | Torus based cryptography - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T09:38:22Z http://mathoverflow.net/feeds/question/58732 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58732/torus-based-cryptography Torus based cryptography Sebastian Petersen 2011-03-17T08:56:21Z 2012-03-20T15:10:24Z <p>In cryptography one needs finite groups $G$ in which the discrete logarithm problem is infeasible. Often they use the multiplicative group $\mathbb{G}_m(\mathbb{F}_p)$ where $p$ is a prime number of bit length $500$, say.</p> <p>Rubin and Silverberg suggested (cf. [1]) to use certain tori instead, if the goal is to minimize the key size. In the easiest case, this comes down to using the group $$T_2(p)=ker(Norm: \mathbb{F}_{p^2}^\times\to \mathbb{F}_p^\times).$$</p> <p>If I understood correctly, then the underlying philosopy seems to be: *The group $T_2(p)$ should be as secure as $\mathbb{F}_{p^2}^\times$, but its size is only $p+1$.* (So, if you use groups of type $T_2(p)$ instead of groups of type $\mathbb{G}_m(\mathbb{F}_p)$, then you can achive the same security with half the key size.)</p> <p><strong>Question.</strong> What are the reasons, be they heurisical or strictly provable, to believe in this philosopy?</p> <p>Denote by $\mathbb{G}'_m$ the quadratic twist of the algebraic group $\mathbb{G}_m$. It is easy to see that $T_2(p)$ is isomorphic to $\mathbb{G}'_m(\mathbb{F}_p)$. (This isomorphism is easy to compute). The philosophy predicts: The quadratic twist of the multiplicative group should be better than the multiplicative group itself.</p> <p>(Compare with elliptic curves: If $E/\mathbb{F}_p$ is an elliptic curve, then I would certainly not expect its quadratic twist to be better than $E$ itself.)</p> <p><strong>Remark:</strong> I concentrated on the simplest case above. One also considers certain groups $T_n(p)$ which are expected to be as secure as $\mathbb{F}_{p^n}^\times$, while their size is only $\approx\varphi(n)p$. Lemma 7 in [1] is meant to explain this. However, I would be keen on a more detailed explanation. </p> <p>[1] Lect. Notes in Comp. Sci. 2729 (2003) 349-365. (available at <a href="http://math.stanford.edu/~rubin/" rel="nofollow">http://math.stanford.edu/~rubin/</a>)</p> http://mathoverflow.net/questions/58732/torus-based-cryptography/58789#58789 Answer by DavidLHarden for Torus based cryptography DavidLHarden 2011-03-17T22:43:18Z 2011-03-17T22:43:18Z <p>The philosophy, as stated, seems off:</p> <p>The multiplicative groups of finite fields have discrete logarithm problems that are vulnerable to the number field sieve. On the other hand, the best known method in an 'abstract' cyclic group is a variant of the Shanks baby-step-giant-step method -- this is much slower than the number field sieve, since its worst-case (and average) running time in a cyclic group of order $n$ are about the size of $\sqrt{n}$ (and this worst case includes the assumption than $n$ is prime).</p> <p>What applies to the multiplicative groups of finite fields carries over to their large quotients as long as computing lifts is easy, and carries over to their large subgroups. (Here, "large" means 'large enough that the number field sieve is faster than any variant of baby-step-giant-step'.)</p> <p>How can baby-step-giant-step be sped up when $n$ is not prime? Let $n = p_{1}^{e_{1}}\ldots p_{k}^{e_{k}}$ be the prime factorization of $n$. Then the idea is to break up the discrete logarithm problem in the cyclic group of order $n$ into $e_{1}$ problems in cyclic groups of order $p_{1}$, $\ldots$, and $e_{k}$ problems in cyclic groups of order $p_{k}$. In more detail:</p> <p>$g^{x} = m$ implies $(g^{L})^{x} = g^{Lx} = m^{L}$ for any integer $L$.</p> <p>Choose $L_{i}$, where $1 \leq i \leq k$, so that $L_{i} \equiv 1 \mod{p_{i}^{e_{i}}}$ and $L_{i} \equiv 0 \mod{p_{j}^{e_{j}}}$ when $j \neq i$. Then $g$ such that $g^{x} = m$ is obtained as $g^{L_{1} + \ldots + L_{k}} = g^{L_{1}}\ldots g^{L_{k}}$.</p> <p>Each $g^{L_{i}}$ solves $( g^{ L_{i} } )^{x} = g^{L_{i} x} = m^{L_{i}}$, which is a discrete logarithm problem in the cyclic subgroup of order $p_{i}^{e_{i}}$ in our cyclic group of order $n$.</p> <p>As for solving the discrete logarithm problem in a cyclic group of order $p_{i}^{e_{i}}$ (here we write $g^{L_{i}} = h$ and $m^{L_{i}} = M$):</p> <p>If $e_{i} > 1$, then $h^{x} = M$ implies $(h^{L})^{x} = h^{Lx} = M^{L}$, where this time $L = p_{i}^{e_{i}-1}$. This is a discrete logarithm problem in a cyclic group of order $p_{i}$, and its solution (call it $x^{'}$) is the solution to $h^{x} = M$, reduced modulo $p_{i}$. Then $h^{x} = M$ implies $h^{x-x^{'}} = M h^{-x^{'}}$, which implies $(h^{p_{i}})^{\frac{x-x^{'}}{p_{i}}} = M h^{-x^{'}}$. This is a discrete logarithm problem in a cyclic group of order $p_{i}^{e_{i}-1}$.</p> <p>Finally, the variant of the basic baby-step-giant-step method when $e_{i} = 1$ (still write this discrete logarithm problem as $h^{x} = M$ for simplicity):</p> <p>Compute $h^{1}, h^{2}, \ldots, h^{S}$ and store that as a sorted list (so that searches can be made using a logarithmic time binary search). Then compute $M, Mh^{-S}, Mh^{-2S}, \ldots$ until a $t$ is found such that $Mh^{-tS}$ is on the list. Then one obtains $Mh^{-tS} = h^{u}$ for some $u$ with $1 \leq u \leq S$, which gives $M = h^{u + tS}$, so that $x = u + tS$. Here, for maximum speed, choose $S$ to be the result of rounding $\sqrt{p_{i}}$ to the nearest integer. </p> http://mathoverflow.net/questions/58732/torus-based-cryptography/58792#58792 Answer by Joe Silverman for Torus based cryptography Joe Silverman 2011-03-17T23:23:18Z 2011-03-17T23:23:18Z <p>The discrete log problem in the multiplicative group of a finite field may be solved using the <em>index calculus</em>, not the number field sieve (although sieves are used to speed the process of checking numbers for smoothness during the index calculus algorithm). Anyway, the idea is as follows. </p> <p>Using index calculus on a field $\mathbb{F}_q^*$ has running time $L(q,1/3)$, which is a subexponential function of $q$ that is approximately $\exp((\log q)^{1/3})$. Now suppose that we work in a subgroup $G\subset \mathbb{F}_q^*$ of order $N$. Then operations in $G$ may be much faster to compute, so we get a more efficient system. And to solve the discrete log problem in $G$, we have two options. We can use a collision algorithm such as Pollard's $\rho$ method, which has running time proportional to $\sqrt{N}$, or we can use the index calculus. But the index calculus doesn't work directly on $G$, so even though we're using elements of $G$, no one knows how to do the index calculus faster than $L(q,1/3)$. </p> <p>So balancing $N$ and $q$ appropriately, one can get a secure cryptosystem that is more efficient than if one worked with arbitrary elements of $\mathbb{F}_q^*$.</p> <p>Note that I'm not claiming that one can't do the DLP in $G$ faster than the minimum of $O(\sqrt{N})$ and $L(q,1/3)$, I'm simply saying that at present, no one knows how to do it. (But that's true of the security of all the problems being used in cryptography, we have no <em>proofs</em> that they are actually difficult.)</p> http://mathoverflow.net/questions/58732/torus-based-cryptography/91726#91726 Answer by Granger for Torus based cryptography Granger 2012-03-20T15:10:24Z 2012-03-20T15:10:24Z <p>Can I refer you to my paper `On the Discrete Logarithm Problem on Algebraic Tori', Advances in Cryptology – CRYPTO 2005, Lecture Notes in Computer Science, 2005, Volume 3621/2005, 66-85, in which myself and Frederik Vercauteren studied this very problem.</p> <p>In particular, we showed that the compression mechanism afforded by the birationality of some algebraic tori may be exploited to obtain a faster discrete logarithm algorithm for some cryptographically practical field sizes. In these instances, attacking the discrete logarithm in $\mathbb{F}_{p^n}^{\times}$ via its decomposition </p> <p>$\prod_{d \mid n} T_d(\mathbb{F}_p)$ is faster than using L[1/3] index calculus techniques.</p> <p>Since then, other work has improved the L[1/3] index calculus techniques. However, our work demonstrates that it is naive to argue that the DLP in algebraic tori must be hard purely because the DLP in the multiplicative group of the extension field is hard, precisely because an attack on the former provides an attack on the latter.</p> | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8114469051361084, "perplexity": 423.7415744772939}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703108201/warc/CC-MAIN-20130516111828-00009-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://ieeexplore.ieee.org/xpl/articleDetails.jsp?reload=true&tp=&arnumber=6278848&contentType=Books+%26+eBooks | We study the dynamic categorization ability of an autonomous agent that distinguishes rectangular and triangular objects. The objects are distributed on a two-dimensional space and the agent is equipped with a recurrent neural network that controls its navigation dynamics. As the agent moves through the environment, it develops neural states which, while not symbolic representations of rectangles or triangles, allow it to distinguish these objects. As a result, it decides to avoid triangles and remain for longer periods of time at rectangles. A significant characteristic of the network is its plasticity, which enables the agent to switch from one navigation mode to another. Diversity of this switching behavior will be discussed. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8574090003967285, "perplexity": 530.8982909109548}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813832.23/warc/CC-MAIN-20180222002257-20180222022257-00750.warc.gz"} |
https://www.computer.org/csdl/trans/tc/1971/12/01671774-abs.html | Issue No. 12 - December (1971 vol. 20)
ISSN: 0018-9340
pp: 1602-1605
C.R. Baugh , IEEE
ABSTRACT
Upper and lower bounds are derived for the number of pseudothreshold functions of n variables. (Pseudothershold logic is a generalization of threshold logic.) It is shown that a lower bound on the number of pseudothreshold functions P(n) of exactly n variables realized by zero-free structures is The number of pseudothreshold functions Q(n) of n variables realized by nontrivial structures is bounded by It is also proven that is a lower bound on the number of positive functions of exactly n variables.
INDEX TERMS
Bounds, positive functions, pseudothreshold functions, separable functions, threshold functions, threshold logic.
CITATION
C.R. Baugh, "Bounds on the Number of Pseudothreshold Functions", IEEE Transactions on Computers, vol. 20, no. , pp. 1602-1605, December 1971, doi:10.1109/T-C.1971.223181 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9231775403022766, "perplexity": 2416.382998953335}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541134.5/warc/CC-MAIN-20161202170901-00021-ip-10-31-129-80.ec2.internal.warc.gz"} |
http://cms.math.ca/10.4153/CJM-2008-007-6 | location: Publications → journals → CJM
Abstract view
# Boundary Structure of Hyperbolic $3$-Manifolds Admitting Annular and Toroidal Fillings at Large Distance
Published:2008-02-01
Printed: Feb 2008
• Sangyop Lee
• Masakazu Teragaito
Features coming soon:
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Format: HTML LaTeX MathJax PDF PostScript
## Abstract
For a hyperbolic $3$-manifold $M$ with a torus boundary component, all but finitely many Dehn fillings yield hyperbolic $3$-manifolds. In this paper, we will focus on the situation where $M$ has two exceptional Dehn fillings: an annular filling and a toroidal filling. For such a situation, Gordon gave an upper bound of $5$ for the distance between such slopes. Furthermore, the distance $4$ is realized only by two specific manifolds, and $5$ is realized by a single manifold. These manifolds all have a union of two tori as their boundaries. Also, there is a manifold with three tori as its boundary which realizes the distance $3$. We show that if the distance is $3$ then the boundary of the manifold consists of at most three tori.
Keywords: Dehn filling, annular filling, toroidal filling, knot
MSC Classifications: 57M50 - Geometric structures on low-dimensional manifolds 57N10 - Topology of general $3$-manifolds [See also 57Mxx] | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9269387722015381, "perplexity": 1622.4549739996412}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00142-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/prooving-crazy-formula.52956/ | # Homework Help: Prooving Crazy Formula
1. Nov 16, 2004
### Ineedphysicshelp
A block slides down a curved frictionless track and then up an inclined plane. the coefficient of kinetic friction between the block and the incline is Uk. Use energy methods to show that the maximum height reached by the block is:
Ymax=
h
-----------------
1 + Uk cot(theta)
2. Nov 16, 2004
### Staff: Mentor
3. Nov 16, 2004
### Ineedphysicshelp
I don't know how to delete the other post. Sorry about that. Ok I know that because it's cot(theta) that is equal to cos(theta)/sin(theta). I've been trying to figure it out in reverse. So I multiplied the right side by Fg/Fg so that I got something that looks like: Fgcos(theta)h / [Fg+UkFgsin(theta)] which looks like Fgyh / (Fg + FgxUk) I don't know where to go from here. I've filled up pages of calculations and it has just stopped making sense. Sorry again about the double posting.
4. Nov 16, 2004
### Gza
It seems like you are making it way too hard for yourself. This is physics, not math. If you understand work and energy, the math will follow. Working backward from the equation will not help you to understand the physics of the situation.
5. Nov 16, 2004
### Staff: Mentor
Don't bother "reverse engineering" the equation...just apply conservation of energy.
Call the height reached Ymax. In reaching that height up the incline: How much work was done against friction? What's the final PE?
Apply conservation of energy: The initial PE = final PE + work done against friction.
6. Nov 16, 2004
### Ineedphysicshelp
Ok, thank you, I realize what I did wrong. I had it as hcos(theta)/[1+Uksin(theta)] which is upside down to what it's supposed to be: hsin(theta)/[1+Uksin(theta)] that's why it wasn't making any sense. :P . Thank you.
7. Nov 16, 2004
### Ineedphysicshelp
I mean hsin(theta)/[1+Ukcos(theta)] | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8649523854255676, "perplexity": 1330.5876891686103}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00325.warc.gz"} |
https://www.physicsforums.com/threads/two-blocks-with-spring-no-friction-force-applied-displacement.527516/ | # Two blocks with spring (no friction), force applied: displacement?
1. Sep 6, 2011
### ddtozone
1. The problem statement, all variables and given/known data
A 2kg block & 3kg block (left & right, respectively) are on a horizontal frictionless surface connected by a spring with K=140 N/m. A 15N force is applied to the 3kg block towards the right. How much does the spring stretch from its equilibrium length?
2. Relevant equations
F=-kx
others, maybe?
3. The attempt at a solution
I simply did force over constant, which is 15/140=0.107m. Is it this simple, or do you use the masses in some way?
Thank you!
2. Sep 6, 2011
### Rayquesto
the gravitational force of the 3kg is going to be more than the 2 kg gravitational force. So, when you pull with 15N to the right, it will not act like a fixed spring constant. So, what you should do is say 3-2kg=1kg so the resultant will include 9.81 newtons in terms of the spring being moved, not the entire system which includes the spring and the two blocks. So, I think the answer is 15N-9.81N then you multiply by its constant.
3. Sep 6, 2011
### ddtozone
Oh ok, that makes more sense. But wouldn't you divide by the constant instead of multiplying by the constant, so the final answer would be N/(N/m)=m?
Similar Discussions: Two blocks with spring (no friction), force applied: displacement? | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9337173700332642, "perplexity": 823.0923265604778}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818687606.9/warc/CC-MAIN-20170921025857-20170921045857-00496.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=Logarithm&diff=121081&oldid=7951 | # Difference between revisions of "Logarithm"
## Powerful use of logarithms
Some of the real powerful uses of logarithms, come down to never having to deal with massive numbers. ex. : would be a pain to have to calculate any time you wanted to use it (say in a comparison of large numbers). its natural logarithm though (partly due to left to right parenthesized exponentiation) is only 7 digits before the decimal point. Comparing the logs of the numbers to a given precision can allow easier comparision than computing and comparing the numbers themselves. Logs also allow (with repetition) to turn left to right exponentiation into power towers (especially useful for tetration (exponentiation repetition with the same exponent)). ex.
Therefore by : and identities 1 and 2 above ( 2 being used twice) we get:
such that :
### Discrete Logarithm
A only partially related value is the discrete logarithm, used in cryptography via modular arithmetic. It's the lowest value such that, for given being integers (as well as the unknowns being integer).
Its related to the usual logarithm, by the fact that if isn't an integer power of then is a lower bound on
## Problems
1. Evaluate .
2. Evaluate .
3. Simplify where .
## Natural Logarithm
The natural logarithm is the logarithm with base e. It is usually denoted , an abbreviation of the French logarithme normal, so that However, in higher mathematics such as complex analysis, the base 10 logarithm is typically disposed with entirely, the symbol is taken to mean the logarithm base e and the symbol is not used at all. (This is an example of conflicting mathematical conventions.)
can also be defined as the area under the curve between 1 and a, or .
All logarithms are undefined in nonpositive reals, as they are complex. From the identity , we have . Additionally, for positive real .
## Problems
### Introductory
• What is the value of for which ?
• Positive integers and satisfy the condition Find the sum of all possible values of .
### Intermediate
• The sequence is geometric with and common ratio where and are positive integers. Given that find the number of possible ordered pairs | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9649744033813477, "perplexity": 786.189892323595}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141737946.86/warc/CC-MAIN-20201204131750-20201204161750-00497.warc.gz"} |
http://darwin.eeb.uconn.edu/eeb348/lecture-notes/inbreeding/node2.html | Next: Partial self-fertilization Up: Inbreeding and self-fertilization Previous: Introduction
# Self-fertilization
Self-fertilization is the most extreme form of inbreeding possible, and it is characteristic of many flowering plants and some hermaphroditic animals, including freshwater snails and that darling of developmental genetics, Caenorhabditis elegans.1It's not too hard to figure out what the consequences of self-fertilization will be without doing any algebra.
• All progeny of homozygotes are themselves homozygous.
• Half of the progeny of heterozygotes are heterozygous and half are homozygous.
So you might expect that the frequency of heterozygotes would be halved every generation, and you'd be right. To see why, consider the following mating table:
Offsrping genotype Mating frequency 1 0 0 0 0 1
Using the same technique we used to derive the Hardy-Weinberg principle, we can calculate the frequency of the different offspring genotypes from the above table.
(1) (2) (3)
I use the to indicate the next generation. Notice that in making this caclulation I assume that all other conditions associated with Hardy-Weinberg apply (meiosis is fair, no differences among genotypes in probability of survival, no input of new genetic material, etc.). We can also calculate the frequency of the allele among offspring, namely
(4) (5) (6) (7)
These equations illustrate two very important principles that are true with any system of strict inbreeding:
1. Inbreeding does not cause allele frequencies to change, but it will generally cause genotype frequencies to change.
2. Inbreeding reduces the frequency of heterozygotes relative to Hardy-Weinberg expectations. It need not eliminate heterozygotes entirely, but it is guaranteed to reduce their frequency.
• Suppose we have a population of hermaphrodites in which and we subject it to strict self-fertilization. Assuming that inbred progeny are as likely to survive and reproduce as outbred progeny, in six generations and in ten generations.
Next: Partial self-fertilization Up: Inbreeding and self-fertilization Previous: Introduction
Kent Holsinger 2012-09-03 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8602075576782227, "perplexity": 3547.76567765975}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119660038.53/warc/CC-MAIN-20141024030100-00274-ip-10-16-133-185.ec2.internal.warc.gz"} |
https://www.collectget.com/questions/925702/am-i-on-the-right-track-to-proving-this-statement-about-composite-function | Latest update
# Am I on the right track to proving this statement about composite function?
2018-03-18 22:43:16
Let $S = \{1,2,3,4\}$. Let $F$ be the sets of all functions from $S$ to $S$.
Now I think this statement is true:
$\forall f \in F , \exists g\in F$ so that $(g \circ f)(1) =2$
I suppose $f \in F$, and I need to show that $\exists g\in F$ so that $(g \circ f)(1) = 2$
Do I just make an example of $g$ so that $(g \circ f)(1) = 2$ ?
Like $f(1) = y$ and then $g (y)=2 ? You could simply design a function$g$such that$g(x)=2$for all$x\in S$. Now since$f\in F$, we know$f(1)\in S$, hence the composition is 2. As f is arbitrary, we have the desired result. Yes, you just define a function$g$which takes$f(1)$to$2$and then define the rest if images to make$g$a well defined function from$S$to$S$• You could simply design a function$g$such that$g(x)=2$for all$x\in S$. Now since$f\in F$, we know$f(1)\in S$, hence the composition is 2. As f is arbitrary, we have the desired result. 2018-03-18 23:40:01 • Yes, you just define a function$g$which takes$f(1)$to$2$and then define the rest if images to make$g$a well defined function from$S$to$S\$
2018-03-19 00:30:22 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9086469411849976, "perplexity": 499.424691134138}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583516003.73/warc/CC-MAIN-20181023023542-20181023045042-00228.warc.gz"} |
https://quantumcomputing.stackexchange.com/questions/9990/is-there-a-lower-bound-on-the-average-diamond-norm-of-two-uniformly-random-unita | # Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?
I could not find any lower bound on the diamond norm for two uniformly random unitaries of dimension D sampled from the haar measure.
This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you translate the question into different terms, as I will describe.
First, suppose that $$\Phi_0(X) = U_0 X U_0^{\dagger}$$ and $$\Phi_1(X) = U_1 X U_1^{\dagger}$$ are any two unitary channels. In this special case, the diamond norm distance between these two channels can be expressed as follows: $$\|\Phi_0 - \Phi_1\|_{\diamond} = 2 \sqrt{1 - \delta(U_0^{\dagger} U_1)^2},$$ where $$\delta(A)$$ denotes the minimum absolute value taken over the numerical range of a given operator $$A$$. That is, $$\delta(A) = \min_{|\psi\rangle} |\langle \psi | A | \psi\rangle|,$$ where the minimum is over all unit vectors $$|\psi\rangle$$.
Now, if you're interested in $$U_0$$ and $$U_1$$ being Haar-random, then you might as well set $$U_0 = I$$ and $$U_1 = U$$ for $$U$$ Haar-random, by the fact that the diamond norm is unitarily invariant. To understand the distribution of the diamond norm distance, you therefore need to understand something about the numerical range of a Haar-random unitary. Because unitary operators are normal, their numerical range is the convex hull of their eigenvalues, so you're essentially trying to understand something about the eigenvalues of a Haar-random unitary. This topic has been studied extensively in random matrix theory.
I do not know enough about this topic to give you precise bounds, but it is clear that as the dimension $$D$$ grows, the probability that the diamond norm distance equals the maximum possible value 2 (which means the two channels can be discriminated perfectly) approaches 1. Indeed, in order to have diamond norm distance strictly less than 2, all of the eigenvalues of $$U$$ (or $$U_0^{\dagger} U_1$$) must fall within an arc on the unit circle having length strictly less than $$\pi$$, which is extremely unlikely for large $$D$$ (surely the probability is exponentially small in $$D$$).
The result you're looking for is effectively Proposition 19 of the paper: Almost all quantum channels are equidistant; which I'm rewriting here for convenience:
Let $$U, V \in \mathcal{U}(d)$$ be two independent random variables, at least one of them being Haar-distributed. Then, with overwhelming probability as $$d \rightarrow \infty$$, the quantum channels $$\Phi(X)=U X U^{\dagger}$$ and $$\Psi(X)=V X V^{\dagger}$$ become perfectly distinguishable: for $$d$$ large enough, $$\mathbb{P}\left[\|\Phi-\Psi\|_{\diamond}=2\right] \geq 1-\exp \left(-\frac{\log 2}{2} d^{2}\right)$$
If you're familiar with typicality results, then this is an example of concentration of measure phenomenon.
Note that this result is general than the one you're looking for since it requires only one of the channels to be Haar. Moreover, it fits nicely with the insights that John Watrous provided above; and the rest of the paper has generalizations to random matrix theory.
• Thank you. The reference has some really nice results. And indeed the unitary invariance property of diamond norm indicates that one needs to choose only one unitary to be haar random while the other can be fixed to show perfect distinguishability with exponentially high probability. Do you know if one can also expect perfect distinguishability when the distinguishability is defined via any other shatten-p norm, instead of diamond norm? Jun 3 '20 at 5:30
• I'd think so, yes. Off the top of my head, some simple bounds on the Schatten 1-norm follow just from the properties relating the 1-norm to the diamond norm (see, for example, Watrous' notes: cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf; page 171). Moreover, the 1-norm can be bounded using the other p-norms in turn. Would that suffice or are you looking for tighter bounds? Jun 8 '20 at 3:42 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 24, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8748131394386292, "perplexity": 217.84287172846825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585348.66/warc/CC-MAIN-20211020183354-20211020213354-00661.warc.gz"} |
https://www.coursehero.com/file/46103702/lec21-slidespdf/ | lec21_slides.pdf - Generalized Linear Models 1 GLMs I We have so far studied linear models We have n observations on a response variable y1 yn and on
# lec21_slides.pdf - Generalized Linear Models 1 GLMs I We...
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Generalized Linear Models 1 April 18, 2019
GLMs I We have so far studied linear models. We have n observations on a response variable y 1 , . . . , y n and on each of p explanatory variables x ij for i = 1 , . . . , n and j = 1 , . . . , p .
GLMs I We have so far studied linear models. We have n observations on a response variable y 1 , . . . , y n and on each of p explanatory variables x ij for i = 1 , . . . , n and j = 1 , . . . , p . I The linear model that we have seen models μ i := E y i = β 0 + β 1 x i 1 + · · · + β p x ip .
GLMs I We have so far studied linear models. We have n observations on a response variable y 1 , . . . , y n and on each of p explanatory variables x ij for i = 1 , . . . , n and j = 1 , . . . , p . I The linear model that we have seen models μ i := E y i = β 0 + β 1 x i 1 + · · · + β p x ip . I What this model implies is that when there is a unit increase in the explanatory variable x j , the mean of the response variable changes by the amount β j .
GLMs I We have so far studied linear models. We have n observations on a response variable y 1 , . . . , y n and on each of p explanatory variables x ij for i = 1 , . . . , n and j = 1 , . . . , p . I The linear model that we have seen models μ i := E y i = β 0 + β 1 x i 1 + · · · + β p x ip . I What this model implies is that when there is a unit increase in the explanatory variable x j , the mean of the response variable changes by the amount β j . I This may not always be a reasonable assumption.
GLMs I For example, if the response y i is a binary variable, then its mean μ i is a probability which is always constrained to stay between 0 and 1.
GLMs I For example, if the response y i is a binary variable, then its mean μ i is a probability which is always constrained to stay between 0 and 1. I Therefore, the amount by which μ i changes per unit change in x j would now depend on the value of μ i (for example, the change when μ i = 0 . 9 may not be the same as when μ i = 0 . 5).
GLMs I For example, if the response y i is a binary variable, then its mean μ i is a probability which is always constrained to stay between 0 and 1. I Therefore, the amount by which μ i changes per unit change in x j would now depend on the value of μ i (for example, the change when μ i = 0 . 9 may not be the same as when μ i = 0 . 5). I Therefore, modeling μ i as a linear combination of x 1 , . . . , x p may not be the best idea always.
GLMs I For example, if the response y i is a binary variable, then its mean μ i is a probability which is always constrained to stay between 0 and 1. I Therefore, the amount by which μ i changes per unit change in x j would now depend on the value of μ i (for example, the change when μ i = 0 . 9 may not be the same as when μ i = 0 . 5). I Therefore, modeling μ i as a linear combination of x 1 , . . . , x p may not be the best idea always. I A more general model might be g ( μ i ) := β 0 + β 1 x i 1 + · · · + β p x ip (1) for a function g that is not necessarily the identity function.
I Another feature of the linear model that people do not always like is that some aspects of the theory are tied to the normal distribution.
I Another feature of the linear model that people do not | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9804040193557739, "perplexity": 327.1259287867784}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487611320.18/warc/CC-MAIN-20210614013350-20210614043350-00632.warc.gz"} |
https://socratic.org/questions/how-do-you-simplify-78-1 | Algebra
Topics
# How do you simplify 78^-1?
Jun 10, 2015
${78}^{- 1} = \frac{1}{78}$
#### Explanation:
Use the negative exponent rule: ${a}^{- n} = \frac{1}{{a}^{n}}$, in which negative exponents in the numerator are moved to the denominator and become positive, and negative exponents in the denominator are moved to the numerator and become positive.
${a}^{- n}$ is understood to be $\frac{{a}^{- n}}{1}$, and ${78}^{- 1}$ is understood to be ${78}^{- 1} / 1$, and $\frac{1}{78}$ is understood to be $\frac{1}{{78}^{1}}$ .
http://www.mesacc.edu/~scotz47781/mat120/notes/exponents/review/review.html
##### Impact of this question
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http://csr.ufmg.br/dinamica/dokuwiki/doku.php?id=patterns_of_change | # Spatial Patterns of Changes
As Dinamica EGO was designed to be a general purpose dynamic modeling software, we present a collection of spatial patterns produced by the various combinations of its transition functions. In order to evaluate the wide range of Dinamicas’s possibilities, a series of simulation were run using synthetic simplified maps. The results for each model are presented as follows:
H0 to H8 are run, for a sole time step, by using a transition matrix 2×2 that models only one transition – from class 2 to 1 – with a rate of 0.01. The landscape map encompasses a matrix of 132 by 132 cells and all cells have equivalent spatial transition probability, which means that ancillary variables are not used to influence the cell allocation process.
H0 There is no spatial arrangement and no patch aggregation. The allocation process takes place randomly and does not interact with the neighborhood. The dynamics is only controlled by the transition matrix. H1 The allocation process is set to form patches with a patch mean size of five cells, patch variance is set to zero. Only the Patcher functor is used. The Patcher isometry factor is set to zero, which means that the patches tend to be most linear as possible. H2 The allocation process is set to form patches with a patch mean size of five cells, patch variance is set to zero. Only the Patcher functor is used. The Patcher isometry factor is set to 1, the patches still take linear form, though shorter. H3 The allocation process is set to form patches with a patch mean size of five cells, patch variance is set to zero. Only the Patcher functor is used. The Patcher isometry factor is set to 1.5, now the patches assume a more isometric form. H4 The allocation process is set to form patches with a patch mean size of 20 cells, patch variance is set to zero. Only the Patcher functor is used. The Patcher isometry factor is set to 1.5. Notice the fewer and larger patches. H5 The allocation process is set to form patches with patch mean size of 200 cells, patch variance is set to zero. Only the Patcher functor is used. The Patcher isometry factor is set to 1.5. Only seven patches are formed. Patches tend to aggregate. H6 The allocation process is set to form patches with patch mean size of 20 cells, patch variance is set to 180. Only the Patcher functor is used. The Patcher isometry factor is set to 1.5. Path size varies greatly. H7 Only the Expander functor is used with patch mean size of 1742 cells, which is tantamount to the expected number of transitions. Patch variance is set to 0. The Expander isometry factor is set to 1.5. Notice the single patch produced around a cell of class 2 located at the center of the map. H8 The transition functions are used in a combination of 0.8 of Expander and 0.2 of Patcher. Patch mean size is set to 600 with patch variance of 0. The isometry factor is set to 1.5. Two more patches are produced around the expanded central cell.
H9: Transitions occur as a function of the spatial probability. DINAMICA sets up a spatial transition probability map for each transition, based on the weights of evidence chosen for specific ranges of each spatial variable stored in the static cube raster dataset. Simulation is run in 15 steps, with a rate of 0.005 per step for transition 2-1. Only the Patcher functor is used with patch mean size of 20 and patch variance of 0. Patch isometry is equal to 2. As a result, patches tend to form around the center of the map. The figures bellow depict, respectively, the static variable map, the calculated spatial transition probability map, and the simulated landscape.
Static Variable MapProbability MapSimulated Landscape
H10: Transitions occur as a function of dynamic spatial probability. DINAMICA calculates dynamic distances to cells in previous state 2 and uses it to set up a spatial transition probability map based on the weights of evidence chosen for specific ranges of distance. Since a dynamic distance map is used, a new transition map is calculated during each iteration. Simulation is run in 15 steps, with a rate of 0.005 per step for the transition 1 to 2. Only the Patcher functor is used with patch mean size of 20 and patch variance of 0. Patch isometry is equal to 2. As a result, patches tend to nucleate around previous patches. The figures bellow depict, respectively, the dynamic distance map, the dynamically calculated spatial transition probability map, and the simulated landscape.
Distance MapProbability MapSimulated Landscape
To test its a ability in simulating multiple transitions, simulations from H11 to H15 are run for a transition matrix 6 by 6. See figure bellow.
H11: There is no spatial arrangement and no patch aggregation. Expander percentage is 0, patch mean size is 1, and patch variance is 0. Transitions take place randomly only obeying the amounts of change set by the transition matrix. Simulations are run for 10 time steps. The figures bellow depict, respectively, the original landscape map, and the simulated landscape.
H12: There is no spatial arrangement but patch aggregation. Patch mean size is set to 5, and simulation is run for 10 time steps.
H13: There is no spatial arrangement but patch expansion produced by the sole use of the Expander functor. Patch mean size is set to 1, and simulation is run for 10 time steps. The left figure depicts the result for patch isometry equal to 1 and the right figure for patch isometry equal to 1.2
Isometry=1Isometry=1.2
H14: There is no patch aggregation but now each transition is influenced by its spatial probability map computed over maps stored in the static raster cube. Expander percentage is 0, patch mean size is 1, and patch variance is 0. Simulations are run for 10 time steps.
Figures bellow in B&W represent the Weights of Evidence function, respectively, for transitions 1-2, 1-3, 1-4, 1-5, 1-6. The color figure represents the transition probability map for 1-2 computed by integrating the single Weights of Evidence functions. The last two color figures depict the simulated landscape maps after 1 and 10 iterations.
Weights Of Evidence Function, Transition 1-2Weights Of Evidence Function, Transition 1-3Weights Of Evidence Function, Transition 1-4Weights Of Evidence Function, Transition 1-5Weights Of Evidence Function, Transition 1-6
Probability Map for Transition 1-2
Simulated Landscape After 1 IterationSimulated Landscape After 10 Iterations
In conclusion, the combination of Dinamica’s transition function presents numerous possibilities with respect to the generation and evolvement of spatial patterns of change. As a result, Dinamica can be considered as a potential tool for the replication of dynamic landscape structures. The calibration of a simulated landscape can be achieved by a series of simulation using varying parameters. An approximated solution can be attained comparing landscape metrics, such as fractal index and mean patch size of the simulated maps with the ones of the reference landscape. We intend to incorporate in the next version of Dinamica an automatic calibration procedure aiming at the match of spatial patterns. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9222618341445923, "perplexity": 1272.1764036593147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964361253.38/warc/CC-MAIN-20211202084644-20211202114644-00228.warc.gz"} |
http://mathhelpforum.com/pre-calculus/130882-resultant-vectors.html | 1. ## Resultant vectors
An airplane is travelling S70W with a resultant ground speed of 414 km/h. The nose of the plane is pointing west with an airspeed of 425 km/h. Determine is the speed and direction of the wind.
how to solve it?
2. Originally Posted by kkaa
An airplane is travelling S70W with a resultant ground speed of 414 km/h. The nose of the plane is pointing west with an airspeed of 425 km/h. Determine is the speed and direction of the wind.
how to solve it?
hi
you will need a diagram to solve this , take a look at the one i attached .
using the cosine rule ,
$\displaystyle |V_w|^2=425^2+414^2-2(425)(414)\cos 20$
$\displaystyle |V_w|=146.09 km/h$
As for the direction , solve for $\displaystyle \alpha$ first
sine rule ,
$\displaystyle \frac{146.09}{\sin 20}=\frac{414}{\sin \alpha}$
$\displaystyle \alpha=75.75^o$
$\displaystyle \theta=90-75.75=14.25$
so the direction is S 14.25 E
Attached Thumbnails
3. how did you know the direction of wind supposes tobe | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8792359232902527, "perplexity": 1242.913880701091}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948126.97/warc/CC-MAIN-20180426105552-20180426125552-00303.warc.gz"} |
http://tex.stackexchange.com/questions/182933/auto-detect-conditional-probabilities-in-macro | auto-detect conditional probabilities in macro
Is it possible to write a macro that detects whether there is a | character in its arguments, and expands that to something else?
In concrete terms, I want to be able to use the following definitions simultaneously:
\def\P(#1){\Pr\left( #1 \right)}
\def\P(#1|#2){\Pr\left( #1 \mid #2 \right)}
Whenever in the text I write a probability $\P(X)$, the first definition should be used, whereas $\P(X|Y)$ should use the second.
If possible, a simple TeX/LaTeX solution is preferred, since many journals put restrictions on usable packages.
I know this could be achieved using simply two different macros. The reason for this question is that the abstract of a paper often ends up printed in plain text (think the arXiv email), and I would like it to be as readable as possible.
-
This can differentiate the argument as you requested.
\documentclass{article}
\def\P(#1){\Phelper#1|\relax\Pchoice(#1)}
\def\Phelper#1|#2\relax{\ifx\relax#2\relax\def\Pchoice{\Pone}\else\def\Pchoice{\Ptwo}\fi}
\def\Pone(#1){\Pr\left( #1 \right)}
\def\Ptwo(#1|#2){\Pr\left( #1 \mid #2 \right)}
\def\Pr{\mathbf{Pr}}
\begin{document}
$\P(A)$ or $\P(A|B)$
\end{document}
If your arguments are always short, then a simpler \def\P(#1){\Pr(#1)} may be sufficient for your need. But since you are using \left(...\right), it might be the case that you want to handle tall arguments. In that case, I would replace \mid with \,\middle|\, so that you could achieve the following, which works in both text style and display style:
\documentclass{article}
\def\P(#1){\Phelper#1|\relax\Pchoice(#1)}
\def\Phelper#1|#2\relax{\ifx\relax#2\relax\def\Pchoice{\Pone}\else\def\Pchoice{\Ptwo}\fi}
\def\Pone(#1){\Pr\left( #1 \right)}
\def\Ptwo(#1|#2){\Pr\left( #1 \,\middle|\, #2 \right)}
\def\Pr{\mathbf{Pr}}
\begin{document}
\centering$\P(A)$ or $\P(A|B)$
$\P(\displaystyle\frac{x}{2}>1 | x>0) = 0.5$
\end{document}
-
Perfect, thank you. It's necessary to differentiate the argument because the non-\mid vertical bar does not look right. – Nicolas Jun 3 at 13:11
I don't think you gain too much by having \P(A|B) in the source rather than \P{A|B}. Also, redefining \P is not a good idea.
Anyway, here's a redefinition of \Pr, which exploits mathtools so it can receive an optional size argument or the * that denotes automatic resizing.
\documentclass{article}
\usepackage{mathtools,xparse}
\RenewDocumentCommand\Pr{sO{}r()}{%
\operatorname{Pr}%
\begingroup
\IfBooleanTF{#1}
{\PrInn*{#3}}
{\PrInn[#2]{#3}}%
\endgroup
}
\DeclarePairedDelimiterX\PrInn[1](){%
\activatebar
#1%
}
\newcommand{\activatebar}{%
\begingroup\lccode~=|
\lowercase{\endgroup\def~}{\;\delimsize\vert\;}%
\mathcode|=\string"8000
}
\begin{document}
$\Pr(a)\quad \Pr(x|y)\quad \Pr[\big](x|y)\quad \Pr*(\frac{X}{2}|\frac{Y}{2})$
\end{document}
-
I wouldn't use \left ..\mid ...\right in this simple case:
\documentclass{article}
\makeatletter
\def\P(#1){\expandafter\P@i#1||\@nil}
\def\P@i#1|#2|#3\@nil{\ifx\relax#2\relax \Pr(#1)\else\Pr( #1|#2 )\fi}
\def\Pr{\mathbf{Pr}}
\makeatother
\begin{document}
$\P(A)$ or $\P(A|B)$
\end{document}
-
This is a nice succinct macro to achieve the desired result, thanks. However, I prefer the extra spacing when using \mid, that's why I did not just write \Pr(A|B) in the first place. – Nicolas Jun 3 at 14:07
Feel free to use it. Changing the above code is easy – Herbert Jun 3 at 14:27
Here is my take on this
\usepackage{mathtools}
\providecommand\given{} % so it exists
\DeclarePairedDelimiterX\PH[1](){
\renewcommand\given{\nonscript\:\delimsize\vert\nonscript\:}
#1
}
\newcommand\Prop{\operatorname{Pr}\PH}
Then use
$\Prop{ A } \qquad \text{or}\qquad \Prop{ A \given B}$
this include auto scaling of () and \vert if you use \Prop*. The very recent mathtools has a macro to make \Prop at one go.
Note I do not make the () a part of the syntax, and I perfer space around the |
-
@ daleif: Why not simply DeclarePairedDelimiter\Prob[1](){\providecommand\given{\nonscript\:\delimsize\vert\nonscript\:}#1 }? – Bernard Jun 3 at 13:48
(1) my bad, has to be \DeclarePairedDelimiterX, (2) yours is not enough, you cannot get the Pr in front. That is what the \DeclarePairedDelimiterXPP from the latest very mathtools` can do. – daleif Jun 3 at 13:52
(1) yes, sorry for the typo. Thanks for the information about \DeclarePairedDelimiterXPP. – Bernard Jun 3 at 14:13 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9695389866828918, "perplexity": 2592.2646778564235}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510259834.25/warc/CC-MAIN-20140728011739-00303-ip-10-146-231-18.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/timelike-v-spacelike-is-it-arbitrary.705544/ | # Timelike v. spacelike, is it arbitrary?
1. Aug 10, 2013
### BruceW
after a bit of searching, I found "An introduction to general relativity and cosmology" jerzy plebanski, andrzej Krasinski, where they don't use different terms for proper time and proper length. They just use the word 'arc length' to mean either. And they use the equation:
$$l = \int \left( | g_{\alpha \beta}( \gamma ) \frac{d x^{\alpha}}{d \gamma} \frac{d x^{\beta}}{d \gamma} | \right)^{1/2} \ d \gamma$$
Where the line | means 'take absolute value'. So maybe I should use the phrase 'arc length' instead. But then it still implies a length, when I might be measuring proper time. I guess at least they are using a different phrase, instead of what I was doing, which was using the old phrase 'proper time' to mean a new thing.
The main reason to use a definition that does not discern between proper time and proper length is that in general relativity, we often don't define a coordinate time, and we can do lots of useful calculations, without ever caring about whether we have a proper length or a proper time. In this sense, the difference between proper time and proper length in general relativity is artificial. (i.e. sure we can add in a coordinate time, but it is not necessary for using relativistic equations). Further, I thought it would be OK to say that the 'arc length' along a null geodesic is just zero. But as wannabenewton said, null worldlines can't be parameterised by proper time. So I think I was wrong to say that the 'arc length' is zero along a null worldline, since it is not possible to define it along a null worldline?
Also, several of our lecturers at undergraduate would use non-standard definitions, to force us to get used to the fact that people don't always play by the same definitions. This is why I have a spirit of using whatever definitions, and making sure I don't assume that someone I am talking to is using a certain definition that I am used to. But I suppose that since I am not an expert in any field of physics, maybe I should not be so carefree with using non-standard definitions myself. And about being homework helper here, if the other helpers think that I should stick to the standard definitions, then I will do that.
Last edited by a moderator: Aug 13, 2013
2. Aug 10, 2013
### Staff: Mentor
I disagree with this as you state it, because the difference between proper time and proper length is independent of coordinates, since the difference between timelike and spacelike curves is independent of coordinates. There may be cases where you don't need to care about the difference, but that doesn't make the difference artificial; it's there whether or not you define coordinates that reflect it, and whether or not you actually make use of it in a particular calculation.
It is possible to define arc length along a null curve; you just can't use proper time as the affine parameter to do so (the $\gamma$ in the equation you gave is an affine parameter, and it can be proper time or proper length, but it doesn't have to be). The arc length is indeed zero along a null curve, so you weren't wrong to say that.
If you're going to do that, you at least need to be explicit about what definition you're using up front, if it's a non-standard one. Otherwise other people are likely to assume you're using the standard definition, which can cause confusion, as in this thread.
It's not so much a matter of being an expert vs. not being an expert, as of trying to facilitate communication in general. Standard definitions of terms exist to help with that. That doesn't mean you *have* to use them, but it means it's easier to communicate if you do.
A more precise way of stating the point I was trying to make here is that, unless you have evidence to the contrary, it's safest to assume that textbooks in the field are using the standard definitions, so if a person who is studying from a textbook asks a question, they're probably used to seeing the standard definitions. But if you see evidence to the contrary, then, as you said, you should make sure to find out what definition they are actually using, to avoid confusion.
3. Aug 12, 2013
### BruceW
True, but only because we define matter to travel along timelike curves. For example, in a universe with radiation but no matter, then surely there is no way to discern which curves are 'timelike' and which curves are 'spacelike'. So then, I would say it is not a relativity concept, but a physics of matter concept. (Since really, it hinges on the assumption that all matter travels along timelike curves, which is experimentally true, but not required by relativity, as far as I know). But then maybe you would group that under a concept of relativity. I would prefer not to.
Oh, right. Thank you for explaining that. I wasn't sure about it. So in my preferred language: it is possible to define the arc length along a null curve, but you can't use the arc length as the affine parameter to do so.
Yeah, that's what I mean. Some of our lecturers would not explicitly tell us which definition they are using, or change notation without telling us (during the lecture), to keep us on our toes / make sure we were don't assume that the standard definition is being used as default, but to think about what the lecturer means. But I appreciate that on a forum, it is probably best to keep to the rule 'assume standard definition unless explicitly stated otherwise'. Also, sorry for derailing this thread a bit.
4. Aug 12, 2013
### Staff: Mentor
Not really; we discover that there are two fundamentally different kinds of objects, those with nonzero rest mass (that travel on timelike curves) and those with zero rest mass (that travel on null curves). Ordinarily we use the term "matter" for the first kind of object and "radiation" for the second. I suppose that counts as a "definition" of the term matter, but I don't think that's the kind of definition you meant.
Sure there is. You can use the null curves, i.e., curves with zero squared length, to define a coordinate grid and construct a metric, and you will find that there are other curves with positive or negative squared length, and the two kinds of curves are clearly distinguished from each other. Or, you could use the null curves to define light cones at each event, and you would find that there were other curves lying inside (timelike) or outside (spacelike) the light cones, with the light cones themselves providing a clear boundary between the two.
What's required by relativity is that objects with nonzero rest mass travel on timelike curves, and objects with zero rest mass travel on null curves. But we don't "assume" that "matter" is the first kind of object; we discover, experimentally, which objects fall into each category, and we adopt the term "matter" to refer to the objects that fall into the first category (nonzero rest mass).
Yes, you could put it that way: since arc length is zero along any null curve, obviously you can't use it to parameterize the curve.
5. Aug 12, 2013
### BruceW
Yes, but this is an arbitrary choice. For a specific physical universe, we still have the choice of what curves are spacelike and what curves are timelike. In other words, for a specific physical universe, if we look at a specific non-null geodesic, and ask 'is it time-like or space-like?' then the answer simply depends on our arbitrary choice.
edit: I wish I could explain better, but as I've said, I'm still a beginner to general relativity. Maybe start with the simple case of special relativity. ds^2 = -dt^2+dx^2+dy^2+dz^2 Now why did you choose 't' to get the negative sign? It is an arbitrary choice. You could instead give 'y' the negative sign, it is just as valid to do that. And this affects which curves are time-like and which curves are space-like.
another edit: the clue is in the names really. relativity puts space and time on an equal footing. The component we choose as 'time' (i.e. to get the negative sign) is an arbitrary choice. There is no non-arbitrary choice of 'space' and 'time', there is only spacetime.
Last edited: Aug 12, 2013
6. Aug 12, 2013
### DrGreg
But however you label your coordinates and whatever metric signature convention you follow, you'll find 3 coordinates with the same sign and one with the opposite sign. The one on its own is timelike and the other 3 are spacelike.
7. Aug 12, 2013
### Staff: Mentor
No, it doesn't. You can choose which sign convention you want to use for the metric, which determines whether the squared length of (timelike, spacelike) vectors is (negative, positive) or (positive, negative). But you can't choose whether a given curve is timelike or spacelike, because the metric only has one timelike dimension and three spacelike ones. The two are not symmetric.
That's just changing the names of the coordinates. It's not changing the physics.
No, it doesn't. It just changes which coordinate labels you put on particular curves. It doesn't change the physical nature of the curves.
Incorrect. See above.
8. Aug 12, 2013
### WannabeNewton
Given a space-time $(M,g_{ab})$, the class of space-like and time-like curves do not depend on any choice of coordinates or frames. Taking the $(-,+,+,+)$ sign convention, we say a curve $\gamma$ is time-like (resp. space-like) if $\xi^a \xi_a < 0$ (resp. $\xi^a \xi_a > 0$) everywhere on $\gamma$ where $\xi^a$ is the tangent vector field to $\gamma$ (in fact for geodesics it is enough to know the sign of $\xi^a \xi_a$ at a single point on the geodesic because $\xi^b \nabla_b (\xi_a \xi^a) = 2 \xi_a \xi^b \nabla_b \xi^a = 0$). This is purely geometric.
Last edited: Aug 12, 2013
9. Aug 12, 2013
### BruceW
I don't see why not. Let's say you have some curve in spacetime. You could define the timelike dimension to lie along that curve (in which case you'd label the curve as 'timelike'). Or you could define one of the spacelike dimensions to lie along the curve, in which case you'd label that curve as spacelike. the spacelike and timelike labels are arbitrary.
10. Aug 12, 2013
### WannabeNewton
No you can't do that. If the tangent vector to a curve satisfies $\xi^a \xi_a < 0$ everywhere on the curve then by definition it is time-like. You cannot change this, it's a geometric property of the curve and the only thing it depends on is the metric tensor.
11. Aug 12, 2013
### BruceW
when you say "no you can't do that" I'm guessing you mean "we have already defined the timelike direction". But it is the definition of the timelike direction that I am saying is arbitrary.
12. Aug 12, 2013
### WannabeNewton
But it isn't arbitrary. The tangent vector lies along the "time-like direction" if $\xi^a \xi_a < 0$ and along a "space-like direction" if $\xi^a \xi_a > 0$. What's arbitrary is the choice of sign convention $(-,+,+,+)$ or $(+,-,-,-)$ in which case the above signs flip but it won't change which curves are time-like and which are space-like.
13. Aug 12, 2013
### Staff: Mentor
You can't do that arbitrarily, because, as I said, the metric has one timelike and three spacelike dimensions.
For example: say I have coordinates $(t, x, y, z)$ set up, with the line element $ds^2 = - dt^2 + dx^2 + dy^2 + dz^2$. Consider the straight line $L1$ that goes from $(0, 0, 0, 0)$ to $(0, 1, 0, 0)$. Its squared length is 1.
Suppose I try to "define" the timelike dimension to lie along this curve $L1$ (i.e,. along the $x$ axis). If your claim is true, we should be able to do this. But can we? No; to see why, consider the following other curves all starting at the origin $(0, 0, 0, 0)$: the straight line $L2$ from the origin to $(0, 0, 1, 0)$, the straight line $L3$ from the origin to $(0, 0, 0, 1)$, and the straight line $L0$ from the origin to $(1, 0, 0, 0)$. The squared lengths of $L2$ and $L3$ are 1; but the squared length of $L0$ is -1.
All four of these lines are orthogonal, as is easily seen by observing that the dot product of any pair of them is zero. But the line $L1$, the one you are claiming can be "defined" to be the timelike dimension, has the *same* sign of its squared length as two other mutually orthogonal lines, $L2$ and $L3$. That can only be true of a spacelike line. The only one of the four lines that can possibly be timelike is $L0$. That's a physical fact, independent of whatever coordinates we choose to adopt.
14. Aug 12, 2013
### BruceW
@WannabeNewton- I agree that flipping the sign convention is not important. I'm saying that we can choose whichever component we want to be the timelike component, and this does change which curves are timelike and which are spacelike.
15. Aug 12, 2013
### BruceW
OK, now the line L1 has squared length -1
edit: and the others will all have squared length +1, under the new definition of the timelike dimension.
16. Aug 12, 2013
### Staff: Mentor
I suppose I should expand on this a bit, in view of the earlier exchange in this thread about deciding to put the minus sign on the $y$ coordinate instead of the $t$ coordinate. Suppose we tried something like that; suppose we said, let's put the minus sign on the $x$ coordinate, i.e., along the direction of the curve $L1$ that we want to define to be the timelike dimension. That is, we want to define the line element to be $ds^2 = dt^2 - dx^2 + dy^2 + dz^2$.
Can we do this? If we keep the coordinate labels of events constant, the answer is no, because it would change a geometric invariant, the squared length of $L1$; with this new line element, the squared length of $L1$ would be -1, not 1. Changing a geometric invariant changes the geometry.
We could, however, change coordinate labels: we could say that we are simply relabeling the coordinates so that $L1$ now goes from the origin to $(t, x, y, z) = (1, 0, 0, 0)$, and $L0$ now goes from the origin to $(0, 1, 0, 0)$. Then we have the same geometry with different coordinate labels on it; but the curve $L0$ is still the one timelike curve out of the four, so we haven't "defined" a different timelike dimension, we've just changed how we label the dimensions with coordinates.
17. Aug 12, 2013
### Staff: Mentor
See my follow-up post; you can't do that, because it's changing the geometry.
18. Aug 12, 2013
### BruceW
I agree that once we have defined a timelike dimension, then it is clear which are the timelike curves and which are the spacelike curves. The point is that our choice of the timelike dimension is arbitrary in the first place. OK, you can call it 'geometry', but it is still arbitrarily defined geometry. It does not correspond to anything physical.
19. Aug 12, 2013
### Staff: Mentor
I don't understand what you mean here. If you mean we can choose which coordinate label to put on the timelike dimension, sure, nobody is questioning that. But if that's all you've been saying, it hasn't been at all clear.
If you mean something else, then you'll need to explain in more detail what you mean, because I'm not getting it.
20. Aug 12, 2013
### Staff: Mentor
Perhaps this will help clarify where you're coming from: are you saying that I can make an arbitrary choice of whether to call the dimension that points into my future (for example, from the event of me right now to the event of me a minute from now) timelike or spacelike? | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8707714080810547, "perplexity": 389.76739315329206}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00475.warc.gz"} |
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Abstract, references, and article information View Article: PDF This article is available free of charge [56] Jacek Dziubanski. Triebel-Lizorkin spaces associated with Laguerre and Hermite expansions. Proc. Amer. Math. Soc. 125 (1997) 3547-3554. MR 1403122. Abstract, references, and article information View Article: PDF This article is available free of charge [57] Yuan Xu. Integration of the intertwining operator for $h$-harmonic polynomials associated to reflection groups. Proc. Amer. Math. Soc. 125 (1997) 2963-2973. MR 1402890. Abstract, references, and article information View Article: PDF This article is available free of charge [58] J. Michael Wilson. A two-parameter Bergman space'' inequality. Proc. Amer. Math. Soc. 125 (1997) 755-762. MR 1415376. Abstract, references, and article information View Article: PDF This article is available free of charge [59] Wo-Sang Young. Almost everywhere convergence of lacunary partial sums of Vilenkin-Fourier series. Proc. Amer. Math. 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https://robotics.stackexchange.com/questions/13947/generating-double-s-curve-velocity-profiles-with-given-time/13950 | # Generating double s-curve velocity profiles with given time
I wonder how to generate double s-curve velocity profile for multiple DOF trajectory. Since there are constraints on initial and final velocities which can be non-zero it is necessary to synchronize each DOF in time. Therefore firstly I would like to compute trajectory for DOF with the largest displacement and then trying to fit other DOFs in the computed execution time for the former. However I was not able to find anything about generating s-curve profile with given time. Having tried to solve it by myself I came up with a conviction that it is an optimization problem. I tried several approaches but they all seemed to have non-convex cost function and hardly could they satisfy constraints on final velocity. Having spent much time I wondered if there is an easy way to synchronize them?
## 2 Answers
Yes, this is indeed a difficult non-convex optimization problem. But if, say, second-order polynomial (parabolic) trajectories work for you, then there are some heuristics that you might want to look into.
For parabolic trajectories, the procedure for computing the time-optimal (1D) trajectory exists (see e.g., this paper). So first you can compute 1D velocity profiles for all DOFs and see which DOF takes the most time. Suppose DOF $i$ takes time $T$ to reach its destination, you need to synchronize other DOFs with time $t =T$, because DOF $i$ cannot go any faster.
Then the next step is kind of velocity profile stretching, i.e., for each DOF $j \neq i$ we stretch the velocity profile such that it has duration $t$ as specified earlier. For this steps, many heuristics can be used. One of them is described in the paper I mentioned above (this method is somewhat constrained due to some assumptions made in the paper). Another one is implemented in OpenRAVE, see this file (the detail of the method implemented was given alongside of the code, this method is less constrained than the first).
Note that given a duration $t$ to synchronize a trajectory with, the problem is not always feasible due to inoperative time interval (this paper mentioned this topic a bit). Basically, due to velocity and acceleration limits of the robot joints, given a set of boundary conditions (such as initial and final velocities), there might exists some time interval $\tau = [t_0, t_1]$ such that there exists no trajectory which satisfies the boundary conditions and has a duration $t \in \tau$.
• Thanks! However I already have solution for trapezoidal profiles which has little improvements of parabolic trajectories as I can see. Unfortunately that is not enough for me. – Long Smith Aug 1 '17 at 10:55
I have found a solution. Quiet simple actually. To generate minimum time double s-curve profile I used algorithm presented in this wonderful book. After computing trajectory for the first DOF I perform the same algorithm for the other DOFs but if trajectory is infeasible with given time, it is appeared to be a good idea to make maximum possible acceleration smaller and smaller with some factor(I use 0.95) which leads to longer trajectory execution. Moreover we can do the same with maximum velocity and jerk constraints which will make more trajectories to be feasible. However this will take significantly more time and since all my tests passed with only decreasing acceleration I am satisfied. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9051693081855774, "perplexity": 515.5933717954296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496672170.93/warc/CC-MAIN-20191122222322-20191123011322-00532.warc.gz"} |
https://www.physicsforums.com/threads/new-physics-in-lhc-data.955077/ | # New Physics in LHC data
• I
• Start date
• #1
140
3
Hello! The attached pic is from this paper. So on the y you have counts and on the x you have a certain parameter used to search for a certain resonance. The colored part is the Standard Model (SM) prediction and the black dots are the obtained data while the dashed lines are the simulated prediction for the contribution of this new resonance. Obviously no deviation is observed in the region of interest. However if you look in the last 5-7 bins (maybe except the last 2), especially in the bottom plot, the data points deviate significantly from the predicted background. Is it because the simulation didn't focus much there (as it is outside the region of interest for the specific resonance) or it can be something deeper (the points seems to be quite a few sigma below the predicted model)? If the simulations are correct over the whole presented range, shouldn't these points be closer (within the error bars) to the predicted model? Thank you!
#### Attachments
• CMS.png
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• #2
mfb
Mentor
35,348
11,680
Probably an issue with the theoretical prediction for the QCD contribution. If you would have systematic uncertainties shown in this plot they would probably cover the observed deviation.
• #3
140
3
Probably an issue with the theoretical prediction for the QCD contribution. If you would have systematic uncertainties shown in this plot they would probably cover the observed deviation.
Thank you for your reply! I am still a bit confused. An issue with the theoretical prediction doesn't mean there is something we don't understand about that particular region of phase space?
• #4
mfb
Mentor
35,348
11,680
It means QCD is complicated. Predictions often have to rely on some assumptions, approximations, experimental data as input and so on. I'm more surprised that most of the region is so well-described. Some deviation at the edge of the phase space with such a steep distribution is perfectly plausible.
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https://www.physicsforums.com/threads/an-easy-question-about-waves.306803/ | • Start date
• #1
36
0
Homework Statement
Hello there,
There is a question that says :
....etc....the metal is then moved towards the hardboard. In moving 6.4cm, four further maxima are observed. Calculate the wavelength of the ........etc........
Now I know how to solve it, and got the answer.
(sine curve)
But why do we count the amplitudes both below in the and above the axis as maxima ?
I thought the maximum points are all the points above the x axis, and the minimum points are the minimum points ??!
The Attempt at a Solution
All I did was : 4 max = 2$$\lambda$$ .. But these 4 maximum points are both ABOVE and BELOW the x axis for a sine curve.
Then 2$$\lambda$$=6.4cm
$$\lambda$$=3.2cm
If you need me to explain my question exactly, please ask me to clarify.
Related Introductory Physics Homework Help News on Phys.org
• #2
140
0
Hi ZaZu,
I'm not 100% of what you're asking so if i'm telling you something you already know - forgive me.
I presume the experiment involves counting how many maxima OR minima of a standing wave propagating between two fixed points, the distance between which you know already.
In this case you're really only after counting either the number of consecutive maxima or minima, not both, for this is the definition of wavelength.
• #3
36
0
Hi ZaZu,
I'm not 100% of what you're asking so if i'm telling you something you already know - forgive me.
I presume the experiment involves counting how many maxima OR minima of a standing wave propagating between two fixed points, the distance between which you know already.
In this case you're really only after counting either the number of consecutive maxima or minima, not both, for this is the definition of wavelength.
Yes thats exactly what I mean, but in our class we did the following :
http://img404.imageshack.us/img404/2170/image352.th.jpg [Broken]
Is this correct ?
Last edited by a moderator:
• #4
Hootenanny
Staff Emeritus
Gold Member
9,622
6
The amplitude is the magnitude of the displacement from the equilibrium position of the oscillating variable. It doesn't matter whether this displacement is positive or negative since the amplitude is the magnitude of the displacement.
For example, consider the function y = sin(x). A maximum value of y occurs as sin(pi/2) = 1 and a minimum value of y occurs at sin(-pi/2) = -1. However, in both cases the amplitude is 1, since A = |y|. Since the amplitude, by definition is non-negative, it's minimum value is clearly the minimum of y > 0.
Do you follow?
• #5
140
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Yes it is.
Note the 2 on the RHS corresponding to 2 consecutive peaks/troughs of maxima OR minima.
• #6
36
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The amplitude is the magnitude of the displacement from the equilibrium position of the oscillating variable. It doesn't matter whether this displacement is positive or negative since the amplitude is the magnitude of the displacement.
For example, consider the function y = sin(x). A maximum value of y occurs as sin(pi/2) = 1 and a minimum value of y occurs at sin(-pi/2) = -1. However, in both cases the amplitude is 1, since A = |y|. Since the amplitude, by definition is non-negative, it's minimum value is clearly the minimum of y > 0.
Do you follow?
So in my question, I can say that both the minimum AND maximum points are the MAXIMA ??
• #7
36
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Yes it is.
Note the 2 on the RHS corresponding to 2 consecutive peaks/troughs of maxima OR minima.
Oh alright, so its concluded that both the crests and troughs can be considered the maximum points ?
• #8
140
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Careful here. They correspond to points of maximum amplitude but in regards to their physical positions, they must be differentiated (i.e. by the use of maxima/minima).
• #9
36
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Careful here. They correspond to points of maximum amplitude but in regards to their physical positions, they must be differentiated (i.e. by the use of maxima/minima).
Oh I see, great !
Its clearer now :)
Thanks alot astrorob and Hootenanny :D :D :D
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1K | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.948880136013031, "perplexity": 800.4763081927848}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145713.39/warc/CC-MAIN-20200222180557-20200222210557-00493.warc.gz"} |
https://www.targetmathematics.com/2020/06/g10-chap5-exercise-55-solving-linear-quadratic-system.html | # Exercise (5.5) : Solving Linear Quadratic Systems Algebraically
1. Find the solution set of each of the systems of equations:
(a) $x^2-y^2=9$
$x+y=1$
Show/Hide Solution
$\begin{array}{l}{{x}^{2}}-{{y}^{2}}=9\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=1\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=1-x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting }y=1-x\text{ in equation }(1)\text{,}\\\\{{x}^{2}}-{{\left( {1-x} \right)}^{2}}=9\\\\{{x}^{2}}-\left( {1-2x+{{x}^{2}}} \right)=9\\\\{{x}^{2}}-1+2x-{{x}^{2}}=9\\\\2x=10\\\\x=5\\\\\text{Substituting }x=5\text{ in equation }(3),\\\\y=1-5=-4\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {5,-4} \right)} \right\}\text{ }\end{array}$
(b) $y=\displaystyle \frac{8}{x}$
$y=7+x$
Show/Hide Solution
$\begin{array}{l}y=\displaystyle \frac{8}{x}\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\y=7+x\ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\\displaystyle \frac{8}{x}=7+x\\\\\therefore \ {{x}^{2}}+7x=8\\\\{{x}^{2}}+7x-8=0\\\\(x+8)(x-1)=0\\\\x=-8\ \text{or}\ x=1\\\\\text{When}\ x=-8,\ y=7-8=-1\\\\\text{When}\ x=1,\ y=7+1=8\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-8,-1} \right),\left( {1,8} \right)} \right\}\text{ }\end{array}$
(c) $x^2+5x+y=4$
$x+y=8$
Show/Hide Solution
$\begin{array}{l}{{x}^{2}}+5x+y=4\cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\\\\x+y=8\ \ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=8-x\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=8-x\text{ in equation }(1),\\\\{{x}^{2}}+5x+8-x=4\\\\{{x}^{2}}+4x+4=0\\\\{{(x+2)}^{2}}=0\\\\x=-2\\\\\text{Substituting}\ x=-2\text{ in equation }(3),\\\\y=8-(-2)=10\\\\\therefore \ \ \text{Solution set}=\left\{ {\left( {-2,10} \right)} \right\}\text{ }\end{array}$
2. The sum of squares of two numbers is 58. If the first number and twice the second add up to 13, find the numbers.
Show/Hide Solution
$\begin{array}{l}\text{Let}\ \text{the two numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\{{x}^{2}}+{{y}^{2}}=58\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\x+2y=13\ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\x=13-2y\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ x=13-2y\text{ in equation }(1),\\\\{{\left( {13-2y} \right)}^{2}}+{{y}^{2}}=58\\\\169-52y+4{{y}^{2}}+{{y}^{2}}=58\\\\169-52y+5{{y}^{2}}=58\\\\\therefore \ \ 5{{y}^{2}}-52y+111=0\\\\(5y-37)(y-3)=0\\\\y=\displaystyle \frac{{37}}{5}\ \text{or}\ y=3\\\\\text{When}\ y=\displaystyle \frac{{37}}{5},x=13-2\left( {\displaystyle \frac{{37}}{5}} \right)\text{=}-\displaystyle \frac{9}{5}\text{ }\\\\\text{When}\ y=3,x=13-2\left( 3 \right)\text{=}7\text{ }\\\\\therefore \ \ \text{The two numbers are }-\displaystyle \frac{9}{5}\ \text{and }\displaystyle \frac{{37}}{5}\ \text{or 7 and 3}\text{.}\end{array}$
3. The sum of the reciprocals of two positive numbers is $\displaystyle \frac{7}{36}$ and the product of the numbers is 108. Find the numbers.
Show/Hide Solution
$\begin{array}{l}\text{Let}\ \text{the two positive numbers be }x\ \text{and }y.\\\\\text{By the problem,}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{y}=\displaystyle \frac{7}{{36}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \ (1)\\\\xy=108\ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\\\\\text{From equation }(2)\text{,we have}\\\\y=\displaystyle \frac{{108}}{x}\ \ \ \ \ \ \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3)\\\\\text{Substituting}\ y=\displaystyle \frac{{108}}{x}\text{ in equation }(1),\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{1}{{\displaystyle \frac{{108}}{x}}}=\displaystyle \frac{7}{{36}}\\\\\displaystyle \frac{1}{x}+\displaystyle \frac{x}{{108}}=\displaystyle \frac{7}{{36}}\\\\\text{Multiplying both sides of equation by}\ 108x,\\\\108+{{x}^{2}}=21x\\\\\therefore \ \ {{x}^{2}}+21x-108=0\\\\(x-9)(x-12)=0\\\\x=9\ \text{or}\ x=12\\\\\text{When}\ x=9,y=\displaystyle \frac{{108}}{9}=12\\\\\text{When}\ x=12,y=\displaystyle \frac{{108}}{{12}}=9\\\\\therefore \ \ \text{The two positive numbers are }9\text{ and 12}\text{.}\end{array}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.832729697227478, "perplexity": 2206.640822084416}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141191692.20/warc/CC-MAIN-20201127103102-20201127133102-00269.warc.gz"} |
https://pureportal.strath.ac.uk/en/publications/preliminary-investigations-of-the-low-velocity-impact-response-of | # Preliminary investigations of the low-velocity impact response of a smart trimorph plate for active damage mitigation
A. Big-Alabo, P. Harrison, M.P. Cartmell
Research output: Contribution to conferenceOtherpeer-review
1 Citation (Scopus)
## Abstract
The elastoplastic impact response of a trimorph plate subjected to low-velocity heavy mass impact has been investigated using analytical models. In formulating the impact model, the displacement of the impactor, vibration of the plate and local contact mechanics were accounted for. The vibration of the trimorph plate was modelled using the classical laminate plate theory, while the local contact mechanics was modelled using a Meyer-type compliance model that accounts for post-yield effects in the loading and unloading stages of the impact. The impact model is a set of coupled nonlinear differential equations and was solved using the NDSolve function in Mathematica™. Both the modelling and solution approach were validated using a benchmark case study. Investigations were carried out for an Al/PVDF/PZT trimorph plate configuration. Contrary to the general position in the literature that the response of a large mass impact is insensitive to the compliance model used to estimate the impact force [1], the simulations of the present impact model show that the indentation history was sensitive to the compliance model used. Also, the elastoplastic compliance model predicted a permanent indentation which the elastic compliance model cannot estimate. Therefore, the use of elastoplastic compliance models in large mass impact analysis is imperative. Finally, the application of a smart trimorph plate with piezoelectric actuator layer for active mitigation of impact damage was discussed.
Original language English 2876-2889 Published - 2015 5th International Conference on Computational Methods in Structural Dynamics and Earthquake Engineering - Crete Island, GreeceDuration: 25 May 2015 → 27 May 2015
### Conference
Conference 5th International Conference on Computational Methods in Structural Dynamics and Earthquake Engineering Greece 25/05/15 → 27/05/15
## Keywords
• investigations
• low-velocity impact
• smart trimorph plate
• active damage mitigation
## Fingerprint
Dive into the research topics of 'Preliminary investigations of the low-velocity impact response of a smart trimorph plate for active damage mitigation'. Together they form a unique fingerprint. | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8336039185523987, "perplexity": 3434.5861805539075}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103922377.50/warc/CC-MAIN-20220701064920-20220701094920-00071.warc.gz"} |
http://mathhelpforum.com/math-challenge-problems/84209-simplify-function.html | # Math Help - Simplify this function
1. ## Simplify this function
Should be a good brain refresher!
$
{4-4\cos^2(x) \cdot \left(1-\frac{1}{\left(\csc^2(x)\right)}\right)^2}+(cot^2( 2x))\frac{(1-2\sin^2(x))^2}{\frac{1}{\frac{1+\cos(2x)}{\cos(2x) }\cdot \frac{1-\cos(2x)}{\cos(2x)}}}
$
The answer should come out to a really nice answer if I did my math right........ After a lot of editing, I hope this comes to the right answer (many mistakes)
THIS IS THE FINALIZED AND EDITED PROBLEM.
2. Hi,
I'm rewriting it in correct latex, because the way you wrote it is not easy to decipher
Tell me if it's correct.
$\frac{1}{\sqrt{4-4\cos^2(x)} \cdot \left(1-\left(\frac{2}{\sec^2(x)}\right)^2\right)}+\frac{1-2\sin^2(2x)}{\frac{1+\cos(2x)}{\cos(2x)}}\cdot \frac{1-\cos(2x)}{\cos(2x)}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9375767111778259, "perplexity": 1607.8917771154604}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1413507442420.22/warc/CC-MAIN-20141017005722-00093-ip-10-16-133-185.ec2.internal.warc.gz"} |
http://mathhelpforum.com/advanced-statistics/222083-finding-cdf-print.html | finding the CDF
• September 19th 2013, 01:09 AM
99.95
1 Attachment(s)
finding the CDF
Please refer to the attached image.
For part a)
when I want to find the CDF, don't I simply take the indefinite integral of e^-|x|, multiply it by c and solve for that = 1?
I am unsure of how to take the integral for this, am i correct in saying it is -e^-x, for all x ?
that would leave me with c* -e^-x = 1, and c = 1/(-e^-x), wolfram says there are two separate results, but i am not sure why. and in that case, I would also have two separate results for c. how can that make sense?
for part b) i am unsure as to how to approach the question. could someone please guide me?
As always, your help is very much appreciated and invaluable.
Thanks
• September 19th 2013, 02:46 AM
Shakarri
Re: finding the CDF
When integrating absolute values you must split up the integration into two parts, the first part where the expression inside the absolute value is always non negative and the second part where the expression inside the absolute value is always non positive.
For example g(x)= |x+2| Find $\int_{-10}^5g(x)dx$
$\int_{-10}^5|x+2|dx=\int_{-10}^{-2}|x+2|dx+\int_{-2}^5|x+2|dx$
$\int_{-10}^{-2}|x+2|dx+\int_{-2}^5|x+2|dx=\int_{-10}^{-2}-(x+2)dx+\int_{-2}^5(x+2)dx$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8702243566513062, "perplexity": 330.8163216001619}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246637445.19/warc/CC-MAIN-20150417045717-00046-ip-10-235-10-82.ec2.internal.warc.gz"} |
https://www.physicsforums.com/threads/analyzing-the-graphs-of-greatest-integer-functions.949820/ | # Analyzing the graphs of Greatest Integer Functions
• Start date
• #1
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## Homework Statement
Consider ##u\left(x\right)=2\left[\frac{-x}{4}\right]##
(a) Find the length of the individual line segments of the function,
(b) Find the positive vertical separation between line segments.
## Homework Equations
The output of Greatest Integer Functions are always integers.
## The Attempt at a Solution
Length:
The text states that the coefficient of x within the greatest integer symbols is the length of the individual line segments of the graph.
In ##u\left(x\right)=2\left[\frac{-x}{4}\right]##, the coefficient of x is ##\frac{-1}{4}##.
However, the solution for the length of the graph states that length=4.
It explains this by stating that there's a decrease of 1 for every increase of 4 in the variable x.
This would make sense if we were talking about the slope of a line, but it doesn't make any sense at all in this context.
And since we're talking about the length of a line segment, does the negation matter?
Vertical Separation:
The text states that the coefficient of the greatest integer function is the positive vertical separation between line segments.
This is a straight forward statement, and the vertical separation=2, but I don't see why this leading coefficient determines this.
Can anyone help me get a better idea of what is going on with the graphs of these functions?
• #2
andrewkirk
Homework Helper
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I haven't seen the notation you are using but I presume that the square brackets [...] denote the function that gives the greatest integer that is less than or equal to the value of the contents of the brackets. In my experience that is usually indicated by ##\lfloor...\rfloor## and is called the 'floor' function.
If that is the case then the function value is piecewise constant and steps down at each multiple of 4, ie at ....,-8, -4, 0, 4, 8, ....
So each line segment is horizontal and extends for the period that x takes to increase by 4. So its length is 4.
How much does it step down by each time? Well ##\lfloor\frac{-x}4\rfloor## always has an integer value and decreases by 1 each time ##x## reaches a new, higher multiple of 4. So that's a step size of 1, but then it is multiplied by 2 - the leading coefficient - so the step size (the vertical separation) is 2.
The impact of the negation is to make the steps go down as ##x## increases, rather than up. But it doesn't affect the step length or height.
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• #3
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Yes we are talking about the same function. My text has the it denoted as something similar to but not exactly like [[x]]. But I looked up on Wikipedia the "floor function" that you stated and we're talking about the same thing.
That was a great explanation, thank you.
Let me see if I understand:
The length is 4 because if we were to create a table of values (leaving the leading coefficient of 2 out for simplicity's sake), it would take 4 integer x inputs for the line segment to be complete and move up or down to the next segment? For example, with an input of x = -8 gives an output of 2, and inputs x = -7, -6, -5, -4 all give an output of 1. Then when x = -3, the output is zero so there's a new line segment vertically shifted.
For the vertical displacement: the greatest integer function output is always an integer. So when we take the output of the function's argument, and multiply it by 2, we are vertically shifting by that value 2.
• #4
andrewkirk
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Yes that is correct!
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• #5
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Thank you for the great explanation!
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3K | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8824185729026794, "perplexity": 497.94744534617615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00543.warc.gz"} |
https://homework.cpm.org/category/CON_FOUND/textbook/caac/chapter/14/lesson/14.3.1.1/problem/3-15 | ### Home > CAAC > Chapter 14 > Lesson 14.3.1.1 > Problem3-15
3-15.
Sketch the shape of the graph of the function $y = b^{x}$ given each of the following values of $b$
1. $b$ is a number larger than $1$.
2. $b$ is a number between $0$ and $1$.
3. $b$ is equal to $1$.
Use the eTool below to solve the parts of the problem.
Click the link at right for the full version of the eTool: 7-14 HW eTool | {"extraction_info": {"found_math": true, "script_math_tex": 9, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8599858283996582, "perplexity": 1162.1776014697512}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00184.warc.gz"} |
http://www.ck12.org/earth-science/Earths-Core/rwa/Discovering-the-Core/r2/ | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Earth's Core
## There is an inner and an outer core made of dense metal; it is known from seismic waves and density calculations.
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Discovering the Core
### Discovering the Core
Credit: Washiucho
Source: http://en.wikipedia.org/wiki/File:Earth-crust-cutaway-English-Large_label.PNG
We may know more about the Moon than Earth’s core, but is that surprising? We can see the Moon and have even visited the far side. It’s a lot harder to learn about the center of the Earth. We can’t go there and we can’t drill there. How can we learn about the core?
#### Amazing But True!
Credit: USGS
Source: http://earthquake.usgs.gov/learn/topics/seismology/keeping_track.php
Travel-time curves on seismograms [Figure2]
• Richard Oldham was the first person to notice the separate arrival times of P-waves, S-waves and surface waves on seismograms.
• From his mathematical analysis he determined that Earth had a core equal to less than 40% its radius.
• Inge Lehman was the first to discover the P wave shadow, indicating the presence of an inner core.
• Lehman later discovered a change in seismic wave velocity in the mantle, which was called the Lehman discontinuity.
#### Explore More
1. Oldham noticed that seismic waves appeared to hit an obstacle, which was the core. What do the P-waves and S-waves do at the core?
2. What did Lehman observe that indicated to her that there were two cores?
3. What are the two main differences between the inner core and the outer core?
4. Why is it important that the outer core is liquid?
### Notes/Highlights Having trouble? Report an issue.
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http://mathhelpforum.com/differential-geometry/107479-norm-uniform-convergence.html | # Math Help - Norm and Uniform Convergence
1. ## Norm and Uniform Convergence
Any hints on how to approach these two questions?
Thanks a lot
2. For a) just observe that A is compact.
b) is immediate from the very definition. Uniform convergence to f certainly is equivalent to the convergence to 0 of
$\{\sup\|f_k(x)-f(x)\|: x\in A\}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9808127880096436, "perplexity": 1371.6568712089845}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-32/segments/1438042989897.84/warc/CC-MAIN-20150728002309-00232-ip-10-236-191-2.ec2.internal.warc.gz"} |
http://www.ma.utexas.edu/mp_arc-bin/mpa?yn=00-436 | 00-436 Palle E. T. Jorgensen
Minimality of the data in wavelet filters (2091K, LaTeX2e amsart class; 69 pages, 2 tables, 4 figures, 12 pages of plots (total 162 EPS graphics)) Nov 7, 00
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. Orthogonal wavelets, or wavelet frames, for $L^{2}\left( \mathbb{R}\right)$ are associated with quadrature mirror filters (QMF), a set of complex numbers which relate the dyadic scaling of functions on $\mathbb{R}$ to the $\mathbb{Z}$-translates. In this paper, we show that generically, the data in the QMF-systems of wavelets is minimal, in the sense that it cannot be nontrivially reduced. The minimality property is given a geometric formulation in the Hilbert space $\ell^{2}\left( \mathbb{Z}\right)$, and it is then shown that minimality corresponds to irreducibility of a wavelet representation of the algebra $\mathcal{O}_{2}$; and so our result is that this family of representations of $\mathcal{O}_{2}$ on the Hilbert space $\ell^{2}\left( \mathbb{Z}\right)$ is irreducible for a generic set of values of the parameters which label the wavelet representations.
Files: 00-436.src( 00-436.comments , 00-436.keywords , datawafi.tex , datawafi-graphics.tar.gz.mm ) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8262283802032471, "perplexity": 1320.3983093460615}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816138.91/warc/CC-MAIN-20180225051024-20180225071024-00099.warc.gz"} |
https://socratic.org/questions/what-fraction-is-halfway-between-1-3-and-1-5 | Algebra
Topics
# What fraction is halfway between 1/3 and 1/5?
Sep 11, 2015
$\frac{4}{15}$
#### Explanation:
A trick question: The obvious, but wrong answer is $\frac{1}{4}$.
Actually $\frac{1}{4}$ is called the harmonic mean of $\frac{1}{3}$ and $\frac{1}{5}$ - being the reciprocal of the average of the reciprocals.
To find the number halfway between $a$ and $b$, calculate $\frac{1}{2} \left(a + b\right)$ ...
$\frac{1}{2} \left(\frac{1}{3} + \frac{1}{5}\right) = \frac{1}{2} \left(\frac{5}{15} + \frac{3}{15}\right) = \frac{1}{2} \left(\frac{8}{15}\right) = \frac{4}{15}$
Note that to add the fractions $\frac{1}{3}$ and $\frac{1}{5}$ we first give them a common denominator $15$ by multiplying them by $\frac{5}{5}$ and $\frac{3}{3}$ respectively. Once we have a common denominator, then we can just add the numerators.
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http://math.stackexchange.com/questions/258635/how-to-solve-2x-12x-2x-3x-4-12/258966 | # How to solve $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$
How many solutions possible for the equation$$2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$$ all x are non-negative integer.
I see these links but I don't know how to solve this problem.(I know how to solve ${x_1}+{x_2}+{x_3}+{x_4}=12$)
How many solutions possible for the equation $x_1+x_2+x_3+x_4+x_5=55$ if
Enumerating number of solutions to an equation
-
Summing up and case-working is inelegant but I think there is no other way. – barto Dec 14 '12 at 12:12
Solve for $2a+b=12$. For each one of the $7$ solutions $(a,b)=(k_1,k_2)$, solve for possibilities which count $(k_1+1)\times(k_2+1)$.
Eg. A solution to $2a+b=12$ is $(a,b)=(5,2)$.
So, $x_1+x_2=5$ while $x_3+x_4=2$. Within themselves;
$(x_1,x_2)=(0,5)$,
$(x_1,x_2)=(1,4)$,
...
$(x_1,x_2)=(5,0)$
while
$(x_3,x_4)=(0,2)$,
$(x_3,x_4)=(1,1)$,
$(x_3,x_4)=(2,0)$
adding up to $(5+1)\times(2+1)=18$ solutions for $(a,b)=(5,2)$.
-
+28-20 for (3,6). Should add up to 140. – ashley Dec 15 '12 at 9:05
we have $(0,12)(1,10)(2,8)(3,6)(4,4)(5,2)(6,0)$ for 2a+b=12 and final answer is 13+22+27+28+25+18+7=140. Thanks. – Sali Me Dec 15 '12 at 9:19
Solve $x_{1}+x_{2}+x_{3}+x_{4}=12-x_{1}-x_{2}$ for all natural $x_{1},x_{2}$ s.t $x_{1}+x_{2}\leq6$ s.t and sum the results.
Can you explain the statement $12-x_1-x_2$ and $x_1+x_2 \le 6$ a bit more? and continue more? – Sali Me Dec 14 '12 at 12:21
If $x_1+x_2\geq 6$ then $2x_1+2x_2>12$ and since $x_3,x_4$ are natural there are no solutions to the equations. Fix any pair $x_1,x_2$ s.t their sum is $\leq 6$ (hence there are solutions) and find how many $x_3,x_4$ there are that solve the equation. Note that for such fixed $x_1,x_2$ you want to count the number of solutions for $x_3+x_4=12-2x_1-2x_2$ which you already know to solve – Belgi Dec 14 '12 at 12:27
It seems you've somehow added an $x_5$ that isn't in the question. – joriki Dec 14 '12 at 12:29 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9266263842582703, "perplexity": 504.06347035517376}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011188282/warc/CC-MAIN-20140305091948-00058-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://planetmath.org/CubeOfAnInteger | # cube of an integer
Any cube of integer is a difference of two squares, which in the case of a positive cube are the squares of two successive triangular numbers.
For proving the assertion, one needs only to check the identity
$a^{3}\;\equiv\;\left(\frac{(a\!+\!1)a}{2}\right)^{\!2}-\left(\frac{(a\!-\!1)a}% {2}\right)^{\!2}.$
For example we have $(-2)^{3}=1^{2}\!-\!3^{2}$ and $4^{3}=64=10^{2}\!-\!6^{2}$.
Summing the first $n$ positive cubes, the identity allows http://planetmath.org/encyclopedia/TelescopingSum.htmltelescoping between consecutive brackets,
$\displaystyle 1^{3}\!+\!2^{3}\!+\!3^{3}\!+\!4^{3}\!+\ldots+\!n^{3}$ $\displaystyle\;=\;[1^{2}\!-\!0^{2}]\!+\![3^{2}\!-\!1^{2}]\!+\![6^{2}\!-\!3^{2}% ]\!+\![10^{2}\!-\!6^{2}]\!+\ldots+\!\left[\!\left(\frac{(n\!+\!1)n}{2}\right)^% {\!2}\!-\!\left(\frac{(n\!-\!1)n}{2}\right)^{\!2}\right]$ $\displaystyle\;=\;\left(\frac{n^{2}\!+\!n}{2}\right)^{\!2},$
saving only the square $\left(\frac{(n+1)n}{2}\right)^{2}$. Thus we have this expression presenting the sum of the first $n$ positive cubes (cf. the Nicomachus theorem).
Title cube of an integer CubeOfAnInteger 2013-03-22 19:34:33 2013-03-22 19:34:33 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 11B37 msc 11A25 NicomachusTheorem TriangularNumbers DifferenceOfSquares | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 9, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9212085604667664, "perplexity": 3747.9644241190244}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039744750.80/warc/CC-MAIN-20181118221818-20181119003818-00444.warc.gz"} |
http://mathhelpforum.com/calculus/31839-calculus-differentiation-print.html | Calculus differentiation
• March 23rd 2008, 06:26 PM
Gorie
Calculus differentiation
f(x) = 3 + the integral from 8 to x cubed of f(cubed root of t) dt
• March 23rd 2008, 06:33 PM
mr fantastic
Quote:
Originally Posted by Gorie
f(x) = 3 + the integral from 8 to x cubed of f(cubed root of t) dt
$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$
Let $u = x^3$ and use the Fundamental Theorem of Calculus and the chain rule:
$= \frac{d}{du} \left[ \int_{8}^{u} f(t^{1/3}) \, dt\right] \, \times \, \frac{du}{dx}$
$= f(u^{1/3}) \, \times \, 3x^2$
$= f(x) \, \times \, 3x^2 = 3x^2 \, f(x)$.
• March 23rd 2008, 06:37 PM
mr fantastic
Quote:
Originally Posted by mr fantastic
$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$
Let $u = x^3$ and use the Fundamental Theorem of Calculus and the chain rule:
$= \frac{d}{du} \left[ \int_{8}^{u} f(t^{1/3}) \, dt\right] \, \times \, \frac{du}{dx}$
$= f(u^{1/3}) \, \times \, 3x^2$
$= f(x) \, \times \, 3x^2 = 3x^2 \, f(x)$.
$\frac{d}{dx} \left[ \int_{8}^{x^3} f(t^{1/3}) \, dt\right]$
Alternatively, make the substitution $t^{1/3} = u \Rightarrow t = u^3$. Then you have
$\frac{d}{dx} \left[ \int_{8}^{x} f(u) \, 3u^2 \, du\right]$
Now apply the Fundamental Theorem of Calculus:
$= f(x) \, 3x^2$. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9894310235977173, "perplexity": 1308.5130426010426}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657120057.96/warc/CC-MAIN-20140914011200-00267-ip-10-196-40-205.us-west-1.compute.internal.warc.gz"} |
http://mathoverflow.net/revisions/118821/list | The integral obviously doesn't depend on $p$, since it must be rotationally invariant, and, using polar coordinates centered on $p$, one obtains $$I(p,\alpha) = \mathrm{vol}(S^{n-1})\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(n-2)/2}dtt}(1{-}t^2)^{(n-2)/2}dt =\frac{2\ \pi^{n/2}}{\Gamma(n/2)}\int_{-1}^1 \cos(\alpha t)(1{-}t^2)^{(n-2)/2}dt.$$ This latter integral can be evaluated by standard techniques, but, for . For odd values of $n$, it is expressed in Bessel functions, while, for even values of $n$ it is expressed in terms of elementary functions. Thus, for $n=2$, one has $$\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(2-2)/2}dt \cos(\alpha t)(1{-}t^2)^{(2-2)/2}dt = \frac{2\sin\alpha}{\alpha},$$ while, for $n=3$, one has $$\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(3-2)/2}dt \cos(\alpha t)(1{-}t^2)^{(3-2)/2}dt = \frac{\pi\ \mathrm{BesselJ}(1,\alpha)}{\alpha}, J_1(\alpha)}{\alpha},$$ where $J_1$ denotes the Bessel function of the first kind, and so on.
The integral obviously doesn't depend on $p$, since it must be rotationally invariant, and, using polar coordinates , you get that centered on $p$, one obtains $$I(p,\alpha) = \mathrm{vol}(S^{n-1})\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(n-2)/2}dt.$$ This latter integral can be evaluated by standard techniques, but, for odd values of $n$, it is expressed in Bessel functions, while, for even values of $n$ it is expressed in terms of elementary functions. Thus, for $n=2$, one has $$\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(2-2)/2}dt = \frac{2\sin\alpha}{\alpha},$$ while, for $n=3$, one has $$\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(3-2)/2}dt = \frac{\pi\ \mathrm{BesselJ}(1,\alpha)}{\alpha},$$ and so on.
The integral obviously doesn't depend on $p$, since it must be rotationally invariant, and, using polar coordinates, you get that $$I(p,\alpha) = \mathrm{vol}(S^{n-1})\int_{-1}^1 e^{i\alpha t}(1-t^2)^{(n-2)/2}dt.$$ This latter integral can be evaluated by standard techniques, but, for odd values of $n$, it is expressed in Bessel functions, while, for even values of $n$ it is expressed in terms of elementary functions. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.99668288230896, "perplexity": 67.200348707833}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703108201/warc/CC-MAIN-20130516111828-00018-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://mathoverflow.net/questions/105222/hales-work-on-kepler-conjecture?sort=votes | # Hales work on Kepler conjecture
It seems that the Fulkerson prize has been attributed to Thomas Hales for this work. What is the present status of the conjecture, then?
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Actually, I should first have read what the Fulkerson prize site has to say : in 2009, they gave the prize to Thomas Hales and Samuel Ferguson ; this should imply their conjoint work i considered as a complete proof. – Feldmann Denis Aug 22 '12 at 10:56
That is an unusual use of the word "attributed". It looks like Hales gave out the prize. – S. Carnahan Aug 23 '12 at 5:57
The original 1998 proof appeared in 2005/6, in an abridged version in Annals of Mathematics:
A proof of the Kepler conjecture. Ann. of Math. (2) 162 (2005), no. 3, 1065–1185.
and unabridged as volume 36 of Discrete & Computational Geometry (2006).
It is my understanding that meanwhile there are no actual doubts in this proof (though, possibly, still some reservations by some, due to the complexity and the computer aid).
Thus, there is an ongoing effort to simplfy/clean up/formalize the proof. For example:
Hales, Thomas C.; Harrison, John; McLaughlin, Sean; Nipkow, Tobias; Obua, Steven; Zumkeller, Roland "A revision of the proof of the Kepler conjecture" Discrete Comput. Geom. 44 (2010), no. 1, 1–34.
From the MR review of that paper (by Uwe Schnell):
Since the original proof makes use of computer programs, the so-called Flyspeck project was started to give a formal proof of the Kepler conjecture. This article summarizes the current status of this initiative and gives a list of minor errata in the original proof.
Regarding this Flyspeck project see for example a recent presentation by Hales (given March 2012, the date on the slides seems wrong), where he says that this project is 80% complete. On the web one can easily find further information related to this project.
OP asks regarding the meaning of the awarding of the Fulkerson prize:
In some sense there is no specific meaning to it (as concerns acceptance of the community of the proof) as the Fulkerson prize is not a prize that is a priori awarded for 'The Kepler Conjecture', as opposed to the situation, say, for the Wolfskehl prize for FLT or the Clay prizes for the seven problems. By contrast 'The Delbert Ray Fulkerson Prize recognizes outstanding papers in the area of discrete mathematics.' So that in principle it would even be possible that the prize would be awarded while people only believe that the papers make significant progress towards a proof. However, this seems not the case as the Notices AMS article on the 2009 awarding of this prize contains 'After four centuries, Ferguson and Hales have now proven Kepler’s assertion.' Thus, yes, this is further evidence that the proof is generally accepted.
However, also note that this was not the first prize they received for their work. They also received the 2007 Robbins prize. The citation contains "The Kepler conjecture asserts [...] The proof of this result is a landmark achievement." but also at the end "Some controversy has surrounded this proof, with its large computer component, concerning its reliable checkability by humans. Addressing this issue, Hales has an ongoing project, called the 'Flyspeck' project, whose object is to construct a 'second-generation' proof which is entirely checkable by computer in a formal logic system."
So, in summary, this prize is a recognition of their work and that it was awarded can be considered as further evidendence that the proof is accepted as correct. However, I would not consider the awarding of this prize as something like the moment when the proof in any sense 'officially' got accepted as such. (Indeed, as far as I know, it was already widely accepted several years before that, e.g., otherwise the Annals would not have published this paper; except for the reservations I mentioned above, but then those also did not get addressed by the prize-awarding.)
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Yes, I am aware of that. But then, what is the meaning of the Fulkerson prize ? – Feldmann Denis Aug 22 '12 at 13:29
I'd say there is no highly specific meaning to it except being additional evidence that there is confidence in the correctness of the proof (possibly up to minor issues, as essentially always). I will edit my answer to include more info on this. – quid Aug 22 '12 at 13:54
Perhaps I should also add this piece of information, which I forgot and I do not want to edit agaon right away: this prize is not awarded each year, but only all three years and for work in the preceeding 6 years, so in some sense 2009 was the first possibly occassion to give them this prize (as it is for published papers). So that it was given in 2009 rather than a year before, say, is merely a technicality of the prize and not evidence that in 2009 something on the perception of the proof changed. – quid Aug 22 '12 at 14:42 | {"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.837956964969635, "perplexity": 505.41855961362296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00111-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://www.nag.com/numeric/nl/nagdoc_27.2/clhtml/s/s21bdc.html | NAG CL Interfaces21bdc (ellipint_symm_3)
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1Purpose
s21bdc returns a value of the symmetrised elliptic integral of the third kind.
2Specification
#include
double s21bdc (double x, double y, double z, double r, NagError *fail)
The function may be called by the names: s21bdc, nag_specfun_ellipint_symm_3 or nag_elliptic_integral_rj.
3Description
s21bdc calculates an approximation to the integral
$RJ(x,y,z,ρ)=32∫0∞dt (t+ρ)(t+x)(t+y)(t+z)$
where $x$, $y$, $z\ge 0$, $\rho \ne 0$ and at most one of $x$, $y$ and $z$ is zero.
If $\rho <0$, the result computed is the Cauchy principal value of the integral.
The basic algorithm, which is due to Carlson (1979) and Carlson (1988), is to reduce the arguments recursively towards their mean by the rule:
$x0 = x,y0=y,z0=z,ρ0=ρ μn = (xn+yn+zn+2ρn)/5 Xn = 1-xn/μn Yn = 1-yn/μn Zn = 1-zn/μn Pn = 1-ρn/μn λn = xnyn+ynzn+znxn xn+1 = (xn+λn)/4 yn+1 = (yn+λn)/4 zn+1 = (zn+λn)/4 ρn+1 = (ρn+λn)/4 αn = [ρn(xn,+yn,+zn)+xnynzn] 2 βn = ρn (ρn+λn) 2$
For $n$ sufficiently large,
$εn=max(|Xn|,|Yn|,|Zn|,|Pn|)∼14n$
and the function may be approximated by a fifth order power series
$RJ(x,y,z,ρ)= 3∑m= 0 n- 14-m RC(αm,βm) + 4-nμn3 [1+ 37Sn (2) + 13Sn (3) + 322(Sn (2) )2+ 311Sn (4) + 313Sn (2) Sn (3) + 313Sn (5) ]$
where ${S}_{n}^{\left(m\right)}=\left({X}_{n}^{m}+{Y}_{n}^{m}+{Z}_{n}^{m}+2{P}_{n}^{m}\right)/2m$.
The truncation error in this expansion is bounded by $3{\epsilon }_{n}^{6}/\sqrt{{\left(1-{\epsilon }_{n}\right)}^{3}}$ and the recursion process is terminated when this quantity is negligible compared with the machine precision. The function may fail either because it has been called with arguments outside the domain of definition or with arguments so extreme that there is an unavoidable danger of setting underflow or overflow.
Note: ${R}_{J}\left(x,x,x,x\right)={x}^{-\frac{3}{2}}$, so there exists a region of extreme arguments for which the function value is not representable.
4References
NIST Digital Library of Mathematical Functions
Carlson B C (1979) Computing elliptic integrals by duplication Numerische Mathematik 33 1–16
Carlson B C (1988) A table of elliptic integrals of the third kind Math. Comput. 51 267–280
5Arguments
1: $\mathbf{x}$double Input
2: $\mathbf{y}$double Input
3: $\mathbf{z}$double Input
4: $\mathbf{r}$double Input
On entry: the arguments $x$, $y$, $z$ and $\rho$ of the function.
Constraint: ${\mathbf{x}}$, y, ${\mathbf{z}}\ge 0.0$, ${\mathbf{r}}\ne 0.0$ and at most one of x, y and z may be zero.
5: $\mathbf{fail}$NagError * Input/Output
The NAG error argument (see Section 7 in the Introduction to the NAG Library CL Interface).
6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 3.1.2 in the Introduction to the NAG Library CL Interface for further information.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 7.5 in the Introduction to the NAG Library CL Interface for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 8 in the Introduction to the NAG Library CL Interface for further information.
NE_REAL_ARG_EQ
On entry, ${\mathbf{r}}=0.0$.
Constraint: ${\mathbf{r}}\ne 0.0$.
On entry, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{y}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{z}}=⟨\mathit{\text{value}}⟩$.
Constraint: at most one of x, y and z is $0.0$.
The function is undefined.
NE_REAL_ARG_GT
On entry, $U=⟨\mathit{\text{value}}⟩$, ${\mathbf{r}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{y}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{z}}=⟨\mathit{\text{value}}⟩$.
Constraint: $|{\mathbf{r}}|\le U$ and ${\mathbf{x}}\le U$ and ${\mathbf{y}}\le U$ and ${\mathbf{z}}\le U$.
NE_REAL_ARG_LT
On entry, $L=⟨\mathit{\text{value}}⟩$, ${\mathbf{r}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{y}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{z}}=⟨\mathit{\text{value}}⟩$.
Constraint: $|{\mathbf{r}}|\ge L$ and at most one of x, y and z is less than $L$.
On entry, ${\mathbf{x}}=⟨\mathit{\text{value}}⟩$, ${\mathbf{y}}=⟨\mathit{\text{value}}⟩$ and ${\mathbf{z}}=⟨\mathit{\text{value}}⟩$.
Constraint: ${\mathbf{x}}\ge 0.0$ and ${\mathbf{y}}\ge 0.0$ and ${\mathbf{z}}\ge 0.0$.
7Accuracy
In principle the function is capable of producing full machine precision. However, round-off errors in internal arithmetic will result in slight loss of accuracy. This loss should never be excessive as the algorithm does not involve any significant amplification of round-off error. It is reasonable to assume that the result is accurate to within a small multiple of the machine precision.
8Parallelism and Performance
s21bdc is not threaded in any implementation.
You should consult the S Chapter Introduction which shows the relationship of this function to the classical definitions of the elliptic integrals.
If the argument r is equal to any of the other arguments, the function reduces to the integral ${R}_{D}$, computed by s21bcc.
10Example
This example simply generates a small set of nonextreme arguments which are used with the function to produce the table of low accuracy results.
10.1Program Text
Program Text (s21bdce.c)
None.
10.3Program Results
Program Results (s21bdce.r) | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 57, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9312921762466431, "perplexity": 822.5522384270246}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00494.warc.gz"} |
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