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http://mathoverflow.net/revisions/13581/list | ## Return to Question
3 deleted 222 characters in body
A quantum channel is a mapping between Hilbert spaces, `$\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$`, where `$L(\mathcal{H}_{i})$` is the family of operators on `$\mathcal{H}_{i}$`. In general, we are interested in CPTP maps. The operator spaces can be interpreted as `$C^{*}$`-algebras and thus we can also view the channel as a mapping between `$C^{*}$`-algebras, `$\Phi : \mathcal{A} \to \mathcal{B}$`. Since quantum channels can carry classical information as well, we could write such a combination as `$\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$` where `$C(X)$` is the space of continuous functions on some set $X$ and is also a `$C^{*}$`-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between `$C^{*}$`-algebras. Note, however, that these are not necessarily the same `$C^{*}$`-algebras. Since the channels are represented by square matrices, the input and output `$C^{*}$`-algebras must have the same dimension, $d$. Thus we can consider them both subsets of some $d$-dimensional `$C^{*}$`-algebra, `$\mathcal{C}$`, i.e. `$\mathcal{A} \subset \mathcal{C}$` and `$\mathcal{B} \subset \mathcal{C}$`. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
Proposition A quantum channel given by `$t: L(\mathcal{H}) \to L(\mathcal{H})$`, together with the $d$-dimensional `$C^{*}$`-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
Proof: Consider the quantum channels
```$\begin{eqnarray*}
r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) &
\qquad \textrm{where} \qquad &
\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\
t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) &
\qquad \textrm{where} \qquad &
\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}
\end{eqnarray*}$```
where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite `$t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$` where
```$\begin{align}
\tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\
& = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\
& = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag
\end{align}$```
and the `$A_{i}$`, `$B_{i}$`, and `$C_{i}$` are Kraus operators.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. `$\sum_{k} C_{k}^{\dagger}C_{k}=1$` (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). $\sum_{k} C_{k}^{\dagger}C_{k}=\mathbf{1}.$\$ For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041).
We take the identity arrow, `$1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$`, to be the time evolution of the state $\rho$ in the absence of any channel. Since this definition is suitably general we have that `$t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$`.
Consider the three unital quantum channels `$r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$`, `$t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$`, and `$v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$` where `$\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$`, `$\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$`, and `$\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$`. We have
```$\begin{align}
v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\
& = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\
& = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag
\end{align}$```
and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either). $\square$
Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.
I'm new here, so feel free to change the tags or point out anything that isn't appropriate.
2 Added a note about Kraus operators.
A quantum channel is a mapping between Hilbert spaces, `$\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$`, where `$L(\mathcal{H}_{i})$` is the family of operators on `$\mathcal{H}_{i}$`. In general, we are interested in CPTP maps. The operator spaces can be interpreted as `$C^{*}$`-algebras and thus we can also view the channel as a mapping between `$C^{*}$`-algebras, `$\Phi : \mathcal{A} \to \mathcal{B}$`. Since quantum channels can carry classical information as well, we could write such a combination as `$\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$` where `$C(X)$` is the space of continuous functions on some set $X$ and is also a `$C^{*}$`-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between `$C^{*}$`-algebras. Note, however, that these are not necessarily the same `$C^{*}$`-algebras. Since the channels are represented by square matrices, the input and output `$C^{*}$`-algebras must have the same dimension, $d$. Thus we can consider them both subsets of some $d$-dimensional `$C^{*}$`-algebra, `$\mathcal{C}$`, i.e. `$\mathcal{A} \subset \mathcal{C}$` and `$\mathcal{B} \subset \mathcal{C}$`. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
Proposition A quantum channel given by `$t: L(\mathcal{H}) \to L(\mathcal{H})$`, together with the $d$-dimensional `$C^{*}$`-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
Proof: Consider the quantum channels
```$\begin{eqnarray*}
r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) &
\qquad \textrm{where} \qquad &
\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\
t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) &
\qquad \textrm{where} \qquad &
\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}
\end{eqnarray*}$```
where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite `$t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$` where
```$\begin{align}
\tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\
& = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\
& = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag
\end{align}$```
and the `$A_{i}$`, `$B_{i}$`, and `$C_{i}$` are Kraus operators.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. `$\sum_{k} C_{k}^{\dagger}C_{k}=1$` (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041).
We take the identity arrow, `$1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$`, to be the time evolution of the state $\rho$ in the absence of any channel. Since this definition is suitably general we have that `$t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$`.
Consider the three unital quantum channels `$r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$`, `$t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$`, and `$v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$` where `$\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$`, `$\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$`, and `$\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$`. We have
```$\begin{align}
v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\
& = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\
& = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag
\end{align}$```
and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).
Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.
I'm new here, so feel free to change the tags or point out anything that isn't appropriate.
1
# Quantum channels as categories: question 1.
A quantum channel is a mapping between Hilbert spaces, `$\Phi : L(\mathcal{H}_{A}) \to L(\mathcal{H}_{B})$`, where `$L(\mathcal{H}_{i})$` is the family of operators on `$\mathcal{H}_{i}$`. The operator spaces can be interpreted as `$C^{*}$`-algebras and thus we can also view the channel as a mapping between `$C^{*}$`-algebras, `$\Phi : \mathcal{A} \to \mathcal{B}$`. Since quantum channels can carry classical information as well, we could write such a combination as `$\Phi : L(\mathcal{H}_{A}) \otimes C(X) \to L(\mathcal{H}_{B})$` where `$C(X)$` is the space of continuous functions on some set $X$ and is also a `$C^{*}$`-algebra. In other words, whether or not classical information is processed by the channel, it (the channel) is a mapping between `$C^{*}$`-algebras. Note, however, that these are not necessarily the same `$C^{*}$`-algebras. Since the channels are represented by square matrices, the input and output `$C^{*}$`-algebras must have the same dimension, $d$. Thus we can consider them both subsets of some $d$-dimensional `$C^{*}$`-algebra, `$\mathcal{C}$`, i.e. `$\mathcal{A} \subset \mathcal{C}$` and `$\mathcal{B} \subset \mathcal{C}$`. Thus a quantum channel is a mapping from $\mathcal{C}$ to itself.
Proposition A quantum channel given by `$t: L(\mathcal{H}) \to L(\mathcal{H})$`, together with the $d$-dimensional `$C^{*}$`-algebra, $\mathcal{C}$, on which it acts, forms a category we call $\mathrm{\mathbf{Chan}}(d)$ where $\mathcal{C}$ is the sole object and $t$ is the sole arrow.
Proof: Consider the quantum channels
```$\begin{eqnarray*}
r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma}) &
\qquad \textrm{where} \qquad &
\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger} \\
t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau}) &
\qquad \textrm{where} \qquad &
\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}
\end{eqnarray*}$```
where the usual properties of such channels are assumed (e.g. trace preserving, etc.). We form the composite `$t \circ r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\tau})$` where
```$\begin{align}
\tau & = \sum_{j}B_{j}\left(\sum_{i}A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger} \notag \\
& = \sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger} \\
& = \sum_{k}C_{k}\rho C_{k}^{\dagger} \notag
\end{align}$```.
Since $A$ and $B$ are summed over separate indices the trace-preserving property is maintained, i.e. `$\sum_{k} C_{k}^{\dagger}C_{k}=1$` (not sure how to do a blackboard 1 on this site - \mathbb doesn't work since it's a package). For a similar methodology see Nayak and Sen (http://arxiv.org/abs/0605041).
We take the identity arrow, `$1_{\rho}: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\rho})$`, to be the time evolution of the state $\rho$ in the absence of any channel. Since this definition is suitably general we have that `$t \circ 1_{A}=t=1_{B} \circ t \quad \forall \,\, t: A \to B$`.
Consider the three unital quantum channels `$r: L(\mathcal{H}_{\rho}) \to L(\mathcal{H}_{\sigma})$`, `$t: L(\mathcal{H}_{\sigma}) \to L(\mathcal{H}_{\tau})$`, and `$v: L(\mathcal{H}_{\tau}) \to L(\mathcal{H}_{\upsilon})$` where `$\sigma=\sum_{i}A_{i}\rho A_{i}^{\dagger}$`, `$\tau=\sum_{j}B_{j}\sigma B_{j}^{\dagger}$`, and `$\eta=\sum_{k}C_{k}\tau C_{k}^{\dagger}$`. We have
```$\begin{align}
v \circ (t \circ r) & = v \circ \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) = \sum_{k}C_{k} \left(\sum_{i,j}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}\right) C_{k}^{\dagger} \notag \\
& = \sum_{i,j,k}C_{k}B_{j}A_{i}\rho A_{i}^{\dagger}B_{j}^{\dagger}C_{k}^{\dagger} = \sum_{i,j,k}C_{k}B_{j}\left(A_{i}\rho A_{i}^{\dagger}\right)B_{j}^{\dagger}C_{k}^{\dagger} \notag \\
& = \left(\sum_{i,j,k}C_{k}B_{j}\tau B_{j}^{\dagger}C_{k}^{\dagger}\right) \circ r = (v \circ t) \circ r \notag
\end{align}$```
and thus we have associativity. Note that similar arguments may be made for the inverse process of the channel if it exists (it is not necessary for the channel here to be reversible). End proof (\Box doesn't seem to work here either).
Question 1: Am I doing the last line in the associativity argument correct and/or are there any other problems here? Is there a clearer or more concise proof? I have another question I am going to ask as a separate post about a construction I did with categories and groups that assumes the above is correct but I didn't want to post it until I made sure this is correct.
I'm new here, so feel free to change the tags or point out anything that isn't appropriate. | {"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": 38, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9424754977226257, "perplexity_flag": "head"} |
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# Mandelbrot set
A rendering of the Mandelbrot set: black points represent the stable points under the iterative map
In mathematics, the Mandelbrot set is a fractal that is defined as the set of points c in the complex plane for which the iteratively defined sequence:
$z_0 = 0\,$
$z_{n+1} = {z_n}^2 + c$
does not tend to infinity.
The sequence is thus expanded mathematically as follows for each point c in the complex plane:
$c=x+iy \,$
$Z_0=0 \,$
$\begin{matrix}Z_1&=&Z_0^2+c \\ \ &=& x+iy\end{matrix} \,$
$\begin{matrix}Z_2&=&Z_1^2+c \\ \ &=&(x + iy)^2+x+iy \\ \ &=&x^2+2ixy-y^2+x+iy \\ \ &=&x^2-y^2+x+(2xy+y)i\end{matrix} \,$
$Z_3=Z_2^2+c=... \,$
and so on.
If we reformulate this in terms of the real and imaginary parts (X and Y coordinates of the complex plane), looking at each iteration, n replacing zn with the point xn + yni and c with the point a + bi, then we get
$x_{n+1} = {x_n}^2 - {y_n}^2 + a \,$
and
$y_{n+1} = 2{x_n} {y_n} + b \,$
The Mandelbrot set can be divided into an infinite set of black figures: the largest figure in the center is a cardioid. There is a (countable) infinity of near-circles (the only one to be actually an exact circle being the largest, immediately on the left of the cardioid) which are in direct (tangential) contact with the cardioid, but they vary in size, tending asymptotically to zero diameter. Then each of these circles has in turn its own (countable) infinite set of smaller circles which branch out from it, and this set of surrounding circles also tends asymptotically in size to zero. The branching out process can be repeated indefinitely, producing a fractal. Note that these branching processes do not exhaust the Mandelbrot set: further upwards in the tendrils, some new cardioids appear, not glued to lower level "circles". The largest of these, and the most easily visible from a view of the entire set, is along the "spike" which follows the negative real axis out, roughly from real values of -1.78 to -1.75
Contents
## History of the Mandelbrot set
The Mandelbrot set was first defined in 1905 by Pierre Fatou, a French mathematician working in the field of complex analytic dynamics. Fatou studied recursive processes like
$z \to z^2 + c\,$
Starting with some point z0 on the complex plane, successive points may be generated by repeatedly applying this formula. The sequence of points thus obtained is called the orbit of z0 under the transformation
$z \to z^2 + c\,$
Fatou realised that the orbit of z0 = 0 under this transformation would provide some insight into the behaviour of such systems. There are an infinite number of such functions - one for each value of c. Fatou did not have access to a computer capable of plotting the orbits of all these functions, but attempted to do so by hand. He proved that once a point moved to a distance greater than 2 from the origin, then the orbit would escape to infinity.
Fatou never saw the image of what we now call the Mandelbrot set as we do because the number of calculations required to generate this is far more than could be calculated by hand. Professor Benoît Mandelbrot was the first person to use a computer to plot the set.
Fractals were popularised by Mandelbrot in 1975 in his book . In this book, Mandelbrot used the term fractal to describe a number of mathematical phenomena that seemed to exhibit chaotic or surprising behaviour. All of these phenomena involved the definition of some curve or set through the use of some recursive functions or algorithms. The Mandelbrot set is one such phenomenon that is named after its discoverer.
## Relationship with Julia sets
The Mandelbrot set was created by Benoît Mandelbrot as an index to the Julia sets: each point in the complex plane corresponds to a different Julia set. The points within the Mandelbrot set correspond precisely to the connected Julia sets, and the points outside correspond to disconnected ones.
Intuitively, the "interesting" Julia sets correspond to points near the boundary of the Mandelbrot set; those far inside tend to be simple geometric shapes, while those well outside look like dust surrounded by blobs of color. Some programs, such as Fractint, let the user choose a point, and jump to the corresponding Julia set, making it easy to find Julia sets that the user is likely to enjoy.
The Mandelbrot set also contains structures that strongly resemble Julia sets; indeed, for any value c, the region of the Mandelbrot set near c resembles the center of the Julia set with parameter c.
## Plotting the set
Sample generated image with colouring given by the rate of divergence to infinity, the brighter the points the slower the divergence
It can be shown that once the absolute value of zn is larger than 2 (in cartesian form , when xn2 + yn2 > 22) the sequence will tend to infinity, and c is therefore outside the Mandelbrot set. This value, known as the bail-out value, allows the calculation to be terminated for points outside the Mandelbrot set. For points inside the Mandelbrot set, i.e. values of c for which zn doesn't tend to infinity, the calculation never comes to such an end, so it must be terminated after some number of iterations determined by the program. This results in the displayed image being only an approximation to the true set. Mandelbrot hobbyists quickly learn to recognize the "blobby" images caused by a program mistakingly placing points in the set, and will then up the iteration count (at the expense of slowing down the calculation of every point actually in the set).
### Adding color
Mathematically speaking, the pictures of the Mandelbrot set and Julia sets are "black and white". Either a point is in the set or it is not. Most computer-generated graphs are drawn in color. Under the most common rendering method, for the points that diverge to infinity, and are therefore not in the set, the color reflects the number of iterations it takes to reach a certain distance from the origin. This creates concentric shapes, each a better approximation to the Mandelbrot set than the last. One possible scheme is that points that diverge quickly are drawn in black; then you have brighter colors for the middle; then you have white for the points in the set, and near-white for the points that diverge very slowly.
In order to determine if the point, Z0, is going to be in the Mandelbrot set (traditionally coloured black) or outside the set (coloured according to the escape speed), the distance of the Zi must be calculated at each iteration in the sequence:
If $Z_i = X + iY \$, then $\|Z_i\| = \sqrt{X^2 + Y^2}$. Note that one of the many possible optimisations for calculating mandelbrots can be applied here. Rather than testing to see if $\sqrt{X^2 + Y^2} > 2$, we can simply test to see if $X^2 + Y^2 > 4 \$ — thus saving the square root operation.
So, if $\|Z_i\|^2 < 4$ then colour the point black, otherwise colour it according to the value of i. Using the number of iterations required to determine that the point has escaped is the easiest and most common way of recording the "speed" of escape. This number can be directly mapped to a colour via a lookup table or palette, or by use of some suitable algorithm.
So far, we have worked out how to tell if the point is outside the set — it moves more than a distance of 2 away from the origin. How do we tell it is inside the set? There are lots of strategies for this — this is not so simple. If it were, there would not be so many different algorithms out there for calculating mandelbrots. This is the essence of the problem. Clearly most points will not actually get to zero in a reasonable number of iterations. The simplest algorithm is to simply limit the number of iterations and if you have not gone outside the boundary by the time you have done your last iteration, then you assume the point is in the set.
Indeed, points nearer the set take longer to escape. This is why plotting slows down near the dark region — because more terms in the expansion are having to be calculated to determine the escape speed. When not near the set, one can "get away with" a lower maximum iteration count. The images below, of the same location, are created with iteration limits of 32, 64, 128, 256, 512, 1024, 2048 and 1000000, respectively. Note that there are diminishing returns in upping the iteration count past a point; these are also good examples of the "blobby" images one gets with a too-low iteration count. These images are centered at -1.7490110509, -0.0004525. The image is roughly 0.000417 of a unit wide, and located about halfway down the "valley" at the +x end of the largest mini-set.
| | | | |
|-------------------------------------------------------------------------------------------------------|------------------------------------------------------------------------------|------------------------------------------------------------------------|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| At a count of 32, the whole image is black, since it is completely inside the false-negative contour. | If we allow 64 iterations, some points are no longer falsely inside the set. | At 128 iterations, the image is blobby, but recognizable as a fractal. | 256 |
| At 512, we get a nice image. The black dots up and to the left of each "wart" contain tiny cardoids. | | | Diminishing returns are quite obvious when we use a million iterations. Even with periodicity checking, this one took 10-15 seconds to generate on an Athlon XP 2000+. |
#### Optimization
One way to improve calculations is to find out beforehand whether the given point lies within the cardioid. The cardioid's polar equation is
$\rho_c = {1 \over 2} - {1 \over 2} \cos \theta$
but with the center of polar coordinates located at (1/4,0) — the cardioid's cusp — instead of at the origin. Given point (x,y), calculate the following
$\rho = \sqrt{\left( x - {1 \over 4} \right)^2 + y^2},$
$\theta = \hbox{atn}_2 \left( y, x - {1 \over 4} \right),$
$\rho_c = {1 \over 2} - {1 \over 2} \cos \theta.$
If $\rho \le \rho_c$ then point (x,y) is in the Mandelbrot set (color it black), and the usual calculations can be skipped. To prevent having to do huge numbers of iterations for other points in the set, one can do "periodicity checking" - which means check if a point reached in iterating a pixel has been reached before. If so, the pixel cannot diverge, and must be in the set.
A portion of the Mandelbrot set centred at (0.282, -0.01)
### Art and the Mandelbrot set
Some people have a hobby of searching the Mandelbrot set for interesting pictures. They have a collection of pictures, along with the coordinates for generating that picture. For example, the image to the right is a closeup centred at coordinates (0.282, -0.01). Below are more examples of cool- and hot-looking Mandelbrot set regions:
There are many free fractal-generating programs available, such as those by Stephen Ferguson (Sterling Fractal and the Tierazon series) and many people write such programs themselves for the highest level of control over their pictures. Below is a min-cardioid, connected by filaments (too thin to be seen) to the main set (and other cardioids, some of which are actually in this image but too small to be seen).
## Other Mandelbrot sets
When people speak of the Mandelbrot set, they usually are referring to the set described above. Any function that maps to and from the complex number plane has a Mandelbrot set, which characterizes whether or not the Julia set corresponding to that function is connected.
Example:
Let fc(z) = z3 + c.
For each value of c, we draw the Julia set Jc of fc(z), and determine if it is connected or not. If Jc is connected, then c is in the Mandelbrot set of {fc}, otherwise c is not in the Mandelbrot set.
This can also be generalized to Julia sets parameterized by more than two real numbers. For example, a collection of Julia sets parametrized by three real numbers will have a three dimensional Mandelbrot set. Of course, only the 2-dimensional case will have an easily viewed picture.
## Mandelbrot set in song
The Australian band GangGajang has a song Time (and the Mandelbrot set) where the term Mandelbrot set is used liberally in the lyrics.
The American singer Jonathan Coulton has a song titled Mandelbrot Set about the history of the Mandelbrot set, and of Benoit Mandelbrot himself.
The Blue Man Group's first album Audio features tracks titled "Mandelgroove", "Opening Mandelbrot", and "Klein Mandelbrot". The album was nominated for a Grammy in 2000.
## External links
• The Chaos Hypertextbook. An introductory primer on chaos and fractals.
• Downloadable Mandelbrot Set Explorer Application
• Mu-Ency - Encyclopedia of the Mandelbrot Set
• Fractal Explorer
• Explore and Discover
• Julia and Mandelbrot Set Explorer, a CGI-based interface created by David Joyce, a Clark University professor.
• The Fractal Microscope provides a nice Java interface (figure to right). Unlike some other Java-based Mandelbrot set interfaces on the internet, this one works for Macintosh browsers (tested on these Mac OS X browsers: Safari, Mozilla, Camino, Internet Explorer).
• Detailed mandelbrot set, see bottom of page
• 3D Mandelbrot Set Explorer Applet
• A gallery of Fractal images and copies of Fractal-exploring programs.
• Realtime Mandelbrot Set Generator - Web based Mandelbrot Set Explorer.
• Color Cycling on the Mandelbrot Set
• Iterations and the Mandelbrot Set
• Mandelbrot and Julia sets
• Xaos, an open source fractal explorer with Mandelbrot/Julia exploration, other fractals, and autopilot mode for easy exploration
• Fract: a free software web zoomer for the Mandelbrot Set
• The Mandelbrot and Julia sets Anatomy
03-10-2013 05:06:04 | {"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": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9013550877571106, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showpost.php?p=3197654&postcount=7 | View Single Post
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Quote by LCKurtz Don't quit yet, you almost have it. Take x > 0 to dispense with the || signs: $$R(x) = e^{\int-\frac 1{2x}dx}= e^{-\frac 1 2 \int\frac 1 x}\, dx = e^{-\frac 1 2 \ln x} = e^{\ln(x^{-\frac 1 2})} = x^{-\frac 1 2}$$ Also, a TeX note, you can use the partial symbol $$\partial Q/\partial x$$ but subscripts are even easier: Qx or $Q_x$.
Quote by pat666 Thanks LCKurtz I now have a general and particular solution, would you mind checking my results: $$u(x,y)=4x^{3/2}+2*\sqrt(x)*y^2+y=c$$ $$c=13/3$$ $$13/3=4x^{3/2}+2*\sqrt(x)*y^2+y$$ THANKS ps should be 4x to the power of 3/2 , doesn't seem to like my tex.
Quote by LCKurtz You should be able to check that yourself. Check that du = Pdx + Qdy with the exact form and that (x,y)=(1,1) works. If not, try to fix it.
Quote by pat666 I don't know what you mean by Check that du = Pdx + Qdy, du as in the differential of $$4x^{3/2}+2*\sqrt(x)*y^2+y$$ also should $$4*1^{3/2}+2*\sqrt(1)*1^2+1$$ = 13/3? because it =s 7?? am I doing something wrong with my checking or is my solution wrong? Thanks
After you multiplied by the integrating factor x-1/2 you should have gotten this DE, which you didn't show:
$$(*)\ (2x^{\frac 1 2}+ 2x^{-\frac 1 2}y^2)dx+4x^{\frac 1 2}y dy = 0$$
Presumably you have checked that this equation is exact and you have a proposed solution u(x,y) written above. The way to check whether your solution is correct is to calculate du = uxdx + uydy and see if it agrees with the exact equation (*) above.
Once you have that you need to check that x = 1, y = 1 works in your final answer. | {"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": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9668077230453491, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/75565/finite-simple-groups-and-conjugacy-classes-with-2p-elements/75672 | ## Finite simple groups and conjugacy classes with 2p elements
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $p$ be an odd prime number. Can a finite simple group have a conjugacy class with $2p$ elements?
-
2
Did you trawl the ATLAS for examples? – Noam D. Elkies Sep 16 2011 at 3:30
Is it known that this can't occur with the alternating groups? Empirically it looks like the sizes of the conjugacy classes of $A_n$ all have at least 3 prime factors once $n>8$. – Alon Amit Sep 16 2011 at 6:17
2
I can see why a conjugacy class $C$ with $p$ elements is impossible: $G$ acts on $C$ by conjugation, faithfully as it is simple, so $G\subset S_p$. Under this embedding, the point-stabilizers $G\cap S_{p-1}$ are then the normalizers of elements of $C$. But every element normalizers itself, so the sizes of these subgroups must be a multiple of $p$, contradiction. This argument does not immediately work for $2p$, but it does put restrictions on $G\subset S_{2p}$ (e.g. it should contain the full $p$-Sylow $C_p\times C_p$ of $S_{2p}$). Maybe it can help to prove that this is impossible? – Tim Dokchitser Sep 16 2011 at 9:24
3
Actually, more generally a conjugacy class with $p^n$ elements is impossible for $n>0$. That's because if $C$ is a conjugacy class and $\chi$ is an irreducible char, and if $(\chi(1),|C|)=1$, then any $g\in C$ is either in the centre of $\chi$ or $\chi(C)=0$. If $G$ is simple, then the centres of characters are trivial, so $\chi(C)=0$ for all $\chi$ with $p\nmid \chi(1)$. The column orthogonality equation between 1 and $g\in C$ now gives a contradiction. This is somewhere in Isaacs, but for me, it's easier to link to my notes: math.postech.ac.kr/~bartel/docs/reptheory.pdf, Thm 6.7. – Alex Bartel Sep 16 2011 at 9:50
Unfortunately, this argument also doesn't immediately generalise to $|C|=2p$. – Alex Bartel Sep 16 2011 at 9:50
show 1 more comment
## 1 Answer
I believe the answer is no. A group with a conjugacy class of degree $2p$ has a transitive permutation action of degree $2p$ on that class, and the stabilizer is the centralizer of an element.
If this action is imprimitive, then the blocks have size $p$ or 2. Wtih blocks of size $p$ there would be a subgroup of index 2, contradicting simplicity. With blocks of size $p$, the group would act faithfully on the blocks. Burnside proved that a permutation group of prime degree $p$ is solvable or 2-transitive. The finite 2-transitive permutation groups are now known (using the classification of finite simple groups). They are listed, for example, in
PETER J. CAMERON, FINITE PERMUTATION GROUPS AND FINITE SIMPLE GROUPS, BULL. LONDON MATH. SOC , 13 (1981),1-22
and it can be checked that none of the examples of prime degree have point stabilisers that have a subgroup of index 2 with nontrivial centre.
So suppose that the action on the conjugacy class is primitive. In the Cameron paper cited above, a result of Wielandt is mentioned that says that primitive permutation groups of degree $2p$ either have rank 3 or are 2-transitive. Using the classification of finite simple groups, it has been checked that the only rank 3 examples are $A_5$ and $S_5$ in degree 10, and $A_5$ does not have a class of degree $2p$. For 2-transitive groups, we can again check the list and observe that none of the groups has point stabilizer with nontrivial centre.
- | {"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": 42, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8953348994255066, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/38448/newton-2nd-law-does-vertical-force-mass-affect-the-horizontal-acceleration/38469 | # Newton 2nd Law: Does vertical force (mass) affect the horizontal acceleration?
I learnt before that if 2 forces are perpendicular to each other, they should not affect each other. However in a recent experiment setup (asked in another question):
I believe the theoratical equation by newton 2nd law is
$$\begin{aligned} F_{horizontal} &= F_{vertical} \\ m_{cart}a_{cart} &= mg \\ a_{cart} &= \frac{mg}{m_{cart}} \\ \end{aligned}$$
Am I right so far? If so, this seem to imply that $m_{cart}$ (vertical force) is somehow affecting acceleration (horizontal)? Why is that?
-
Assuming no friction? – ja72 Sep 27 '12 at 17:35
## 1 Answer
What the diagram doesn't show is the force on the pulley:
It's the vector sum of this force and the force due to the weight that gives a horizontal force on the cart.
The tension in the string must be constant, because if it varied along the string the string would strtech or contract until the tension was constant, so $F$ is the tension in the string times $\sqrt{2}$.
-
hmm, I still don't get why forces acting perpendicular will affect each other? Does that have to do with the pulley? I know the tension in string is constant. For now, I think friction/mass of pulley is ignored ... not sure how `F=sqrt(2)`T? – Jiew Meng Sep 27 '12 at 9:49
try resolving the diagonal force into orthogonal components. – Nic Sep 27 '12 at 10:21
The pulley is exerting a force $F$ to the string. If you add (vector addition) the force $F$ to the downward force caused by the weight you'll find it's equal to the horizontal force on the cart. – John Rennie Sep 27 '12 at 10:26 | {"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": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9197506904602051, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/29249/quick-finite-intersection-property-question | ## Quick Finite Intersection Property Question [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I just need a quick clarification:
Given a sequence of sets `$\{a_n\}_{n \in \mathbb{N}}$` in some field $\mathbb{K}$, is saying that it satisfies the finite intersection property equivalent to saying $(\forall n\in \mathbb{N})(\exists x\in \mathbb{K})(x \in \cap_{i=1}^n a_n)$
If the previous statement is true, then it seems almost reasonable to say that, because `$\{\cap_{i=1}^n a_n\}_{n\in\mathbb{N}}$` is a nested sequence of nonempty sets (because of the f.i.p), `$\cap_{i=1}^{\infty} a_n \neq \emptyset$`, but I know that is not necessarily true
Thanks
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1
Think about `$A_n=\{m\in\mathbb{N}:m>n\}$`. This is really a bit too basic for MO. – Robin Chapman Jun 23 2010 at 15:51
Seconded (see also the answer below). Voting to close. – Yemon Choi Jun 23 2010 at 16:16
Closed. See wikipedia. – S. Carnahan♦ Jun 23 2010 at 16:56
1
I did look at wikipedia, I just wanted to ensure that what is stated on wikipedia is the same as the formal statement above – thedude Jun 23 2010 at 17:01
@thedude: I think everyone missed your point because the actual question was poorly emphasized. – François G. Dorais♦ Jun 23 2010 at 18:12
show 1 more comment
## 1 Answer
What you quote is the finite intersection property.
The point of confusion is in what follows. There is no reason why a nested sequence of nonempty sets can't have empty interesection. Just because we can find an $x_n\in\cap_{i=1}^n a_n$ for every $n$, there is no reason to expect that this can be done uniformly in $n$.
The intervals $[n,\infty)$ also form a nested sequence of nonempty sets with empty intersection.
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1
But there are some reasons to expect that this can be done uniformly in $n$. Namely, when the $a_n$ are compact sets in some topological space and, in particular, when the $a_n$ are finite sets. – François G. Dorais♦ Jun 23 2010 at 16:33 | {"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": 9, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9284203052520752, "perplexity_flag": "head"} |
http://nrich.maths.org/2292 | ### Weekly Problem 14 - 2012
Weekly Problem 14 - 2012
### Lost
Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses.
### Square Coordinates
A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?
# Coordinate Patterns
##### Stage: 3 Challenge Level:
Charlie and Alison have been drawing patterns on coordinate grids. You may want to choose just one to explore or you may like to try all three.
#### Charlie's Squares
Charlie has been drawing squares.
What will the coordinates of the centre of square number 3 be?
How do you know?
Charlie wants to know where the centre of square number 20 will be.
Can you use the diagram above to help you to work this out?
Can you suggest a quick and efficient strategy for working out the coordinates of the centre of any square?
Would your strategy work if Charlie's sequence extended to the left? $$....-2, -1, 0, 1, 2, 3....$$
Can you adapt your strategy to work out the coordinates of the corners of any square?
#### Alison's Triangles
Alison has been drawing triangles.
She wants to know where the vertices of triangle number 23 will be.
Can you use the diagram to work it out?
Can you suggest a quick and efficient strategy for working out the coordinates of the vertices of any triangle?
Would your strategy work if Alison's sequence extended to the left? $$....-2, -1, 0, 1, 2, 3....$$
#### More Squares from Charlie
Charlie has been drawing more squares.
He wants to know what the coordinates of the centre of square 22b will be.
Can you use the diagram to work it out?
Can you suggest a quick and efficient strategy for working out the coordinates of the vertices of any square?
The ideas for these problems originally came from the SMP11-16 booklets on Coordinate Patterns published by CUP.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8934177160263062, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-geometry/171541-show-these-intervals-open-closed-sets-print.html | # Show these intervals are open and closed sets.
Printable View
• February 16th 2011, 04:09 PM
seamstress
Show these intervals are open and closed sets.
Show that the intervals (a, ∞) and (-∞, a) are open sets and that [b, ∞) and (-∞, b] are closed sets.
Thank you!
• February 16th 2011, 04:18 PM
Plato
Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on this problem or explain what you do not understand about the question.
• February 16th 2011, 04:31 PM
seamstress
It's not homework, actually. I'm studying for a test and trying to understand something my teacher said in class the other day. I do not understand how to do the question, or why the answer is what it is, and I would like to understand open and closed sets. I thought maybe by understanding this basic question, I would be able to figure out other things that stem off of this one. If somebody could explain why the first two are open and why the second two are closed, and possibly show me how I would go about proving it, I would be helped tremendously in understanding this question. Thank you.
• February 16th 2011, 05:01 PM
Tinyboss
Closed is usually defined as "its complement is open", so it's enough to prove that the open intervals are indeed open. Try showing that for any point in one of the open intervals, you can find a smaller open interval around that point and contained in the original interval.
• February 16th 2011, 09:29 PM
seamstress
I still do not know where to get started. I recall that definition of closed sets is that, but I feel like I don't have enough information to properly prove these statements. Any suggestions on how to get started? Thanks
• February 17th 2011, 02:59 AM
DrSteve
For the first interval, let $x\in (a,\infty )$, so that $x>a$. Let $\epsilon = \frac{x-a}{2}$ (half the distance from $x$ to $a$). Then the open interval centered at $a$ with radius $\epsilon$ is contained in the original interval.
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http://mathhelpforum.com/differential-equations/156947-partial-differential-equations-problems.html | # Thread:
1. ## Partial Differential Equations Problems
Hello.
Question 1
Find the temperature distribution T(x,t) in a long thin bar $-a \leq x \leq a$ with given initial temperature $u(x,0) = f(x)$
The side walls of the bar are insulated, while heat radiates from the ends into the surrounding medium whose temperature is $T=0$.
The radiation at the ends is taken to obey Newton's Law.
In particular, find the Fourier coefficients in terms of $f(x)$
Here's what I did so far:
I set the boundary condition to be $u(-a) = 0, u(a) = 0$, initial condition to be $u(x,0) = f(x)$.
I am not sure what to do with the "radiation" part of the problem, and thus i don't really know what my partial differential equation should look like...
Question 2
A wave system that includes damping and dispersion is represented by
$u_{tt} + 2au_t + bu - c^2u_{xx} = 0$
where $a, b, c$ are positive constants.
Solve by separation of variables
$u_{tt} + 2au_t + bu - cu_{xx} = 0, 0 < x < \pi, t > 0$
$u(0,t) = u(\pi, t) = 0, t > 0$
$u(x,0) = 0, u_t(x,0) = g(x)$
For simplicity assume that $a^2 - b - c^2 < 0$.
And again here's what I did:
$u = X(x)T(t)$
$XT'' +2aXT' + bXT - c^2X''T =0$
$X(T'' + 2aT' + bT) = c^2X''T$
$\frac{T'' + 2aT' + bT}{c^2T} = \frac {X''}{X}$
Please help...and in the mean time, I'll try to work something out with my classmates...
Thank you very much.
2. Question 2:
You're doing great so far. What's the next step? You've got all $T$ on the LHS, and all $X$ on the RHS. What can you say?
3. I can say that I have separated the variables. Should I make the equation equal to $\lambda$?
And I'm not too sure what to do with the $T$ part.
I know what to do for X now.
Nvm...i get it now. This problem isn't that hard after all...probably because I had a limited lunch break.
Still working on Question 1 though.
4. Glad you got Question 2. As for Question 1, I'm not so sure your boundary conditions are correct. You're supposed to assume Newtonian cooling there, not necessarily equilibrium with ambient temperature. So I'd say your initial conditions would be $u(-a,0)=f(-a),$ and $u(a,0)=f(a).$ The values at the endpoints then obey Newton's Law of Cooling compared with the ambient environment. Perhaps Danny or CB could correct me if I'm wrong, but that's how I read it. Not sure how you're going to get the heat transfer coefficient. I suppose you could always just assume it's a constant, and have it show up in your final answer.
Those are my thoughts.
5. Following from Ackbeet, from what I see, your radiating BC at the endpoints is
$u_x = - k\left( u - u_0\right)$ where $u_0 = 0$.
6. Or would that be
$u_{t}=-k(u-u_{0})=-ku$?
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http://mathhelpforum.com/calculus/148312-instantaneous-rate-word-problem.html | # Thread:
1. ## instantaneous rate word problem
So I was able to do the first part (hopefully it's right), but I'm having trouble trying to figure out what it's asking in the last parts. Here's my problem:
A business is prospering in such a way that its total (accumulated) profit after t years is $1000t^2$ dollars.
a) How much did the business make during the third year?
so, $f(3)=1000(3)^2=9000$ hopefully its right.
b) What was its average rate of profit during the first half of the third year, between t=2 and t=2.5?
c) What was its instantaneous rate of profit at t=2?
Do I plug in the values for t in part b and c or is there something else invovled? Any help is greatly appreciated!
2. Originally Posted by ascendancy523
b) What was its average rate of profit during the first half of the third year, between t=2 and t=2.5?
$\frac{f(2.5)-f(2)}{2.5-2}$
Originally Posted by ascendancy523
c) What was its instantaneous rate of profit at t=2?
$f'(2)$ | {"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": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9857778549194336, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/tagged/dimensional-analysis+thermodynamics | # Tagged Questions
2answers
302 views
### Why are expressions such as $\operatorname{ln}T$ used in thermodynamics where $T$ is not dimensionless?
In all thermodynamics texts that I have seen, expressions such as $\operatorname{ln}T$ and $\operatorname{ln}S$ are used, where $T$ is temperature and $S$ is entropy, and also with other thermodynamic ...
5answers
435 views
### Why isn't temperature measured in units of energy?
Temperature is the average of the kinetic energies of all molecules of a body. Then, why do we consider it a different fundamental physical quantity altogether [K], and not an alternate form of ... | {"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": 6, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9414421319961548, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/210614/proving-irrationals-are-dense-in-0-1 | # Proving irrationals are dense in (0, 1)
The definition of dense I'm using is the following: a subset $S$ of $(0, 1)$ is dense in $(0, 1)$ if for every $x \in (0, 1)$ and every $\epsilon > 0$ there exists $s \in S$ with $|s - x| < \epsilon$.
I'm trying to show that the set of irrationals in $(0, 1)$ (i.e. the set $\{x \in \mathbb{R} \setminus \mathbb{Q} : 0 < x< 1\}$ is dense in $(0, 1)$.
My idea is this: given $x \in (0, 1)$ and $N \in \mathbb{N}$, try to find an irrational number $\alpha \in (0, 1)$ that depends on $N$ and $x$ such that $|\alpha - x| < 1/N$. I'm not sure how to construct such an $\alpha$ though. I'm thinking I need to to take some irrational multiple of $\lfloor{Nx\rfloor}$ or something but I can't quite work it out. Can anyone help?
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1
If it wasn't the case there would exist a whole open interval $(a,b)$ without irrationals. – Martino Oct 10 '12 at 18:57
## 1 Answer
Hint:
Take any irrational number $\alpha$; now given $x$ then for every $\varepsilon$ take a rational number $q$ such that $\alpha\cdot q<\varepsilon$ and consider $s=x+\alpha\cdot q$.
(Such $q$ can be always taken $\frac1n$ for a large enough $n$)
- | {"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": 28, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9580705165863037, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/38885/how-do-astronomers-measure-the-distance-of-the-planets-from-its-star/38887 | # How do astronomers measure the distance of the planets from its star?
I know they have accurate means of measuring the distance of the planets in the Solar System from the Sun. I'm skeptical how can they use the same tools or techniques for other systems so far away as 22 light years.
According to this link the planet GJ667Cc resides in the "habitable zone". How are they able to measure the distance of something from something while observing it from a distance of 22 light years?
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– nico Oct 2 '12 at 6:01
Yes, possibly triangulation. But what I actually meant to ask is, do they really have the tools to accurately use triangulation as far as 22 light years? I mean, we can surely measure the distance of something using a ruler but if you're going to use it for long distances, the magnitude of error will surely increase (of course I'm just oversimplifying here but you get the idea). And for a distance of 22 light years I'm skeptical about its accuracy to be any meaningful at all. – supertonsky Oct 2 '12 at 6:09
1
There's no triangulation, since indeed that would be unusably imprecise. John Rennie's answer explains it all. However, you should take claims of "habitable zone" with a grain of salt. Everyone is very eager to find the first Earth analogue, and so sometimes claims of this nature get blown out of proportion. – Chris White Oct 2 '12 at 8:04
## 1 Answer
The Doppler shift in the light from the star tells you the period of the planet's orbit and also the velocity the star moves. You need to know the mass of the star, but this can be estimated to good accuracy from the star brightness and type. Once you know the mass of the star you can calculate the distance of the planet from it's period using:
$$r^3 = \frac{GM}{4\pi^2}P^2$$
where $M$ is the mass of the star and $P$ is the period of the oscillation.
Not that it's directly relevant to your question, but from the velocity of the star's oscillation we can calculate the minimum mass of the planet, because the velocity of the stars displacement depends on the gravitational force between the two. We can only calculate a minimum planet mass because if the plane of the system it tilted relative to us the true mass is higher than the one we calculate.
Having said this, these days most extrasolar planets are discovered because they transit their star, and these systems are not tilted relative to us (otherwise they wouldn't transit!). That means we can calculate an accurate mass for the planet.
In practice we normally turn the calculation over to a computer model (called a Bayesian Kepler periodogram if you want to Google it) because there are usually several planets and the oscillation is not a simple sine wave. We use a numerical fit to work out how many planets there are and how far from the star they are.
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– supertonsky Oct 2 '12 at 9:06
1
– John Rennie Oct 2 '12 at 9:33
I mentioned the planet mass only because estimating the planet mass is an important part of finding Earth-like planets. As you say, it doesn't affect the calculation of the star-planet distance. – John Rennie Oct 2 '12 at 9:34 | {"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}, "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}, "english_confidence": 0.9361861348152161, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/152021-indefinite-integral.html | # Thread:
1. ## Indefinite integral
Find the following indefinite integral, identifying any rules of calculus that you use:
$\int1/\sqrt{(1-x^2)*\arcsin(x)}$
I think I need to use integration by substitution, so have rearranged to give:
$\int(1-x^2)^{\-1/2} * (sin(x))^{1/2}$
I am not quite sure what to do next. The formula I have for integration by substitution is:
$\int{f(g(x))g'(x)dx} = \int{f(u)du}$, where u = g(x)
Would I then get
f(x) = $(1-x^2)^{\-1/2}$
and
g(x) = $(sin(x))^{1/2}$, so u = -cosx
or do I need to integrate both parts first as they are both composite integrals and then use the above formula on the answers I get?
Thanks for any help in advence
2. Originally Posted by cozza
Find the following indefinite integral, identifying any rules of calculus that you use:
$\int1/\sqrt{(1-x^2)*\arcsin(x)}$
$\displaystyle \int \frac{1}{\sqrt{\arcsin{x}}} \cdot \frac{1}{\sqrt{1-x^2}} \, dx$
$u = \arcsin{x}<br />$
$\displaystyle du = \frac{1}{\sqrt{1-x^2}} \, dx$
substitute ...
$\displaystyle \int \frac{1}{\sqrt{u}} \, du<br />$
finish it
3. I'll give you a huge hint. You will have to use the u-substitution, this identity:
$\displaystyle{\frac{d}{dx} arcsin(x) = \frac{1}{\sqrt{1-x^2}} = (1-x^2)^{-1/2}}$
and the integral rewritten as: $\displaystyle{\int\frac{1}{\sqrt{(1-x^2)\cdot arcsin(x)}}dx=\int (1-x^2)^{-1/2}\cdot(arcsin(x))^{-1/2}dx}$
This should work.
4. Hi, thanks for the help. Can I just check that the answer I have is right:
$1/(\sqrt{1-x^2})+c$
5. I think the answer is:
2sqrt(arcsinx) + c
6. Originally Posted by Also sprach Zarathustra
I think the answer is:
2sqrt(arcsinx) + c
Originally Posted by cozza
Hi, thanks for the help. Can I just check that the answer I have is right:
$1/(\sqrt{1-x^2})+c$
On rechecking my answer I don't think the square root should be in there. I think (hope) it is:
$1/(1-x^2)$ ?
Where does the 2 come from? Have I missed something?
7. Originally Posted by cozza
On rechecking my answer I don't think the square root should be in there. I think (hope) it is:
$1/(1-x^2)$ ?
Where does the 2 come from? Have I missed something?
yes, you have.
as stated earlier ...
$\displaystyle \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C<br />$
back substitute $u = \arcsin{x}$ ...
$2\sqrt{\arcsin{x}} + C$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9288584589958191, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/33504?sort=oldest | ## Degree sequences of multigraphs with bounded multiplicity
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I got to thinking about this problem while sifting through the math puzzles for dinner thread. There's a fun puzzle by rgrig which asks the guests to prove that when they came to dinner two of them shook hands the same amount of times. The solution is to make the handshake graph, and apply the pigeonhole principle on the vertex degrees. The key observation is that if there are $n$ guests, then the degrees 0 and $n-1$ cannot both occur.
Yaakov Baruch made the astute comment that the result is false if the forgetful mathematicians shake hands twice with each other. However, note that mathematicians are probably not so forgetful to shake hands three times with someone. This leads us to the following questions:
Question 1. Is it true that for large $n$, every loopless multigraph on $n$ vertices with at most two parallel edges between any two vertices has two vertices of the same degree? This is false for $n=3,4,5,6,7$.
Question 2. Is there a nice characterization of such degree sequences?
I'll finish with what is known. If we allow loops, then the problem is trivial, because all sequences whose sum is even are degree sequences. To see this, put a matching on the odd degree vertices, and then add loops. For simple graphs, there are many nice characterizations. For example, the Erdos-Gallai theorem says that a decreasing sequence $(d_1, \dots, d_n)$ is a degree sequence of a simple graph if and only if $\sum_i d_i$ is even and for all $k \in [n]$
$\sum_{i=1}^k d_i \leq k(k-1)+ \sum_{i=k+1}^n \min (k, d_i).$
This sort of answers Question 2, since a degree sequence of a multigraph with multiplicity at most 2 is the sum of two degree sequences of simple graphs. However, this is a rather convoluted characterization, and it is unclear how to answer Question 1 from it.
I'll end by mentioning that if we do not bound the multiplicity of edges, then there is a nice characterization of degree sequences of multigraphs by Hakimi (1962).
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## 1 Answer
The answer to Question 1 is no. Here's a construction. First we construct a simple graph $G_n$ on $n$ vertices for each $n\geqslant3$ which has every degree from $1$ to $n-1$ inclusive occurring, with the degree $n-1$ occurring only once. You do this recursively: to construct $G_n$, just take the complement of $G_{n-1}$ and add a vertex joined to all the others.
Now to construct the desired multigraph: label the vertices of $G_n$ as $v_1,\dots,v_n$ in such a way that the degrees of these vertices are increasing; then these degrees are $1,2,\dots,i,i,i+1,\dots,n-1$ for some $i < n-1$. Now just duplicate the edge from $v_j$ to $v_n$ for $j=i+1,\dots,n-1$, and you have your multigraph with distinct degrees $1,2,\dots,n-1,2n-i-2$.
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Thanks! Nice argument. – Tony Huynh Jul 27 2010 at 19:47 | {"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": 25, "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}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9474908113479614, "perplexity_flag": "head"} |
http://motls.blogspot.com.au/2012/06/have-starkman-and-pals-shown-that.html | # The Reference Frame
## Sunday, June 24, 2012
... /////
### Have Starkman and pals shown that the hierarchy problem doesn't exist?
The Higgs mass around $125\GeV$ together with the Higgs vev around $246\GeV$ and the masses of related particles such as the W-bosons and Z-bosons are much smaller than some other energy scales, e.g. the GUT scale and the Planck scale, where some new interesting effects almost certainly occur.
If we calculate the correction to the Higgs mass from quantum loops of virtual particles whose energy may be very high because we know that physics should make sense up to these very high energy scales, we obtain a quadratically divergent term, too:\[
\delta m_h^2 = \mp C\cdot \Lambda^2.
\] Here, $\Lambda$ is the maximum energy that particles may carry; of course, you would like to send it to infinity or at least $m_\text{GUT}\sim 10^{16}\GeV$ or $m_\text{Planck}\sim 10^{19}\GeV$ except that you can't as it creates havoc in the equations. Dimensional analysis shows that there's enough potential for such terms to arise and unless there is an explanation why they shouldn't arise, Gell-Mann's totalitarian principle pretty much guarantees that they will arise.
So the tree-level Higgs squared mass has to be finely adjusted to a huge number of order $\pm C\cdot \Lambda^2$ that almost completely (but not quite completely) cancels against the loop corrections that are equally high to yield a result that is as small as experiments suggest – and, incidentally, as small as we need for the existence of stars and life. The probability that the tree-level Higgs mass is so accurately adjusted to lead a tiny leftover after the loop corrections are added is small so theories of this kind apparently predict that our world with a light Higgs boson should be unlikely. That's too bad because if theories predict that the reality is unlikely, they don't look like too good theories. This problem with the "unnaturally low" observed Higgs mass is known as the hierarchy problem.
In April, I have explained that the paragraphs above only represent a technologically shallower way to discuss what the hierarchy problem is. A better way is to discuss a full-fledged theory that is valid up to the very high energy scales. This theory has lots of its own parameters – parameters appropriate for a high-energy scale – and these have to be carefully and unnaturally adjusted for a very light Higgs to emerge.
Recently, we've seen several papers arguing that the hierarchy problem is an illusion. Let me say in advance that I find it plausible that this is what physics will ultimately learn about the hierarchy problem; I just don't think that either of these physicists has actually found the right argument that would show such a thing.
First, John Ralston published a very fresh hep-ph preprint
Where and How to find susy: The auxiliary field interpretation of supersymmetry
Writing susy in between two dollar signs looks rather stupid but it's not the main problem with the paper. ;-)
Ralston presents examples that demonstrate that the Taylor expansions may be a lousy guide when we want to learn what happens with a function for very high values of the argument. The general claim is right but the examples he chooses have something illogical in them. For example, his example with a Gamma function is one in which the perturbative expansion actually correctly indicates how the function grows for very large values of the argument. It would probably be more logical to choose functions that converge to zero at infinity although the power laws indicate that they diverge.
But I don't want to get into vague interpretations or obvious mathematical tasks. Instead, let's get to the point. The point – the correct point, not Ralston's point – is that we don't have to rely on what Ralston calls the "perturbative" part of the calculation and what sane physicists call "low-energy expansions". Instead, we may and we should directly discuss the models that are valid up to very high energy scales. When we do so, the would-be "non-perturbative" (we actually mean "high-energy-physics-related") corrections are included. And when we do so, an extremely fine adjustment of the parameters of the high-energy theory is needed to yield light particles such as the Higgs boson. That's the hierarchy problem.
We can show that this problem exists in theories that look like ordinary enough theories at the high scale. This proof has no errors in which some vital "non-perturbative" effects would be neglected. We know it's right. The only non-anthropic way to fix the hierarchy problem is to change the theory, to switch to a new theory in which the sensitivity disappears. The known strategies include supersymmetry and its cancellations; new strongly coupled QCD-dynamics at a lower energy scale whose smallness may be explained by the very slow logarithmic running of the coupling; and, less usually, completely new theories resembling the string-theoretical "unexpectedly ultrasoft" short-distance behavior.
But one has to change the laws of physics. The laws of physics have to contain something unusual that actually invalidates the low-energy power-law arguments based on the quadratic divergences. If the theory has no special features, we know from "exact" high-energy calculations that the hierarchy problem is there and that the simple guess based on the quadratic divergences may indeed be extrapolated and produces the right conclusion.
Also, if Ralston's toy model were relevant, you would need not only quadratic but also quartic, sixth-order, and all higher-order divergent terms, too. That would make the theory really pathological and totally non-renormalizable as a quantum field theory although it could be a result of some stringy effects. But almost no one expects these intrinsically stringy effects to kick in below energies close to the GUT scale or Planck scale so it seems unlikely to the physicists that they could have something to do with the solution to the hierarchy problem. There may be a loophole in these expectations but once again, it doesn't seem as though Ralston has found such a loophole.
Starkman et al.
Four days ago, Glenn Starkman wrote a guest blog for the Scientific American whose goal was extremely similar but the strategy was a bit different (or, more precisely, completely different):
Beyond Higgs: On Supersymmetry (or Lack Thereof)
He argues that supersymmetry or any other new physics isn't needed as a solution of the hierarchy problem because the problem doesn't exist. He has (and his co-authors have) a different rhetorical strategy to show that the hierarchy problem doesn't exist: the Higgs mass may be linked to the masses of the Goldstone bosons – the angular directions of the Higgs doublet – which have a reason to remain massless, namely the Goldstone theorem.
Except that this claim is wrong at many levels. First of all, the Goldstone bosons don't stay massless in the Standard Model because they're eaten by the W-boson and Z-boson fields and become the gauge bosons' (equally massive) longitudinal polarizations. But even if you considered the spontaneous breaking of a global symmetry and not a local one, it wouldn't work because only the 3 Goldstone bosons – the "angular" components of the Higgs doublet, one for each generator of the symmetry group that is broken – have a reason to remain massless.
So any attempt to claim that the Higgs itself – the radial component – remains light or massless or unaffected by the quadratic divergences is an artifact of circular reasoning. The circular reasoning may have various forms. These guys, for example, decided to describe the system in terms of a low-energy theory based on pions etc. However, if the Higgs isn't made light to start with, the dynamics of this theory isn't described by this approximate low-energy effective theory at all. There aren't any low energies. You can't absorb divergences to the pion masses because the pion masses aren't among the degrees of freedom of the right description.
Just to be sure, I am talking about the following paper:
A Goldstone "Miracle": The Absence of a Higgs Fine Tuning Problem in the Spontaneously Broken $O(4)$ Linear Sigma Model
by Bryan W. Lynn, Glenn D. Starkman, Katherine Freese, Dmitry I. Podolsky. As far as I know, I only know Kathy Freese in person and I only know Dmitry Podolsky from intense communications on the Internet (he has a blog, nonequilibrium.net).
Under Starkman's essay in the Scientific American, I left the following comment.
Exactly one year ago, Dmitry Podolsky, a co-author of yours, had a correspondence – about 10 e-mails – with me. He argued in favor of a simpler “argument” than your paper’s argument why there were no quadratic divergences to the Higgs mass.
The claim was that they only affected the mass parameters so the quartic coupling wasn’t affected by them and the influence of the quadratic divergences on the Higgs mass is actually linked to the correction to the tadpole trying to change the vev (location of the minimum of the potential) which has to be zero for stability. So even the former thing is zero.
For hours if not days, I was a bit confused but then I came back to my senses and wrote him an explanation – one that he never replied to again. The explanation is that the correction of the tadpole is indeed linked to the correction of the Higgs mass and the vev. That’s OK but *none* of these mutually related things is guaranteed to be zero by any principle. In fact, we know in particular theories it’s not zero. The quadratic divergences reflect the big sensitivity of the Higgs mass/vev on all the detailed parameters of any high-energy theory with start with. We may choose to say whether it’s the Higgs mass or the Higgs vev that is threatened by these divergent terms but both of them are!
Without a principle that guarantees a cancellation, both the Higgs vev and the Higgs mass – together with masses of the W-bosons, Z-bosons, and some leptons and quarks – would be close to the GUT scale or any other high-energy scale we find in the full theory. Their fates and values could be correlated but there exists no argument for either of these quantities why they should be small.
Now, the final paper of yours talks about the pions etc. It is a bizarre treatment. From the viewpoint of the fundamental theory, pions are composite objects and their properties are derived quantities. We don’t really have special counterterms for pions or even technipions in the Standard Model. They’re not fundamental parameters. So at most, you link the divergences to yet another quantity that isn’t guaranteed to be zero by any principle. At most, you are supplying a new principle but this principle is equivalent to saying “there shouldn’t be a hierarchy problem”. You don’t have any independent justification why it’s not there.
It’s only the Goldstone bosons that have a reason to be massless, the Goldstone theorem, but there are only 3 of them, one for each broken generator, and the physical Higgs boson simply isn’t one of them. That’s why I believe that your paper boldly claiming that the quadratic divergences aren’t really there is wrong.
So I don't think that either of these authors has addressed the actual hierarchy problem. They have just addressed oversimplified beginners' caricatures of the problem. If they actually understood how the low-energy Higgs physics emerges from some complete theories of the known types that work up to very high energy scales, they would notice how ludicrous their "solutions" are.
We can explicitly see that the fine-tuning of the high-energy-based theory is needed to produce the light particles so hopes inspired by some unusual extrapolations of the incomplete, low-energy approximation of the situation have nothing to do with the reality. In other words, you can't solve the hierarchy problem by looking at it through strong and dirty eyeglasses and by claiming that it doesn't look too sharp in this way.
And that's the memo.
Posted by Luboš Motl
|
Other texts on similar topics: string vacua and phenomenology
#### snail feedback (9)
:
reader Dilaton said...
So with my very bad myopia, I probably should not look at the hierarchy problem without using any optical devices ... :-P
Nice article, I`ve already seen the corresponding question deiscussing the Starkman paper at physics.SE :-)
reader anynamenottaken said...
Assuming then that hierarchy problem does exist, do you (personally) believe the Randall+Sundrum model is a solution of it? After all what is above the cut-off of their effective theory... Would the Higgs not still receive quadratically divergent corrections from some UV physics above the cut-off, such as from string theory?
If so, does that leave SUSY as the only solution? I would be interested to know if you found plausible any alternative solutions to the hierarchy problem.
reader Physics Junkie said...
I read somewhere that the unification of the couplings is predominantly influenced by higgisinos. Is it plausible to postulate that the unification of couplings, a dark matter candidate and a solution to the hierachy problem can be solved with and absolute minimum extension of the standard model with
a supersymmetry of only the higgisinos?
reader Luboš Motl said...
Dear Physics Junkie, the reason why higgsinos are important for coupling unification is that leptons and quarks come in complete representations of the grand unified group so they don't affect the unification.
Otherwise no, you can't say what you said because there is no supersymmetry if you cherry-pick particle species one by one. Supersymmetry is a symmetry that - if it exists at all - must constrain the whole spectrum. The right unification of the MSSM is also obtained for higgsinos from two doublets and the fact that one needs two electroweak Higgs doublets (supermultiplets) and not just one has reasons that lie outside coupling unification: it's about the anomaly cancellation and the need to produce masses of both upper and lower quarks.
Moreover, more generally, if one were able to adjust gauge coupling unification by cherry-picking the field content, it wouldn't be impressive at all. The agreement in the MSSM is that the unification is achieved by doing nothing, just postulating SUSY and the minimal content. So even if your "higgsino only" theory would achieve coupling unification, it would be a far less interesting achievement than what one gets in the MSSM.
reader Luboš Motl said...
Hi, I think that RS is a very serious theory but I personally consider the probability that it's relevant in Nature - at accessible energies - to be just 1%-2%.
And no, once you get above the scale where string theory starts to be relevant, there are no quadratic divergences anymore. The dependence on energy gets hugely softer - a typical property of string theory. String theory has no UV divergences, after all.
I am not 100% sure whether SUSY is the right solution to the hierarchy problem. If it is, it should be ultimately be seen at the LHC, kind of. I guess that the probability of either of these statements is about 60%, far below 100% although above 50%.
There have been other proposals to solve the hierarchy problem that seemed worth considering to me. But instead of the technical ones that seem much less well-motivated to me than SUSY (and even than RS), i would mention the anthropic "solution": a light Higgs is simply needed for life so we find outselves in a world where the HIggs is light although there are many more worlds where it is not.
I dislike the anthropic principle but I do admit it's plausible that it will be ultimately needed to explain the lightness of the Higgs. 10% probability or so.
reader Luboš Motl said...
Right, Dilaton, I've seen it, too.
reader guest said...
I completely agree with Lubos but want to add an extra comment. Having light gauginos also plays an important role in unification as their contributions to the one-loop beta function coefficients decrease the slopes of the lines, thereby pushing the unification scale from 10^14 to 10^16 GeV, which helps avoid the experimental constraints from the proton decay.
reader Dmitry Podolsky said...
Also reposting my reply from Scientific American:
Dear Lumo,
My memory tells me that it was me who made the last reply in that
discussion but if I am wrong, I ask you to forgive me. My answer to your
observation was more or less as follows.
1. Stability
Whether renormalized tadpole is zero on not shows one whether Higgs
particles can be spontaneously produced from the vacuum or not. If they
can be produced, then the vacuum is not stable. The particles will be
produced from vacuum and the physical Higgs’ VEV will change until the
renormalized tadpole becomes zero.
A simple observation which goes back to work done by Bryan Lynn long
time ago (and which is actually present in the textbook by Peskin and
Schroder citing Bryan’s work) is that the quadratically divergent
contribution to the Higgs’ mass coincides identically with the
expression for the renormalized tadpole. Hence the first statement of
the paper – if renormalized vacuum is stable w.r.t. spontaneous particle
production, quadratic divergences are absent in the Higgs’ mass. This
observation is actually well known to CMT physicists. There are plenties
of phenomena in condensed matter physics involving spontaneously broken
global symmetries, and none of them feature quadratic (or linear –
because they are (2+1)d) divergences in the effective mass of
quasiparticles.
2. Massless modes
I am afraid you cannot simply say that there are several independent
generators of global symmetries, so divergences in Goldstone’s masses
are unrelated to divergences in Higgs’ mass. Even if the global symmetry
is broken spontaneously, effective potential should respect it in some
way or another – that’s why the word “spontaneously” (instead of
“explicitly”) is used. Whether pions are composite or fundamental
objects again does not matter – once you have an effective
renormalizable lagrangian for low energy degrees of freedom, you can and
should study the fate of divergences appearing in the effective theory.
It so happens that global symmetry relates some contributions to the
masses of Goldstones and the Higgs, namely, quadratically divergent
contributions. This statement is again known for years and can be found
for example in Peskin and Schroder. So, if you want to see that the
Goldstone theorem holds explicitly at a given energy scale (i.e,
renormalized mass of Goldstones is zero), you should conclude that the
quadratically divergent contribution to the Higgs’ mass also vanishes.
I share very much your sentiment about the sensitivity of the Higgs’
mass to parameters of HE theories (that’s why we always thought we need
SUSY, Technicolor, etc., etc., etc), but that’s unfortunately not how it
works in these theories.
Thanks for the link by the way!
Cheers,
Dmitry.
reader Luboš Motl said...
Dear Dmitry, thanks for your comment but I think that you're just repeating the very same observation you sent in the very first e-mail in our conversation, with all the sloppiness, too.
Yes, the renormalization of the Higgs mass is linked to the tadpole - which changes the values of the vev, the point where the vacuum is stable - but *none* of these two related quantities is protected by any principle against huge corrections in the Standard Model. Instead, the huge corrections want to make the Higgs mass very large; and they also want to make the Higgs vev - the expectation value around which we have to expand in order to get a vanishing tadpole - very large. Haven't I explained this in the blog entry above, too?
Concerning your second point, it's just completely wrong because the Higgs mode isn't related to the Goldstone mode by any symmetry - neither in the broken phase nor in the unbroken phase. Goldstone modes are related to each other by the symmetry - they're the angular modes. But the Higgs boson is the radial component and there's simply no symmetry that exchanges e.g. "r" and "phi" in polar coordinates.
Whether you may find "some of the same terms" in two expressions is totally irrelevant. What's important is that this is not true for *all* the terms - which could only be true because of a symmetry or "something equally strong" but there simply isn't anything of the sort.
Note that none of your argument above carries any resemblance to your paper with Starkman and others - you don't talk about any pions etc. in this comment although this is the focus of your paper. So it suggests that even you don't believe the paper you have co-authored. To summarize, there are just several sloppy arguments with clear errors in them. But several sloppy arguments can't replace a valid argument.
My e-mail archive shows that the last message of the conversation was written by me, on June 7th, 2011, and it did find the very same point. I wrote:
Dear Dmitry,I forgot to write that I was silly and I don't think that there is any confusion about your question on the Higgs divergences. (the correction to the Higgs' mass quadratic in cutoff has the same structure as tadpole corrections and the latter are zero by stability of the effective potential)This is a loaded interpretation. While the corrections to the mass are linked to the tadpole, and the tadpole is indeed the same thing as the stability of the effective potential around the same point, NONE of these three things is guaranteed to be zero by any principle. So yes, the quadratic divergences change the Higgs mass and they also change the Higgs vev (the value of the Higgs field at which the potential has a minimum). At any rate, these terms are there and they affect both quantities.
Cheers
LM | {"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": 7, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9450902938842773, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-equations/58911-initial-value-problem.html | # Thread:
1. ## Initial value problem
Hi.
Solve the Initial value problem:
$\dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t$
(theres nothing said about 'a' and 'b')
does anyone know how to solve it, because i dont
Thanks so much!
Rapha
2. Originally Posted by Rapha
Hi.
Solve the Initial value problem:
$\dot{u}(t) = \begin{pmatrix} a & 1 \\ 0 & b \end{pmatrix}\cdot u(t) \ , \ t\in [0,T] \ , \ u(0) = (1,1)^t$
(there's nothing said about 'a' and 'b')
Let $u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}$. Then $\begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri x}$. This says that $\dot{x} = ax+y,\ \dot{y} = by$, with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve. Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).
3. Hey Opalg,
I really appreciate your answer, thank you
Originally Posted by Opalg
Let $u(t) = \begin{bmatrix}x(t)\\y(t)\end{bmatrix}$. Then $\begin{bmatrix}\dot{x}(t)\\\dot{y}(t)\end{bmatrix} = \begin{bmatrix}a & 1 \\ 0 & b\end{bmatrix}\begin{bmatrix}x(t)\\y(t)\end{bmatri x}$. This says that $\dot{x} = ax+y,\ \dot{y} = by$, with initial conditions x(0)=y(0)=1. The equation for y should be easy to solve.
Indeed, $y(t) = e^{bt} \Rightarrow \dot{y} = y'(t) = b*e^{bt}$
Originally Posted by Opalg
Then substitute the formula for y into the equation for x. Again, it will be a reasonably easy ordinary differential equation (but there will be two different forms of the solution, depending on whether a=b or not).
$\dot{x} = ax(t)+e^{bt}$
I dont get it
4. Originally Posted by Rapha
$\dot{x} = ax(t)+e^{bt}$
I don't get it
If a ≠ b then you should be able to find a solution of the form $x(t) = Ae^{at} + Be^{bt}$, for suitable constants A and B.
If a = b then the solution will be of the form $x(t) = (At+B)e^{at}$. | {"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": 13, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8945712447166443, "perplexity_flag": "middle"} |
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http://www.sagemath.org/doc/reference/matrices/sage/matrix/matrix_symbolic_dense.html | # Symbolic matrices¶
Symbolic matrices
Matrices with symbolic entries. The underlying representation is a pointer to a Maxima object.
EXAMPLES:
```sage: matrix(SR, 2, 2, range(4))
[0 1]
[2 3]
sage: matrix(SR, 2, 2, var('t'))
[t 0]
[0 t]
```
Arithmetic:
```sage: -matrix(SR, 2, range(4))
[ 0 -1]
[-2 -3]
sage: m = matrix(SR, 2, [1..4]); sqrt(2)*m
[ sqrt(2) 2*sqrt(2)]
[3*sqrt(2) 4*sqrt(2)]
sage: m = matrix(SR, 4, [1..4^2])
sage: m * m
[ 90 100 110 120]
[202 228 254 280]
[314 356 398 440]
[426 484 542 600]
sage: m = matrix(SR, 3, [1, 2, 3]); m
[1]
[2]
[3]
sage: m.transpose() * m
[14]
```
Computing inverses:
```sage: M = matrix(SR, 2, var('a,b,c,d'))
sage: ~M
[-b*c/((b*c/a - d)*a^2) + 1/a b/((b*c/a - d)*a)]
[ c/((b*c/a - d)*a) -1/(b*c/a - d)]
sage: (~M*M).simplify_rational()
[1 0]
[0 1]
sage: M = matrix(SR, 3, 3, range(9)) - var('t')
sage: (~M * M).simplify_rational()
[1 0 0]
[0 1 0]
[0 0 1]
sage: matrix(SR, 1, 1, 1).inverse()
[1]
sage: matrix(SR, 0, 0).inverse()
[]
sage: matrix(SR, 3, 0).inverse()
Traceback (most recent call last):
...
ArithmeticError: self must be a square matrix
```
Transposition:
```sage: m = matrix(SR, 2, [sqrt(2), -1, pi, e^2])
sage: m.transpose()
[sqrt(2) pi]
[ -1 e^2]
```
.T is a convenient shortcut for the transpose:
```sage: m.T
[sqrt(2) pi]
[ -1 e^2]
```
Test pickling:
```sage: m = matrix(SR, 2, [sqrt(2), 3, pi, e]); m
[sqrt(2) 3]
[ pi e]
sage: TestSuite(m).run()
```
Comparison:
```sage: m = matrix(SR, 2, [sqrt(2), 3, pi, e])
sage: cmp(m,m)
0
sage: cmp(m,3) != 0
True
sage: m = matrix(SR,2,[1..4]); n = m^2
sage: (exp(m+n) - exp(m)*exp(n)).simplify_rational() == 0 # indirect test
True
```
Determinant:
```sage: M = matrix(SR, 2, 2, [x,2,3,4])
sage: M.determinant()
4*x - 6
sage: M = matrix(SR, 3,3,range(9))
sage: M.det()
0
sage: t = var('t')
sage: M = matrix(SR, 2, 2, [cos(t), sin(t), -sin(t), cos(t)])
sage: M.det()
sin(t)^2 + cos(t)^2
```
Permanents:
```sage: M = matrix(SR, 2, 2, [x,2,3,4])
sage: M.permanent()
4*x + 6
```
Rank:
```sage: M = matrix(SR, 5, 5, range(25))
sage: M.rank()
2
sage: M = matrix(SR, 5, 5, range(25)) - var('t')
sage: M.rank()
5
.. warning::
:meth:`rank` may return the wrong answer if it cannot determine that a
matrix element that is equivalent to zero is indeed so.
```
Copying symbolic matrices:
```sage: m = matrix(SR, 2, [sqrt(2), 3, pi, e])
sage: n = copy(m)
sage: n[0,0] = sin(1)
sage: m
[sqrt(2) 3]
[ pi e]
sage: n
[sin(1) 3]
[ pi e]
```
Conversion to Maxima:
```sage: m = matrix(SR, 2, [sqrt(2), 3, pi, e])
sage: m._maxima_()
matrix([sqrt(2),3],[%pi,%e])
```
class sage.matrix.matrix_symbolic_dense.Matrix_symbolic_dense¶
Bases: sage.matrix.matrix_generic_dense.Matrix_generic_dense
See Matrix_generic_dense for documentation.
TESTS:
We check that the problem related to Trac #9049 is not an issue any more:
```sage: S.<t>=PolynomialRing(QQ)
sage: F.<q>=QQ.extension(t^4+1)
sage: R.<x,y>=PolynomialRing(F)
sage: M = MatrixSpace(R, 1, 2)
sage: from sage.matrix.matrix_generic_dense import Matrix_generic_dense
sage: Matrix_generic_dense(M, (x, y), True, True)
[x y]
```
arguments()¶
Returns a tuple of the arguments that self can take.
EXAMPLES:
```sage: var('x,y,z')
(x, y, z)
sage: M = MatrixSpace(SR,2,2)
sage: M(x).arguments()
(x,)
sage: M(x+sin(x)).arguments()
(x,)
```
charpoly(var='x', algorithm=None)¶
Compute the characteristic polynomial of self, using maxima.
EXAMPLES:
```sage: M = matrix(SR, 2, 2, var('a,b,c,d'))
sage: M.charpoly('t')
t^2 + (-a - d)*t + a*d - b*c
sage: matrix(SR, 5, [1..5^2]).charpoly()
x^5 - 65*x^4 - 250*x^3
```
TESTS:
The cached polynomial should be independent of the var argument (trac ticket #12292). We check (indirectly) that the second call uses the cached value by noting that its result is not cached:
```sage: M = MatrixSpace(SR, 2)
sage: A = M(range(0, 2^2))
sage: type(A)
<type 'sage.matrix.matrix_symbolic_dense.Matrix_symbolic_dense'>
sage: A.charpoly('x')
x^2 - 3*x - 2
sage: A.charpoly('y')
y^2 - 3*y - 2
sage: A._cache['charpoly']
x^2 - 3*x - 2
```
eigenvalues(solution_set=False)¶
Compute the eigenvalues by solving the characteristic polynomial in maxima
EXAMPLES:
```sage: a=matrix(SR,[[1,2],[3,4]])
sage: a.eigenvalues()
[-1/2*sqrt(33) + 5/2, 1/2*sqrt(33) + 5/2]
```
eigenvectors_left()¶
Compute the left eigenvectors of a matrix.
For each distinct eigenvalue, returns a list of the form (e,V,n) where e is the eigenvalue, V is a list of eigenvectors forming a basis for the corresponding left eigenspace, and n is the algebraic multiplicity of the eigenvalue.
EXAMPLES:
```sage: A = matrix(SR,3,3,range(9)); A
[0 1 2]
[3 4 5]
[6 7 8]
sage: es = A.eigenvectors_left(); es
[(-3*sqrt(6) + 6, [(1, -1/5*sqrt(6) + 4/5, -2/5*sqrt(2)*sqrt(3) + 3/5)], 1), (3*sqrt(6) + 6, [(1, 1/5*sqrt(6) + 4/5, 2/5*sqrt(2)*sqrt(3) + 3/5)], 1), (0, [(1, -2, 1)], 1)]
sage: eval, [evec], mult = es[0]
sage: delta = eval*evec - evec*A
sage: abs(abs(delta)) < 1e-10
abs(sqrt(144/25*(sqrt(2)*sqrt(3) - sqrt(6))^2 + 1/25*(3*(sqrt(6) - 2)*(2*sqrt(2)*sqrt(3) - 3) + 16*sqrt(2)*sqrt(3) + 5*sqrt(6) - 54)^2 + 1/25*(3*(sqrt(6) - 4)*(sqrt(6) - 2) + 14*sqrt(2)*sqrt(3) + 4*sqrt(6) - 42)^2)) < (1.00000000000000e-10)
sage: abs(abs(delta)).n() < 1e-10
True
```
```sage: A = matrix(SR, 2, 2, var('a,b,c,d'))
sage: A.eigenvectors_left()
[(1/2*a + 1/2*d - 1/2*sqrt(a^2 - 2*a*d + 4*b*c + d^2), [(1, -1/2*(a - d + sqrt(a^2 - 2*a*d + 4*b*c + d^2))/c)], 1), (1/2*a + 1/2*d + 1/2*sqrt(a^2 - 2*a*d + 4*b*c + d^2), [(1, -1/2*(a - d - sqrt(a^2 - 2*a*d + 4*b*c + d^2))/c)], 1)]
sage: es = A.eigenvectors_left(); es
[(1/2*a + 1/2*d - 1/2*sqrt(a^2 - 2*a*d + 4*b*c + d^2), [(1, -1/2*(a - d + sqrt(a^2 - 2*a*d + 4*b*c + d^2))/c)], 1), (1/2*a + 1/2*d + 1/2*sqrt(a^2 - 2*a*d + 4*b*c + d^2), [(1, -1/2*(a - d - sqrt(a^2 - 2*a*d + 4*b*c + d^2))/c)], 1)]
sage: eval, [evec], mult = es[0]
sage: delta = eval*evec - evec*A
sage: delta.apply_map(lambda x: x.full_simplify())
(0, 0)
```
This routine calls Maxima and can struggle with even small matrices with a few variables, such as a $$3\times 3$$ matrix with three variables. However, if the entries are integers or rationals it can produce exact values in a reasonable time. These examples create 0-1 matrices from the adjacency matrices of graphs and illustrate how the format and type of the results differ when the base ring changes. First for matrices over the rational numbers, then the same matrix but viewed as a symbolic matrix.
```sage: G=graphs.CycleGraph(5)
sage: am = G.adjacency_matrix()
sage: spectrum = am.eigenvectors_left()
sage: qqbar_evalue = spectrum[2][0]
sage: type(qqbar_evalue)
<class 'sage.rings.qqbar.AlgebraicNumber'>
sage: qqbar_evalue
0.618033988749895?
sage: am = G.adjacency_matrix().change_ring(SR)
sage: spectrum = am.eigenvectors_left()
sage: symbolic_evalue = spectrum[2][0]
sage: type(symbolic_evalue)
<type 'sage.symbolic.expression.Expression'>
sage: symbolic_evalue
1/2*sqrt(5) - 1/2
sage: qqbar_evalue == symbolic_evalue
True
```
A slightly larger matrix with a “nice” spectrum.
```sage: G=graphs.CycleGraph(6)
sage: am = G.adjacency_matrix().change_ring(SR)
sage: am.eigenvectors_left()
[(-1, [(1, 0, -1, 1, 0, -1), (0, 1, -1, 0, 1, -1)], 2), (1, [(1, 0, -1, -1, 0, 1), (0, 1, 1, 0, -1, -1)], 2), (-2, [(1, -1, 1, -1, 1, -1)], 1), (2, [(1, 1, 1, 1, 1, 1)], 1)]
```
eigenvectors_right()¶
Compute the right eigenvectors of a matrix.
For each distinct eigenvalue, returns a list of the form (e,V,n) where e is the eigenvalue, V is a list of eigenvectors forming a basis for the corresponding right eigenspace, and n is the algebraic multiplicity of the eigenvalue.
EXAMPLES:
```sage: A = matrix(SR,2,2,range(4)); A
[0 1]
[2 3]
sage: right = A.eigenvectors_right(); right
[(-1/2*sqrt(17) + 3/2, [(1, -1/2*sqrt(17) + 3/2)], 1), (1/2*sqrt(17) + 3/2, [(1, 1/2*sqrt(17) + 3/2)], 1)]
```
The right eigenvectors are nothing but the left eigenvectors of the transpose matrix:
```sage: left = A.transpose().eigenvectors_left(); left
[(-1/2*sqrt(17) + 3/2, [(1, -1/2*sqrt(17) + 3/2)], 1), (1/2*sqrt(17) + 3/2, [(1, 1/2*sqrt(17) + 3/2)], 1)]
sage: right[0][1] == left[0][1]
True
```
exp()¶
Return the matrix exponential of this matrix $$X$$, which is the matrix
$e^X = \sum_{k=0}^{\infty} \frac{X^k}{k!}.$
This function depends on maxima’s matrix exponentiation function, which does not deal well with floating point numbers. If the matrix has floating point numbers, they will be rounded automatically to rational numbers during the computation.
EXAMPLES:
```sage: m = matrix(SR,2, [0,x,x,0]); m
[0 x]
[x 0]
sage: m.exp()
[1/2*(e^(2*x) + 1)*e^(-x) 1/2*(e^(2*x) - 1)*e^(-x)]
[1/2*(e^(2*x) - 1)*e^(-x) 1/2*(e^(2*x) + 1)*e^(-x)]
sage: exp(m)
[1/2*(e^(2*x) + 1)*e^(-x) 1/2*(e^(2*x) - 1)*e^(-x)]
[1/2*(e^(2*x) - 1)*e^(-x) 1/2*(e^(2*x) + 1)*e^(-x)]
```
Exp works on 0x0 and 1x1 matrices:
```sage: m = matrix(SR,0,[]); m
[]
sage: m.exp()
[]
sage: m = matrix(SR,1,[2]); m
[2]
sage: m.exp()
[e^2]
```
Commuting matrices $$m, n$$ have the property that $$e^{m+n} = e^m e^n$$ (but non-commuting matrices need not):
```sage: m = matrix(SR,2,[1..4]); n = m^2
sage: m*n
[ 37 54]
[ 81 118]
sage: n*m
[ 37 54]
[ 81 118]
sage: a = exp(m+n) - exp(m)*exp(n)
sage: a.simplify_rational() == 0
True
```
The input matrix must be square:
```sage: m = matrix(SR,2,3,[1..6]); exp(m)
Traceback (most recent call last):
...
ValueError: exp only defined on square matrices
```
In this example we take the symbolic answer and make it numerical at the end:
```sage: exp(matrix(SR, [[1.2, 5.6], [3,4]])).change_ring(RDF)
[346.557487298 661.734590934]
[354.500673715 677.424782765]
```
Another example involving the reversed identity matrix, which we clumsily create:
```sage: m = identity_matrix(SR,4); m = matrix(list(reversed(m.rows()))) * x
sage: exp(m)
[1/2*(e^(2*x) + 1)*e^(-x) 0 0 1/2*(e^(2*x) - 1)*e^(-x)]
[ 0 1/2*(e^(2*x) + 1)*e^(-x) 1/2*(e^(2*x) - 1)*e^(-x) 0]
[ 0 1/2*(e^(2*x) - 1)*e^(-x) 1/2*(e^(2*x) + 1)*e^(-x) 0]
[1/2*(e^(2*x) - 1)*e^(-x) 0 0 1/2*(e^(2*x) + 1)*e^(-x)]
```
expand()¶
Operates point-wise on each element.
EXAMPLES:
```sage: M = matrix(2, 2, range(4)) - var('x')
sage: M*M
[ x^2 + 2 -2*x + 3]
[ -4*x + 6 (x - 3)^2 + 2]
sage: (M*M).expand()
[ x^2 + 2 -2*x + 3]
[ -4*x + 6 x^2 - 6*x + 11]
```
factor()¶
Operates point-wise on each element.
EXAMPLES:
```sage: M = matrix(SR, 2, 2, x^2 - 2*x + 1); M
[x^2 - 2*x + 1 0]
[ 0 x^2 - 2*x + 1]
sage: M.factor()
[(x - 1)^2 0]
[ 0 (x - 1)^2]
```
fcp(var='x')¶
Return the factorization of the characteristic polynomial of self.
INPUT:
• var - (default: ‘x’) name of variable of charpoly
EXAMPLES:
```sage: a = matrix(SR,[[1,2],[3,4]])
sage: a.fcp()
x^2 - 5*x - 2
sage: [i for i in a.fcp()]
[(x^2 - 5*x - 2, 1)]
sage: a = matrix(SR,[[1,0],[0,2]])
sage: a.fcp()
(x - 2) * (x - 1)
sage: [i for i in a.fcp()]
[(x - 2, 1), (x - 1, 1)]
sage: a = matrix(SR, 5, [1..5^2])
sage: a.fcp()
(x^2 - 65*x - 250) * x^3
sage: list(a.fcp())
[(x^2 - 65*x - 250, 1), (x, 3)]
```
is_simplified()¶
Return True if self is the result of running simplify() on a symbolic matrix. This has the semantics of ‘has_been_simplified’.
EXAMPLES:
```sage: var('x,y,z')
(x, y, z)
sage: m = matrix([[z, (x+y)/(x+y)], [x^2, y^2+2]]); m
[ z 1]
[ x^2 y^2 + 2]
sage: m.is_simplified()
doctest:...: DeprecationWarning: is_simplified is deprecated
See http://trac.sagemath.org/6115 for details.
False
sage: ms = m.simplify(); ms
[ z 1]
[ x^2 y^2 + 2]
sage: m.is_simplified()
False
sage: ms.is_simplified()
False
```
number_of_arguments()¶
Returns the number of arguments that self can take.
EXAMPLES:
```sage: var('a,b,c,x,y')
(a, b, c, x, y)
sage: m = matrix([[a, (x+y)/(x+y)], [x^2, y^2+2]]); m
[ a 1]
[ x^2 y^2 + 2]
sage: m.number_of_arguments()
3
```
simplify()¶
Simplifies self.
EXAMPLES:
```sage: var('x,y,z')
(x, y, z)
sage: m = matrix([[z, (x+y)/(x+y)], [x^2, y^2+2]]); m
[ z 1]
[ x^2 y^2 + 2]
sage: m.simplify()
[ z 1]
[ x^2 y^2 + 2]
```
simplify_rational()¶
EXAMPLES:
```sage: M = matrix(SR, 3, 3, range(9)) - var('t')
sage: (~M*M)[0,0]
-(3*((6/t + 7)/((t - 3/t - 4)*t) + 2/t)*((6/t + 5)/((t - 3/t - 4)*t) + 2/t)/((6/t + 5)*(6/t + 7)/(t - 3/t - 4) - t + 12/t + 8) - 1/t - 3/((t - 3/t - 4)*t^2))*t + 3*(6/t + 7)*((6/t + 5)/((t - 3/t - 4)*t) + 2/t)/((t - 3/t - 4)*((6/t + 5)*(6/t + 7)/(t - 3/t - 4) - t + 12/t + 8)) + 6*((6/t + 5)/((t - 3/t - 4)*t) + 2/t)/((6/t + 5)*(6/t + 7)/(t - 3/t - 4) - t + 12/t + 8) - 3/((t - 3/t - 4)*t)
sage: expand((~M*M)[0,0])
1
sage: (~M * M).simplify_rational()
[1 0 0]
[0 1 0]
[0 0 1]
```
simplify_trig()¶
EXAMPLES:
```sage: theta = var('theta')
sage: M = matrix(SR, 2, 2, [cos(theta), sin(theta), -sin(theta), cos(theta)])
sage: ~M
[-sin(theta)^2/((sin(theta)^2/cos(theta) + cos(theta))*cos(theta)^2) + 1/cos(theta) -sin(theta)/((sin(theta)^2/cos(theta) + cos(theta))*cos(theta))]
[ sin(theta)/((sin(theta)^2/cos(theta) + cos(theta))*cos(theta)) 1/(sin(theta)^2/cos(theta) + cos(theta))]
sage: (~M).simplify_trig()
[ cos(theta) -sin(theta)]
[ sin(theta) cos(theta)]
```
variables()¶
Returns the variables of self.
EXAMPLES:
```sage: var('a,b,c,x,y')
(a, b, c, x, y)
sage: m = matrix([[x, x+2], [x^2, x^2+2]]); m
[ x x + 2]
[ x^2 x^2 + 2]
sage: m.variables()
(x,)
sage: m = matrix([[a, b+c], [x^2, y^2+2]]); m
[ a b + c]
[ x^2 y^2 + 2]
sage: m.variables()
(a, b, c, x, y)
```
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http://math.stackexchange.com/questions/3750/aunt-and-uncles-fuel-oil-tank-dip-stick-problem?answertab=votes | # Aunt and Uncle's fuel oil tank dip stick problem
This problem first came to me in high school, and a couple times since, and I even assigned it for extra credit in one of my calculus classes after I became a teacher. So I know the solution. What I am looking for is other WAYS to obtain the solution. I've been told there exists a solution using only arithmetic, but have never figured it out. Other solutions using ordinary calculus, trigonometry, algebra of conic sections, and so on are also possible.
The problem is usually stated in the form of a letter from an Aunt and Uncle:
Dear niece/nephew, How are things going for you and your folks? We hear you are doing quite well it school. Keep it up! Given this success, we were hoping you could help us figure out a little dilemma. As you know, our home is heated by fuel oil, and we have a big tank buried in the side yard. The tank is a cylinder, 20 feet long and 10 feet in diameter, lying on its side five feet deep, with a narrow tube coming to a fill cap at ground level. Your uncle has a 15 foot length of old pipe that we'd like to utilize as a dip stick in order to know when we are getting close to needing a fill-up. We know that 0 feet is empty, 5 feet is half full, and 10 feet is completely full. Trouble is, we don't know how to mark any other points. We are pretty sure they will not be uniformly spaced. What we really want is to know, within the nearest 0.01 foot, where to mark the dip stick for every multiple of 10% from 0% to 100%. Can you figure this out for us? Of course, we will want to see details of your solution and check it ourselves, and it would especially help if you could draw us a scale model of the dip stick. Love, Auntie Flo and Uncle Jim
That last sentence shows the teacher influence on the problem. So, my challenge to this community is not to find any old solution, but to find the solution at the lowest possible grade level, so to speak.
Thanks.
UPDATE: To those who are focusing in on the .01 feet accuracy, I apologize. The intent was merely to state, it is acceptable to estimate. If the exact answer is sqrt(2)*pi/2 or some other silly thing, go ahead and just write 2.22 feet, for example.
-
12
Empty tank. Add 10% capacity of oil, dip pipe, mark pipe just above the black stuff. Repeat. How's that for "lowest possible grade level"? You don't even have to clean off the pipe in between steps. :) (Offered as comment, since it doesn't satisfy the requirements of the question.) – Blue Sep 1 '10 at 1:01
I think my car must have been designed by these people because the digital odometer is more accurate than the analog fuel gauge. – Dan Brumleve Sep 1 '10 at 1:13
– Dan Brumleve Sep 1 '10 at 2:50
I've updated my answer to the +/-.01 feet accuracy range. – Américo Tavares Sep 2 '10 at 19:09
All: We have a few excellent responses so far, and I have up voted them. Please allow me to leave this one open for a while before accepting an answer. I really am curious what variety of response this will generate. – cobaltduck Sep 6 '10 at 21:54
## 4 Answers
Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.
Lower figure: graph of oil volume/max. volume (in %) versus oil level $l$ (in feet). The horizontal straight lines represent the area/volume ratio $A(l)/A(10)=V(l)/V(10)$ (in %) for every multiple of 10% from 0% to 100%.
--
Since the tank radius is $5$, the oil level with respect to the bottom of the tank is given by $l=5-5\cos \frac{\theta }{2}$, where $\theta$ is the central angle as shown in the figure. The area of the tank cross section filled with oil is
$A(\theta )=\frac{25}{2}\theta -\frac{25}{2}\sin \theta$
or
$A(l)=25\arccos (\frac{5-l}{5})-\frac{25}{2}\sin (2\arccos (\frac{5-l}{5}))$
The area ratio $A(l)/A(10)=V(l)/V(10)$ where $V(l)$ is the oil volume.
Let $f(l)$ denote this area ratio in percentage:
$f(l)=\frac{100}{\pi }\arccos \left( 1-\frac{1}{5}l\right) -\frac{50}{\pi }% \sin \left( 2\arccos \left( 1-\frac{1}{5}l\right) \right)$
Here are the the sequence of $f(l)$ for $l=0,1,2,\ldots ,10$. The graph of $% f(l)$ is shown above.
$f(0)=0$, $f(1)=5.2044$, $f(2)=14.238$, $f(3)=25.232$, $f(4)=37.353$, $f(5)=50$,
$f(6)=62.647$, $f(7)=74.768$, $f(8)=85.762$, $f(9)=94.796$, $f(10)=100$
Edit: One still needs to solve the nonlinear equation $f(l)-10k=0$ for $k=1,2,3,4,6,7,8,9$, e. g. by the Secant Method.
Edit 2: The problem of solving graphically, as shown in the secong figure, is that it would be very difficult, or impossible, to get the required accuracy of 0.01 (feet).
Update: The oil level marks (in feet) should be placed at
$0,1.57,2.54,3.40,4.21,$
$5,5.79,6.60,7.46,8.44,10$
corresponding to the oil volume percentage of
$0,10,20,30,40,$
$50,60,70,80,90,100$.
This calculation was based on the following $f$ function values:
$f(0)=0.0$, $f(1.5648)=10.0$, $f(2.5407)=20.0$, $f(3.40155)=30.0$, $f(4.21135)=40.0$
$f(5)=50.0$, $f(5.7887)=60.0$, $f(6.59845)=70.000$, $f(7.4593)=80.000$,
$f(8.4352)=90.000$, $f(10)=100.0$
Update 2 Figure of marks:
[Rearranged to show the sequence of editions and updates.]
-
The graph looks nicely linear in the middle; maybe interpolation might more expediently give intermediate values than having to solve a nonlinear equation. – J. M. Sep 1 '10 at 12:44
J. M. Yes, more expediently, but the question asks for an accuracy of 0.01 (feet). – Américo Tavares Sep 1 '10 at 13:30
1
+1: For finding f(i). – Aryabhata Sep 1 '10 at 16:24
@Moron: Thanks! I have added the "scale model of the dip stick". – Américo Tavares Sep 16 '10 at 21:20
Correction: "Here is the sequence of" instead of "Here are the the sequence of" – Américo Tavares Sep 17 '10 at 20:32
It is sufficient to consider the circular cross-section of the tank and volumes below 50% (marks for those above 50% are the reflection image of those below 50% across the 50% mark). Consider the radius of the tank to be 1 unit. For some amount of oil in the tank, consider the central angle formed by the points on the circular cross-section at the top of the oil and the center of the circle--call this $\alpha$, measured in radians.
The area of the cross-section of the oil is the area of the sector determined by $\alpha$ ($\frac{\alpha}{2\pi}\pi r^2=\frac{\alpha}{2}$) minus the area of the triangle that is part of the sector but not part of the cross-section of oil ($\frac{1}{2}ab\sin C=\frac{\sin\alpha}{2}$), $\frac{1}{2}(\alpha-\sin\alpha)$. The portion of the circular cross-section (and hence the portion of the volume) corresponding to this angle is $\frac{\alpha-\sin\alpha}{2\pi}$. Setting this expression equal to 0.1, 0.2, 0.3, and 0.4 and solving for $\alpha$ will give the values of $\alpha$ corresponding to 10%, 20%, 30%, and 40% full--solving here is done numerically/graphically, as there is no algebraic method to solve these equations. For each value of $\alpha$, the distance from the center of the circle to the oil level is $\cos\frac{\alpha}{2}$, so the depth of the oil is $1-\cos\frac{\alpha}{2}$. (Note that these are for a radius of 1 unit and need to be rescaled for the original problem's specific numbers.)
-
@Isaac: I think you have wanted to write $\frac{\alpha}{2}r^2$, in your 1st formula (anyway this doesn't matter since you are using the unitary radius). – Américo Tavares Sep 1 '10 at 10:58
@Américo Tavares: My intent was that I was substituting in 1 for r across that = in the first formula, hence no more $r^2$ term. – Isaac Sep 1 '10 at 11:25
@Isaac: I was able to see it, but only in your note at the bottom line. – Américo Tavares Sep 1 '10 at 15:06
+1 for your elegant solution. – Américo Tavares Sep 1 '10 at 19:36
Flo and Jim say: This answer is really getting to the heart of the problem (i.e. reducing volume in the cylinder to area under a chord of a circle, reducing to the simplest case then going back and substituting actual values, etc.) But can you tell us how your angle alpha relates to h, the height of the shaded region along a vertical radius? That is how the dipstick will be marked. – cobaltduck Sep 1 '10 at 23:45
show 1 more comment
I will give the calculus-based solution myself just for the sake of argument, taking note that I am still hoping to obtain a wide variety of other solutions, if possible.
For this, I am going to mentally rotate the tank (or the oil within it) $90^\circ$ clockwise, cut it in half, and center it at $(0,0)$, so that the upper half of the tank is represented by $y=\sqrt{25-x^2}$ and the volume of the oil by $\int_{-5}^h \sqrt{25-x^2}\mathrm{d}x$.
I know from simple $A=\pi r^2$ that the total cross-section area of the tank is $25\pi$ and thus for the upper half from -5 to +5 is $12.5\pi$. I will set the result of the integral to the various 10% proportions of this value, knowing that 10% of the upper half will occur at the same position as 10% of the entire circle etc.
The integral is $\frac12\left(x\sqrt{25-x^2} + 25\arcsin\left(\frac{x}{5}\right)\right)$ evaluated from -5 to h, so the equation we need to solve is: $\frac12\left(h\sqrt{25-h^2}+25\arcsin(\frac{h}{5})\right)+\frac{25\pi}{4}=12.5\pi P$
Substituting values of .1, .2, .3, .4, and .5 successively for $P$ and using various tools to estimate $h$, my results are -3.43424, -2.45931, -1.59846, -0,788681, 0. Adding 5 to account for the central displacement, rounding off, then reflecting these values for the right (upper) values, confirms the values given earlier by Americo.
-
1
– J. M. Sep 7 '10 at 0:30
+1 for a completely different solution. – Américo Tavares Sep 12 '10 at 19:00
Looking @Isaac's answer, the appearance of "$\alpha - \sin\alpha$" and "$1-\cos\frac{\alpha}{2}$" made me think that power series would come in handy, with their respective initial terms vanishing. They didn't turn out to be as handy as I might have hoped.
Write $w$ for the ratio of oil to the tank's capacity, and $d$ for the ratio of the oil's depth to the radius of the tank. Then, by Isaac's work ...
$$2\pi w = \alpha - \sin\alpha = \alpha-\sum_{m=0}^\infty\frac{(-1)^m}{(2m+1)!}\alpha^{2m+1}= \sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m+1)!}\alpha^{2m+1}$$ $$d = 1 - \cos\frac{\alpha}{2}=1-\sum_{m=0}^\infty\frac{(-1)^m}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}=\sum_{m=1}^\infty\frac{(-1)^{m+1}}{(2m)!}\left(\frac{\alpha}{2}\right)^{2m}$$
Write $w_n$ and $d_n$ for the values associated with lopping off the power series at the $n$-th term (that is, from $m=1$ to $m=n$). In particular, we have
$$2\pi w_1 = \frac{\alpha^3}{3!}$$ $$d_1 = \frac{\alpha^2}{2^2 2!}$$
Eliminating $\alpha$ yields the relation
$$32 d_1^3 = 9 w_1^2 \pi^2$$
This relation, though, isn't very accurate (as shown by the $n=1$ curve in the diagram).
If we take $n=2$ ...
$$2\pi w_2 = \frac{\alpha^3}{3!}-\frac{\alpha^5}{5!}$$ $$d_2 = \frac{\alpha^2}{2^2 2!}-\frac{a^4}{2^4 4!}$$
... then eliminating $\alpha$ isn't quite so easy unless you do something like invoke Mathematica's `Resultant` function, which yields
$$4718592 d_2^5 - 13762560 d_2^4 + 10035200 d_2^3 - 768000 d_2^2 w_2^2 \pi^2 + 3024000 d_2 w_2^2 \pi^2 - 2822400 w_2^2 \pi^2 + 1875 w_2^4 \pi^4 = 0$$
Also not terribly accurate at we get close to $w=0.5$ (which should have $d=1.0$). With $n=3$ we're very close, and with $n=4$ and above, we've pretty much converged on our target (at least according to the graph). I won't give the $n=4$ equation (the coefficient on $d^9$ is $2^{71} 5^4 7^2$), but here are some data points:
$$(0.0, 0.000000)$$ $$(0.1, 0.312953)$$ $$(0.2, 0.508164)$$ $$(0.3, 0.680452)$$ $$(0.4, 0.842809)$$ $$(0.5, 1.001730)$$
Of course, power series and resultants aren't exactly "lowest possible grade level", but this is the best I can do at this point.
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I guess this rigorously confirms Américo's note that approximating by a linear function mid-range won't give the required accuracy. – J. M. Sep 1 '10 at 21:29 | {"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": 87, "mathjax_display_tex": 14, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9477295875549316, "perplexity_flag": "middle"} |
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http://mathoverflow.net/questions/24697?sort=newest | ## Isomorphism between direct sum of modules
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Let $M$, $N$ be two modules over ring $A$. If $M\oplus M\cong N\oplus N$, can we conclude $M\cong N$? In the case that $M$, $N$ are completely decomposable (e.g. finite-length module by Krull-Schmidt Theorem), it is easy to show this must be true. Does the general case also hold?
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12
False, even in the locally free case: if $A$ is a Dedekind domain and $I$ is a nonzero ideal of $A$ which is 2-torsion in the class group then $I \oplus I \simeq A \oplus I^2 = A \oplus A$. So if the class group of $A$ is finite with even order then any $I$ representing an element of exact order 2 in the class group provides a counterexample. – BCnrd May 15 2010 at 5:15
@BCnrd: Why don't you put this as an answer. – Chandrasekhar Feb 3 2011 at 2:50
## 1 Answer
There are even counterexamples in the case $A = {\mathbb Z}$: at the end of B. Jónsson’s paper “On direct decompositions of torsion-free abelian groups,” Math. Scand. 5 (1957), 230–235, an example is given of torsion-free, finite-rank abelian groups $B \not\cong C$ such that $B \oplus B \cong C \oplus C$.
A further counterexample, which I believe has been pointed out independently by L. S. Levy, R. Wiegand, and R. G. Swan: let $A$ be the coordinate ring of the real 2-sphere and ${}_AM$ the module for the tangent bundle; then $M \oplus M$ is free of rank $4,$ but $M$ is not free of rank $2$.
In the positive direction, K. R. Goodearl has proved (“Direct sum properties of quasi-injective modules,” Bull. Amer. Math. Soc. 82 (1976), no. 1, 108–110, Theorem 3) that if $M$ and $N$ are quasi-injective modules over a ring (commutative or not), then $M^n \cong N^n$ implies $M \cong N$ for any positive integer $n$.
Your question is related to an important open problem in noncommutative ring theory, the “separativity” problem for von Neumann regular rings: if $R$ is a von Neumann regular ring (or more generally an exchange ring), and $A$ and $B$ are finitely generated projective left $R$-modules with the property that $A \oplus A \cong A \oplus B \cong B \oplus B$, must we have $A \cong B$? An affirmative answer would resolve several major open problems, as explained in P. Ara, K. R. Goodearl, K. C. O’Meara, and E. Pardo’s paper “Separative cancellation for projective modules over exchange rings,” Israel J. Math. 105 (1998), 105–137.
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## Z-Boson reasonance and the number of neutrino varieties
I've read that the experimental results of Z-Boson resonance confirm the theoretical expectations that there are 3-types of Neutrinos, not 2 or 4.
How are these theoretical expectations calculated? I.e. how does the number of neutrino varieties affect Z-boson resonance?
Thank you
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Mentor Does http://en.wikipedia.org/wiki/Neutrino#Types answer your question? If not, can you be more specific as to what you are looking for?
At the LEP experiment Z bosons were produced through colliding electrons and positrons. This was a very clean experiment, unlike the LHC, allowing the properties of the Z boson to be studied to high precision. The presence of the Z boson shows up as a peak in the total cross-section as a function of center of mass energy with the peak located at ~91 GeV corresponding to the mass of the Z boson (there is a nice plot of this on p711 fig.20.5 of Peskin & Schroeder). By measuring the Z peak very carefully they can determine the total width (##\Gamma_{tot}##) and the cross-section at the peak (##\sigma_{peak}##). The total Z width ##\Gamma_{tot}## is actually made up of two parts. The first is the visible part ##\Gamma_{vis}## made up of decays to charged leptons and hadrons and this is related to the observed peak cross-section ##\sigma_{peak}##. The second is the invisible part ##\Gamma_{inv}## made up of decays to neutrinos which are not observed in the experiment. This second part can be determined by taking ##\Gamma_{tot}-\Gamma_{vis}=\Gamma_{inv}=N_\nu\Gamma(Z\to \nu\overline{\nu})## where ##N_\nu## is the number of active neutrinos. Notice it is "active" neutrinos that matter not the number of neutrinos as there might be "sterile" neutrinos which the Z doesn't decay to. Putting it all together we can get ##N_\nu## from $$N_\nu = (\Gamma_{tot}-\Gamma_{vis})/\Gamma(Z\to \nu\overline{\nu})$$ and the result is very close to ##N_\nu=3##. You can try to look at chapter 20 of Peskin & schroeder and in particular do the problems 20.2 and 20.3 on p 728 which will give you a good understanding of the Z resonance and how it relates to the number of active neutrinos.
## Z-Boson reasonance and the number of neutrino varieties
This does help a lot. Thanks Vanadium and jkp.
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High Energy, Nuclear, Particle Physics 4
Introductory Physics Homework 5
High Energy, Nuclear, Particle Physics 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": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.8848477602005005, "perplexity_flag": "middle"} |
http://www.citizendia.org/Hexagonal_number | A hexagonal number is a figurate number, The nth hexagonal number will be the number of points in a hexagon with n regularly spaced points on a side, as shown in [1]. A figurate number is a number that can be represented as a regular and discrete geometric pattern (e
The formula for the nth hexagonal number is:
$h_n= n(2n-1)\,\!$
The first few hexagonal numbers (sequence A000384 in OEIS) are:
1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780, 861, 946
Every hexagonal number is a triangular number, but not every triangular number is a hexagonal number. The On-Line Encyclopedia of Integer Sequences ( OEIS) also cited simply as Sloane's, is an extensive searchable Database of Integer sequences Mathematics For any number x: x ·1 = 1· x = x (1 is the multiplicative identity In mathematics Six is the second smallest Composite number, its proper Divisors being 1, 2 and 3. 28 ( twenty-eight) is the Natural number following 27 and preceding 29. 45 ( forty-five) is the Natural number following 44 and followed by 46. 66 ( sixty-six) is the Natural number following 65 and preceding 67. 91 ( ninety-one) is the Natural number following 90 and preceding 92. 190 is the natural number following one hundred [and] eighty-nine and preceding one hundred [and] ninety-one. Four hundred ninety-six is the Natural number following four hundred ninety-five and preceding four hundred ninety-seven 500 ( five hundred) is the Natural number following 499 and preceding 501. A triangular number is the sum of the n Natural numbers from 1 to n. Like a triangular number, the digital root in base 10 of a hexagonal number can only be 1, 3, 6, or 9. The digital root (also repeated digital sum) of a number is the number obtained by adding all the digits then adding the digits of that number and then continuing until a single-digit
Any integer greater than 1791 can be expressed as a sum of at most four hexagonal numbers, a fact proven by Adrien-Marie Legendre in 1830. Adrien-Marie Legendre ( September 18 1752 – January 10 1833) was a French Mathematician. For the game see 1830 (board game. Year 1830 ( MDCCCXXX) was a Common year starting on Friday (link will display
Hexagonal numbers can be rearranged into rectangular numbers n long and 2n−1 tall (or vice versa).
Hexagonal numbers should not be confused with centered hexagonal numbers, which model the standard packaging of Vienna sausages. A centered hexagonal number, or hex number, is a centered Figurate number that represents a Hexagon with a dot in the center and all other dots A Vienna sausage is a wiener. The word wiener means Viennese in German. To avoid ambiguity, hexagonal numbers are sometimes called "cornered hexagonal numbers".
## Test for hexagonal numbers
One can efficiently test whether a positive integer x is an hexagonal number by computing
$n = \frac{\sqrt{8x+1}+1}{4}.$
If n is an integer, then x is the nth hexagonal number. If n is not an integer, then x is not hexagonal. | {"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": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8652400970458984, "perplexity_flag": "middle"} |
http://katlas.math.toronto.edu/0506-Topology/index.php?title=Classnotes_for_January_10&oldid=765 | # Classnotes for January 10
### From 0506Topology
Revision as of 14:18, 6 Feb 2006; view current revision
## Group-Theoretical Defintions
The goal of this lecture is to state van Kampen's theorem (also known as the Seifert-van Kampen theorem), give a brief explanation of its plausibility, and supply several examples of its use. In order to do this, we need to first define the group-theoretical notion of the free product of two groups.
Definition: Let (G1, * 1) and (G1, * 2) be groups. Then the free product of (G1, * 1) and (G2, * 2), denoted by $G_{1} \star G_{2}$, is defined to be the group whose elements are words with letters in $G_{1} \sqcup G_{2}$ (i.e. the disjoint union of G1 and G2) and whose multiplication is the concatenation of words, with the empty word being the identity e, together with the relations:
1. $e_{G_{1}} = e = e_{G_{2}}$, where $e_{G_{i}}$ is the identity of (Gi, * i),i = 1,2;
2. for i = 1,2, if $a,b \in G_{i}$ then a * b = a * ib, where * without subscript denotes the multiplication of $G_{1} \star G_{2}$.
This definition can be extended readily to arbitrary finite free products by induction. One can alternatively define free products by means of a universal property (with uniqueness up to group isomorphism), and provide a rigorous construction to prove their existence. For this approach, see here (http://planetmath.org/encyclopedia/FreeProduct.html).
To see what a free product looks like, let us consider $\mathbb{Z} \star \mathbb{Z}$. If we call the generator of the first copy of $\mathbb{Z}$ by a, and the generator of the second copy by b, an example of an element of $\mathbb{Z} \star \mathbb{Z}$ would be a2ba − 3a, and an example of a calculation following the relations outlined in the definitions of a free product is a2ba − 3a * a2ba = a2ba − 1ba.
As shall soon be explained, it turns out that the concept of the free product is not quite sufficient for the statement of van Kampen's theorem. What we shall need instead is a slight modification, called the free product with amalgamated subgroup:
Defintion: Let G1, G2, and H be groups such that there exist injective group homomorphisms (i.e. monomorphisms) $\phi_{1} : H \rightarrow G_{1}$ and $\phi_{2} : H \rightarrow G_{2}$. Then the free product of G1 and G2 with amalgamated subgroup H, denoted by $G_{1} \star_{H} G_{2}$, is defined to be $G_{1} \star G_{2}$ together with the additional relation that for every $h \in H$, φ1(h) = φ2(h).
Once again, one can define free products with amalgamated subgroup by means of a universal property (with uniqueness up to group isomorphism), and provide a rigorous construction to prove their existence. For this approach, see here (http://planetmath.org/encyclopedia/FreeProductWithAmalgamatedSubgroup.html).
## Van Kampen's Theorem
We are now in a position to finally state van Kampen's theorem:
Theorem (van Kampen): Let X be a topological space, let $U_{1},U_{2} \subset X$ be open, and suppose that $X = U_{1} \cup U_{2}$ and that $U_{1} \cap U_{2}$ is path-connected. Let $b \in U_{1} \cap U_{2}$, and for i = 1,2 let $\iota_{i} : U_{1} \cap U_{2} \rightarrow U_{i}$ be the inclusion map. Then $\pi_{1}(X,b) \approx \pi_{1}(U_{1},b) \star_{\pi_{1}(U_{1} \cap U_{2},b)} \pi_{1}(U_{2},b)$, where, for i = 1,2, the monomorphism from $\pi_{1}(U_{1} \cap U_{2},b)$ to π1(Ui,b) is taken to be (ιi) * .
Let us see how this is plausible on an intuitive level. If we are given a loop in X with basepoint b in $U_{1} \cap U_{2}$, then, since $U_{1} \cap U_{2}$ is path-connected, one can apply a homotopy to the portion of the loop within $U_{1} \cap U_{2}$ to decompose the loop, as it were, into the concatenation of several loops, each lying entirely within U1 or U2. This motivates π1(X,b) being some sort of free product of π1(U1,b) and π1(U2,b). If we are given a loop which lies entirely within $U_{1} \cap U_{2}$, then we would like for it to correspond to a single element of π1(X,b), whether we view the loop as residing in U1 or in U2. In this manner, we see that the additional relations which distinguish the free product with amalgamated subgroup from the free product are entirely natural and, in fact, forced upon us, if we wish to avoid double counting of homotopy classes of loops residing entirely in the intersection. Thus, van Kampen's theorem is seen to be an entirely plausible result, but, as we shall see, this does not mean that the proof will be all that easy.
## Examples
To provide further motivation for study of van Kampen's theorem, we now examine a number of examples of its application.
### The Figure Eight $X = S^{1} \vee S^{1}$
Our open subsets U1 and U2 are provided by taking the one or the other copy of S1 together with a small neighbourhood of the join point b. Thus U1 and U2 are both readily homotopic to S1, the homotopy consisting of the retraction of the "antennae", whilst $U_{1} \cup U_{2}$, which is path-connected and readily homeomorphic to $\times$, and thus homotopic to a point by the obvious retraction. Thus, for i = 1,2, $\pi_{1}(U_{i},b) \approx \pi_{1}(S^{1}) \approx \mathbb{Z}$, whilst $\pi_{1}(U_{1} \cap U_{2},b)$ is trivial. Hence $\pi_{1}(U_{1} \cap U_{2},b)$ cannot impose any new relations on the free product $\pi_{1}(U_{1},b) \star \pi_{1}(U_{2},b)$, so, by van Kampen's theorem, $\pi_{1}(X,b) \approx \pi_{1}(U_{1},b) \star_{\pi_{1}(U_{1} \cap U_{2},b)} \pi_{1}(U_{2},b) = \pi_{1}(U_{1},b) \star \pi_{1}(U_{2},b) \approx \mathbb{Z}\star\mathbb{Z}$
## Record of the Lecture
### Scanned lecture notes
Available in PDF format.
### Audio recording
1st hour lecture recording
2nd hour lecture recording | {"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": 36, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9241499900817871, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/66830/list | ## Return to Answer
2 added 2 characters in body
Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum A_i\cos(\lambda_i t+\phi_i)\,dt$$ t+\phi_i)\sum_j A_j\cos(\lambda_j t+\phi_j)\,dt is $A_i/2$. Clearly the average is bounded by the supremum of the function.
1
Yes, this is true. It follows from the fact that the average $$\lim_{T\to\infty} {1\over T}\int_0^T\cos(\lambda_i t+\phi_i)\sum A_i\cos(\lambda_i t+\phi_i)\,dt$$ is $A_i/2$. Clearly the average is bounded by the supremum of the 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": 3, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9507707357406616, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/221887/probability-involving-two-poisson-constants?answertab=votes | # Probability involving two Poisson constants.
A discrete stochastic variable $X$ is defined as the number of eggs laid by a bird at a certain time. Depending on the specie of the bird, the Poisson constant for this distribution is $\lambda = \lambda_1$ or $\lambda_2$. What is $\text {Pr}(X = k)$?
Am I crazy to think that it is the sum of the Poisson functions on both constants?
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## 2 Answers
The question requires interpretation. One reasonable thing is to suppose that the bird laying the eggs is of Species $1$ with probability $p$, and of Species $2$ with probability $1-p$. Perhaps, quite unreasonably, you are expected to assume that $p=\frac{1}{2}$.
Then if $X$ is the number of eggs, $$\Pr(X=k)=pe^{-\lambda_1}\frac{\lambda_1^k}{k!}+(1-p)e^{-\lambda_2}\frac{\lambda_2^k}{k!}.$$ If $\lambda_1\ne \lambda_2$, this is not Poisson for any "mixed" $\lambda$, except in the extreme cases $p=0$ or $p=1$.
Remark: I am not sure what you meant by the sum of the Poisson functions. Maybe you meant $\Pr(X=k)=e^{-\lambda_1}\frac{\lambda_1^k}{k!}+e^{-\lambda_2}$. If so, that is not quite right, though if we take $p=\frac{1}{2}$, as you may be intended to do, it is very close. But we need to divide by $2$. In general $\Pr(X=k)$ is a weighted average of the two mass functions.
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a bird can't be of two different species – OlayinkaSF Oct 27 '12 at 4:05
True, I changed the wording. The bird that laid the eggs might be of Species $1$ or it might be of Species $2$. – André Nicolas Oct 27 '12 at 4:08
Thanks you very much, again! – OlayinkaSF Oct 27 '12 at 4:24
In a general case, you can always write the interested probability as follows $${\rm{Pr}}(X=k) = {\rm{Pr}}(X=k|\lambda_1){\rm{Pr}}(\lambda=\lambda_1)+{\rm{Pr}}(X=k|\lambda_2){\rm{Pr}}(\lambda=\lambda_2),$$ which uses the total probability formula.
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http://mathoverflow.net/questions/67562/weak-convergence-of-measures-on-non-metrizable-spaces/67577 | ## Weak convergence of measures on non-metrizable spaces
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(ZF + Countable Choice)
Let $\langle X,\mathcal{T} \hspace{.06 in} \rangle$ be a second-countable Hausdorff space. Let $\mu$ be a Borel measure on $X$.
Let $\langle I,\leq_I \rangle$ be a directed set, and let `$\{\mu_i : i\in I\}$` be a collection of Borel measures on $X$.
Are the following are equivalent?
$1. \qquad$ For all open subsets $U$ of $X$, $\hspace{.12 in} \mu(U) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(U)$
$2. \qquad$ For all closed subsets $C$ of $X$, $\hspace{.12 in} \displaystyle\limsup_i \hspace{.06 in} \mu_i(C) \hspace{.08 in} \leq \hspace{.08 in} \mu(C)$
Assume $\delta$ is a proximity relation on $X$ that induces $\mathcal{T} \hspace{.05 in}$. $\hspace{.04 in}$ Are the following equivalent?
For all Borel subsets $A$ and $B$ of $X$, if $\quad A \; \; \delta \; \; (X-B) \quad$ is false, then
$3. \qquad \mu(A) \hspace{.08 in} \leq \hspace{.08 in} \displaystyle\liminf_i \hspace{.06 in} \mu_i(B)$
$4. \qquad \displaystyle\limsup_i \hspace{.06 in} \mu_i(A) \hspace{.08 in} \leq \hspace{.08 in} \mu(B)$
Is (1 and 2) equivalent to (3 and 4) ?
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Another setting: omit hypothesis of second-countable, but use general completely regular Hausdorff space. In place of closed sets use zero sets (of continuous real-valued functions); in place of open sets use cozero sets; in place of Borel sets use Baire sets (sigma-algebra generated by zero sets). Good reference: Gillman & Jerison, Rings of Continuous Functions – Gerald Edgar Jun 12 2011 at 12:25
## 1 Answer
Allowing infinite measures will defeat equivalence of 1 and 2. Say Hausdorff measures of various dimensions $<1$ on $\mathbb R$. All nonempty open sets have measure $\infty$. But closed sets can have interesting values.
On the other hand, for finite measures, then of course 1 and 2 are equivalent, by taking complements.
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http://mathoverflow.net/questions/121340/plus-construction-considerations/121368 | ## Plus construction considerations.
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In order to realise the K-groups of a ring as the homotopy groups of some space associated to that ring, Quillen proposed the following (roughly-sketched) construction:
Recall that $K_1(R) = GL(R)/E(R)$, so we're, at least, looking for a space $X$ with $\pi_1(X) = K_1(R)$. The classifying space of $K_1(R)$ is obviously not a serious candidate, but we can start with the classifying space of $GL(R)$ (given the discrete topology), $BGL(R)$. Then, choosing representative loops whose classes generate $E(R) \subset \pi_1(BGL(R))$, and then attaching 2-cells using these loops on the boundaries, we end up with something that has a fundamental group of $K_1(R)$. Furthermore, now Quillen adjoins 3-cells essentially to correct the homology back to that of $BGL(R)$, which was messed up in the addition of those 2-cells. We end up with a space denoted $BGL(R)^+$, on which we define $K_i(R) := \pi_i(BGL(R)^+)$.
More specifically, Quillen sought to find a space $BGL(R)^+$ for which $(BGL(R), BGL(R)^+)$ was an acyclic pair (that is, the induced map $H_*(BGL(R), M) \to H_{*}(BGL(R)^+,M)$ is an isomorphism for all $K_1(R)$-modules $M$). My question is
In search of a space on which to define K-groups, why was it desirable to find something satisfying the above condition on homology?
My best guess is that it was observed that $K_1(R) = GL(R)/E(R) = GL(R)_{ab} = H_1(GL(R), \mathbb{Z})$, and $K_2(R) = H_2(E(R), \mathbb{Z}) = H_2([GL(R), GL(R)], \mathbb{Z})$, and so it seemed reasonable that all K-groups should be related to the homology of $GL(R)$ - and so the above is a stab at preserving that homology.
I am trying to learn about K-theory, but, as with most presentations of math, I'm sure, I'm coming across too many cleanly-shaven formulas and propositions, entirely divorced from any sorts of thought-processes, big-pictures, or mentions of what is trying to be done and to what ends. Please share with me what you think is going on; here, with the +-construction, that is. (And, if and only if it is not a completely unrelated question, what sort of K-theoretical phenomena suggested that these groups should be homotopy groups?)
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## 3 Answers
Here are some thoughts, gathered from reading many texts about algebraic K-theory. Let me start with some historical remarks, then try to give a more revisionist motivation of the plus construction.
First of all, it's true as you say that the already-divined definitions of the lower K-groups made it seem like the higher K-groups, whatever they might be, bear the same relation to the homology of GL(R) as the homotopy groups of an H-space bear to its homology groups; thus one is already looking for an H-space K(R) whose homology agrees with that of GL(R), in order to define the higher algebraic K-groups as its homotopy groups. This line of thought is amplified by the observation that the known partial long exact sequences involving lower K-groups seemed like they could plausibly arise as the long exact sequences on homotopy groups associated to fibrations between these hypothetical H-spaces.
So it may already at this point be natural to try to turn GL(R) into a H-space while preserving its homology, which is what the + construction does. However, the facts that:
1) the + construction ignores the admittedly crucial K_0-group; and
2) in any case, at the time there were very few lower K-groups to extrapolate from in the first place
mean that perhaps the above is insufficient motivation for defining and investigating such a seemingly ad hoc construction as the + construction.
However, we can bear in mind that Quillen's definition of the + construction came on the heels of his work on the Adams conjecture, during the course of which --- using his expertise on the homology of finite groups --- he was able to produce a (mod l) homology equivalence
BGL(F) --> BU
when F is the algebraic closure of a finite field of characteristic different from l. Now, BU is a classifying space for complex K-theory (in positive degrees), so its homotopy groups provide natural definitions of the (topological) algebraic K-theory of the complex numbers. Furthermore, by analogy with the theory of etale cohomology (known to Quillen), it is not entirely unreasonable to guess that, from the (mod l) perspective, all algebraically closed fields of characteristic different from l should behave in the same way as the topological theory over C. (This was later borne out in work of Suslin.) Then the above (mod l) homology equivalence adds further weight to the idea that the hypothetical K-theoretic space K(F) we're searching for should have the same homology as BGL(F). But what's more, Quillen also calculated the homology of BGL(F) when F is a finite field, and found this to be consistent with the combination of the above and a Galois descent'' philosophy for going form the algebraic closure of F down to F.
That said, in the end, there is good reason why the plus construction of algebraic K-theory is difficult to motivate: it is, in fact, less natural than the other constructions of algebraic K-theory (group completion, Q construction, S-dot construction... applied to vector bundles, perfect complexes, etc.). This is partly because it has less a priori structure, partly because it ignores K_0, partly because it has narrower applicability, and partly because it is technically inconvenient (e.g. for producing the fiber sequences discussed above). Of course, Quillen realized this, which is why he spent so much time working on the other constructions. Probably the only claim to primacy the + construction has is historical: it was the first construction given, surely in no small part because of personal contingencies --- Quillen was an expert in group homology.
In fact, probably the best motivation for the + construction -- ahistorical though it may be -- comes by comparison with another construction, the group completion construction (developed by Segal in his paper "on categories and cohomology theories"). Indeed, Segal's construction is very well-motivated: it is the precise homotopy-theoretic analog of the classical procedure of going from isomorphism-classes of f.g. proj. modules to the Grothendieck group K_0 by formally turning direct sum into a group operation. To get this homotopy theoretic analog, one "simply" carries along the isomorphisms in this construction (c.f. also Grayson's article "higher algebraic k-theory II"). The connection with the plus construction comes from the group completion theorem'' (see the McDuff-Segal article on this subject), which, under very general conditions, allows to calculate the homology of such a homotopy-theoretic group completion in terms of the homology of the relevant isomorphism groups. If you look at the group completion theorem in the case of the space of f.g. proj. modules over a ring, you'll see the connection with the plus construction immediately.
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@Dustin: Thank you for such a detailed response, and for the history. This is exactly the sort of answer I was hoping for. – Joshua Seaton Feb 9 at 22:04
I recommend Quillen's 3 page ICM address "Cohomology of groups" for the connection between cohomology of $GL_n(F_q)$, topological K-theory, and algebraic K-theory. It's true that other approaches have better formal properties, but those formal properties are mainly used to reduced to computations in group cohomology, which is one of the only ways to get started on calculation. (and, yes, group completion is the best point of view, motivating both plus and S) Quillen 1970: imu2.zib.de/ICM-search/answer.cgi?first=quillen – Ben Wieland Feb 9 at 23:20
@Dustin: regarding "...bear the same relation to the homology of GL(R) as the homotopy groups of an H-space bear to its homology groups...": may you please elaborate on what the general relation is between the homology and homotopy of an H-space? – Joshua Seaton Feb 14 at 1:09
Well, in general it's not deterministic, even if you know Steenrod actions on the homology. I had in mind whatever you can get in low degrees from considerations as in Goodwillie's answer. In our case that gives K_1 = H_1(GL(R)) and K_2 = H_2([GL(R),GL(R)]) and K_3 = H_3(universal central extension of [GL(R),GL(R)]), but there we stop because H_3 doesn't have as nice a group theory interpretation. Another fact (at least provided the H-space is actually a loop space) is that the rational homotopy maps isomorphically to the primitives of the Hopf algebra structure on the rational homology. – Dustin Clausen Feb 14 at 16:21
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I don't really know if this helps, but you can in effect give the plus-construction of $K$-groups without explicitly mentioning homotopy groups, and without ever doing the plus construction:
$H_1BGL(R)$ is $K_1(R)$. Map $BGL(R)$ to an Eilenberg-MacLane space BK_1(R) and consider the homotopy fiber.
This space has trivial $H_1$. Its $H_2$ is $K_2(R)$. Map it to an Eilenberg-MacLane space B^2K_2(R) and consider the homotopy fiber.
This space has trivial $H_2$. Its $H_3$ is $K_3(R)$. Map it to an Eilenberg-MacLane space B^3K_3(R) and consider the homotopy fiber.
And so on.
-
@Tom: Thank you. It certainly helps. – Joshua Seaton Feb 9 at 22:02
For convenience (at least my own) and completeness, I want to give an explanation of Tom Goodwillie's answer, as it was not obvious to me how to prove the statement he makes. I wanted to leave it as a comment to his answer, but it became way too long. In summary, the point is that applying the plus construction to the spaces defined by Tom gives the connective covers (i.e. the Whitehead tower) of $BGL(R)^+$. This follows from the fact that the plus construction preserves certain fibre sequences.
Abbreviate $X=BGL(R)$. Let $F_1=X=BGL(R)$, and denote by $F_{i+1}$ the $i$-th space constructed by Tom, so that his claim becomes $H_i(F_i) = \pi_i(X^+) = K_i R$. Explicitly, the spaces $F_k$ are defined inductively by a fibre sequence $$F_{i+1}\longrightarrow F_i\longrightarrow K(H_i(F_i),i)$$ for each $i\geq 1$, where the map on the right induces the "identity" on $H_i$. Applying the plus construction to this fibre sequence, we get a new sequence $$(FS^+): \qquad\qquad (F_{i+1})^+ \longrightarrow (F_i)^+ \longrightarrow K(H_i(F_i),i)^+ \simeq K(H_i(F_i),i) \hphantom{\qquad\qquad\qquad}$$ where we have noted that the natural map $K(H_i(F_i),i) \to K(H_i(F_i),i)^+$ is a weak equivalence. Importantly, the new sequence $(FS^+)$ is again a fibre sequence. This result is an instance of proposition 3.D.3-(2) in page 74 of Dror Farjoun's book "Cellular spaces, null spaces and homotopy localization": applying the plus construction to a fibre sequence of path connected spaces gives a fibre sequence as long as the homotopy type of the base space is unchanged by the plus construction. In fact, the book states this for any nullification functor, and the plus construction is one such functor.
The fundamental group of $(F_1)^+=X^+$ is abelian (it is $K_1 R$, after all). From the fibration sequence $(FS^+)$ for $i=1$ we then conclude that $(F_2)^+$ is simply connected. Moreover, that fibre sequence is just $(F_2)^+\to X^+\to B(\pi_1(X^+))$, and it realizes $(F_2)^+$ as the universal cover of $X^+$.
We can proceed inductively in this manner. More precisely, knowing that $\pi_1((F_1)^+)=\pi_1(X^+)$ is abelian, we can show by induction that for all $i\geq 1$:
• The space $(F_i)^+$ is $(i-1)$-connected, and the Hurewicz theorem implies $\pi_i((F_i)^+)=H_i((F_i)^+)=H_i(F_i)$.
• Hence, the second map in the fibre sequence $(FS^+)$ above is an isomorphism on $\pi_i$. In other words, that fibre sequence is simply killing $\pi_i$ of $(F_i)^+$.
• Finally, there is a canonical equivalence $(F_i)^+ \simeq (X^+)^{\geq i}$ from $(F_i)^+$ to the $(i-1)$-connected cover of $X^+$ (also known as the $i$-th stage of the Whitehead tower of $X^+$).
Consequently, $H_i(F_i)=H_i((F_i)^+)=H_i((X^+)^{\geq i})=\pi_i((X^+)^{\geq i})=\pi_i(X^+)$ as desired.
Obviously, the above argument is not specific to $X=BGL(R)$. In fact, we only require that the pointed space $X$ is path connected, and $\pi_1(X^+)$ is abelian.
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http://mathhelpforum.com/calculus/79019-solved-calculating-limit-sequence.html | # Thread:
1. ## [SOLVED] Calculating limit of a sequence
Can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What I have been trying to do to first take the limit. If i got an indeterminate, I changed it to a function and took the derivative, but I'm not sure that is the way I need to evaluate all of these. Any help is greatly appreciated!
1) $a_n = 1 - (0.1)^n<br />$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
2) $a_n = \frac{n^3}{3n + 1}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
3) $a_n = e^{7/(n+9)}$
$<br /> \lim_{n\rightarrow \infty}$ $a_n = ?$
4) $a_n = \frac{3^{n+1}}{5^n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
5) $a_n = tan (\frac{6n\pi}{4 + 24n})$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
6) $a_n = n^2e^{-5n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
2. Originally Posted by mollymcf2009
Can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What I have been trying to do to first take the limit. If i got an indeterminate, I changed it to a function and took the derivative, but I'm not sure that is the way I need to evaluate all of these. Any help is greatly appreciated!
1) $a_n = 1 - (0.1)^n<br />$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
Hint: $\lim a^n=0$ when $\left|a\right|<1$
2) $a_n = \frac{n^3}{3n + 1}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
Hint: Divide the numerator and denominator through by $n^3$ to get $\lim\frac{1}{\frac{3}{n^2}+\frac{1}{n^3}}$
Can you take it from here?
3) $a_n = e^{7/(n+9)}$
$<br /> \lim_{n\rightarrow \infty}$ $a_n = ?$
$\lim e^{\frac{7}{n+9}}=e^{\lim\frac{7}{n+9}}$
Can you take it from here?
4) $a_n = \frac{3^{n+1}}{5^n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
$\lim\frac{3^{n+1}}{5^n}=3\lim\frac{3^n}{5^n}=3\lim \left(\frac{3}{5}\right)^n$. Now see my hint for #1
5) $a_n = tan (\frac{6n\pi}{4 + 24n})$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
$\lim\tan\left(\frac{6n\pi}{24n+4}\right)=\tan\left (\lim\frac{6\pi n}{24n+4}\right)$
Can you take it from here?
6) $a_n = n^2e^{-5n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$
$\lim n^2e^{-5n}=\lim\frac{n^2}{e^{5n}}$
Can you take it from here?
(Note that $\lim$ is analogous with $\lim_{n\to\infty}$)
3. Originally Posted by mollymcf2009
can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What i have been trying to do to first take the limit. If i got an indeterminate, i changed it to a function and took the derivative, but i'm not sure that is the way i need to evaluate all of these. Any help is greatly appreciated!
1) $a_n = 1 - (0.1)^n<br />$
$\lim_{n\rightarrow \infty}$ $a_n = ?$ 1
2) $a_n = \frac{n^3}{3n + 1}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$ diverges
3) $a_n = e^{7/(n+9)}$
$<br /> \lim_{n\rightarrow \infty}$ $a_n = ?$ 1
4) $a_n = \frac{3^{n+1}}{5^n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$ 0
5) $a_n = tan (\frac{6n\pi}{4 + 24n})$
$\lim_{n\rightarrow \infty}$ $a_n = ?$ 1
6) $a_n = n^2e^{-5n}$
$\lim_{n\rightarrow \infty}$ $a_n = ?$ 0
correct answers are in red above for future reference | {"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": 64, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9379574060440063, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?t=288086 | Physics Forums
## Reflection and Refraction, much index of refraction
1. The problem statement, all variables and given/known data
1. The index of refraction for water is 1.33 and that of glass is 1.50.
a. What is the critical angle for a glass-water interface?
b. In which medium is the light ray incident for total internal reflection?
2. Relevant equations
nisin$$\vartheta$$i=nrsin$$\vartheta$$r
3. The attempt at a solution
a. I think the answer to a. is 62.46 degrees, but I am not sure.
b. I think the answer is glass, simply because it is going to be moving from a less dense area to a more dense area.
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Use Snell's law: $$n_1\sin\theta_1 = n_2\sin\theta_2\ .$$ Note: It's not additive like you suggested. At the critical angle, $$\theta_2$$ is 90 degrees (the light refracts along the boundary). That is to say, it's sin is 1. We can therefore rearrange for the critical angle: $$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$ Now we have two refractive indices, the glass and the water. If you plug them in incorrectly, you're going to end up with trying to find the inverse sin of a value > 1, which isn't possible as this has no solution.
Quote by astrorob Use Snell's law: $$n_1\sin\theta_1 = n_2\sin\theta_2\ .$$ Note: It's not additive like you suggested. At the critical angle, $$\theta_2$$ is 90 degrees (the light refracts along the boundary). That is to say, it's sin is 1. We can therefore rearrange for the critical angle: $$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$ Now we have two refractive indices, the glass and the water. If you plug them in incorrectly, you're going to end up with trying to find the inverse sin of a value > 1, which isn't possible as this has no solution.
Quite sorry for the mistake in Snell's Law. I knew what it meant, I just messed up when writing it in the forum.
Anyway, I appreciate the input, but if you are using
$$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{n_2}{n_1} \right)$$, would it not be possible to put 1.33 (n2) over 1.5 (n1)? This would look like
$$\theta_{\mathrm{crit}} = \sin^{-1} \left( \frac{1.33}{1.5} \right)$$ and if plugged into a calculator, would return 62.45732485 degrees.
## Reflection and Refraction, much index of refraction
For this problem I calculated 62.4 degrees, the same thing you got.
As for B, I put water.
Yes, that's correct as you've stated. It also gives you the answer to your second question as $$n_2$$ represents the refractive index of the medium that the light travelling in medium $$n_1$$ is incident on.
Quote by patriots1049 For this problem I calculated 62.4 degrees, the same thing you got. As for B, I put water.
Ah, thank you. I was actually thinking that it would be glass because my book details the fact that "Total internal reflection occurs when light passes from a more optically dense medium to a less optically dense medium at an angle so great that there is no refracted ray." This would mean that glass-water would be a more optically dense to a less optically dense scenario.
Quote by astrorob Yes, that's correct as you've stated. It also gives you the answer to your second question as $$n_2$$ represents the refractive index of the medium that the light travelling in medium $$n_1$$ is incident on.
Thank you very much. You have been a big help today, astrorob.
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http://mathoverflow.net/questions/39791?sort=oldest | ## Explicit metrics
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Every surface admits metrics of constant curvature, but there is usually a disconnect between these metrics, the shapes of ordinary objects, and typical mathematical drawings of surfaces.
Can anyone give an explicit and intuitively meaningful formulas for negatively curved metrics that are related to an embedding of a surface in space?
There is an easy way to do this for an open subset of the plane. If the metric of the plane is scaled by a function that is $\exp$ of a harmonic function, the scaling factor is at least locally the norm of the derivative of a complex analytic function, so the resulting metric is still flat; the converse is true as well. Therefore, the sign of curvature of a conformally modified metric $\exp(g)$ depends only on the sign of the laplacian of the $g$. If the value of $g$ at a point is less than the average value in a disk centered at that point, then the metric $\exp(g) ds_E$ is negatively curved, where $ds_E = \sqrt(dx^2 + dy^2)$ is Euclidean arc length.
For example, in a region $R$, if we impose a limit that speed is not to exceed the distance to the complement of $R$, this defines a non-positively curved metric. (The metric is 1/(distance to boundary)$ds_E$). In this metric, geodesics bend around corners: it doesn't pay to cut too close, it's better to stay closer to the middle. If the domain is simply-connected, you see one and only one image of everything, no matter where you are.
There are a number of other ways to write down explicit formulas for negatively curved or non-positively curved metrics for a subset of the plane, but that's not the question: what about for closed surfaces in space? Any closed surface $M^2$ has at least a total of $4 \pi$ positive curvature, where the surface intersects its convex hull. If $M$ is a double torus, how can this be modified to make it negative? It would be interesting to see even one good example of a negatively curved metric defined in terms of Euclidean geometry rather than an indirect construction. (In particular: it can be done by solving PDE's, but I wwant something more direct than that.)
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1
This is a great question, and it is indeed somewhat surprising that nobody has figured anything like this out yet. I find it rather difficult to envision a negatively curved metric on the double torus, given what it looks like when embedded in $R^3$. We know that it is possible to change the "visual" metric into a negatively curved metric via a conformal factor, but it is highly unclear how to do this without using heavy analytic machinery. I imagine that Bill and other visually oriented mathematicians have already tried to do this, so I doubt it's an easy question to answer. – Deane Yang Sep 23 2010 at 20:40
2
Although it doesn't answer your question, I've been looking at a related thing recently. Given a finite group $G$ acting on a surface $\Sigma$, try to find an embedding of $\Sigma$ in $S^3$ such that the action of $G$ on $\Sigma$ extends to an action of $G$ on $S^3$. So there's all kinds of restrictions on when this is possible, but looking at examples can be quite pleasant. – Ryan Budney Sep 23 2010 at 20:46
1
@Ryan: Yes, that's a fun problem. There are many possible subgroups of $O(3)$, and many possible constructions. An interesting extra geometric condition: when are they equivariant minimal surfaces? Sometimes they're not equivariantly compressible, which I believe means they can be made into minimal surfaces. – Bill Thurston Sep 23 2010 at 21:38
1
A slightly cheezy modification of Deane Yang's idea would be to remove a large disc from a hyperbolic surface, so large that the complement is a thin regular neighbourhood of a graph in the surface. Embed that regular neighbourhood in Euclidean space much like how one constructs zero Gauss curvature embeddings of cylinders and Moebius bands in Euclidean space. – Ryan Budney Sep 24 2010 at 0:13
1
Dear Bill, I like very much the picture and the idea of this non-positively curved metric (though, funny enough in the beginning I thought, that these circles are wholes in the paddle), and I have a question. Does this type of construction works for domains in higher-dimensional spaces? For some domains in R^n (maybe convex)? Or. for example, suppose we want to prove that a complement to some hyperplane arrangement in $C^n$ is $K(\pi,1)$, is there a chance that by some similar kind of method we could find a complete, non-positively curved metric on its complement? (this would do the job). – Dmitri Sep 24 2010 at 22:15
show 13 more comments
## 3 Answers
I'm thinking that the trouble with the metric in the positively curved parts of the surface comes from the fact that when building these things in $R^3$ out of a polygon in a Euclidean or hyperbolic plane, we need to do a bit of stretching, because we run out of dimensions for just rolling - in $R^4$ a torus can be flat, yes? So how about if we fix a particular curvature for a curve on the surface, say the one corresponding to the shortest (in the embedded, Euclidean sense) generator of the fundamental group for a basepoint in the middle of the picture (again, in the embedded, Euclidean sense). Then we can scale tangents by the inverse of the curvature in the direction they specify, so that the directions that were the most stretched if we think of the surface as coming from a glued polygon are the easiest to move along. This could be done so that the osculating circles, once scaled, all end up the same size as our particular circle. This would have the effect, for example, of "tightening the belt" around the outside of the positively curved region, so that Deane Yang's flipping of the cylinder would happen.
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1
 I've thought a little more about this, but still not enough to give an explicit answer as was originally asked for. But I thought I'd share this picture I drew of geometry being put on a genus two surface. From this picture, my idea is to choose a cycle in homology that looks nice on the given surface in space corresponding to each of the heavy dashed lines, which are lines in the Poincare model shown. Then use the embedding to find nearby cycles for the thin equidistant curves, and arrange – Zachary Treisman Oct 27 2010 at 21:38
(continued) the tangents so that the time it takes to traverse the cycles for the equidistant curves matches the time it takes to traverse the originally chosen cycle. I feel like this would give a metric at least on much of the surface - at some point the equidistant curves start bumping into each other... – Zachary Treisman Oct 27 2010 at 21:42
### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
This is not a complete answer either -- in particular, I can't write down any formulas yet, but I wanted to share some pictures I made to help build my intuition. Zachary Treisman's construction may be related.
As Bill Thurston's illustration of the (1/d)-metric taught me, by drawing disks living on the surface which represent how far I can get in a certain constant time, I can recover a great deal of tactile intuition for a metric on a surface, even if my eyes are showing me a different one.
So let me try that out in a really simple and special case: flattening a torus, that is visualizing a flat metric on an embedded torus.
The embedding I considered is parametrized by coordinates (u,v) which each live in [0,2π]:
$x(u,v)=(c+\cos(v))\cos(u),$
$y(u,v)=(c+\cos(v))\sin(u),$
$z(u,v)=\sin(v),$
where the radius of a meridian circle is 1 and c is the distance from the center of a meridian circle to the center of the "outer" hole of the torus. If c>1, then the torus will not self-intersect.
I first placed disks in a triangular lattice in a rectangle with aspect ratio c. Here c=2. These are the disks of constant speed:
I rescaled the rectangle so that it had dimensions [0,2π]x[0,2π] and plotted the disks on the surface of the torus. Here are the results (the disks seem to be peeling off the torus because I shrunk the torus so that it wouldn't intersect the disks and then didn't do enough fiddling to make it perfect):
c=1.2
c=2
c=5
I found it useful to imagine the disks moving on the embedded torus by isometries (which are just translations in the rectangle picture). You can see how the disks get sheared on the embedding if we take v to v+c so that the meridians of the torus all rotate -- this is an effect of the differing principal curvatures of the embedding. The outer disks move much faster than the inner ones if we take u to u+c; this rotates the embedding about a central vertical axis.
How hard would it be to make these pictures for negatively curved surfaces? I don't know a nice parametrization for an embedded double torus, let alone one that plays nicely with a negatively curved metric. I also haven't been able to pick out the right features to focus on in these pictures in order to imagine what happens in other cases, so any guidance or comments would be appreciated!
Code:
u3[r_, [Theta], u0] := r Cos[[Theta]] + u0
v3[r_, [Theta], v0] := r Sin[[Theta]] + v0
cent[u0_, v0_] := Module[{rad = c + Cos[v0]}, {rad Cos[u0], rad Sin[u0], Sin[v0]}]
c = 2; xx = 16; yy = 8; max = (.8 [Pi])/xx;
lattice = Flatten[2 [Pi]/ xx Mod[Table[{m + 1/2. n, Sqrt[3]/2. n}, {m, 1, xx}, {n, 0, yy - 1}], xx], 1];
toruspic = ParametricPlot3D[Module[{rad = 1.01 c + .96 Cos[v]} (* perturb so that disks won't intersect torus *), {rad Cos[u], rad Sin[u], .96 Sin[v]}], {u, 0, 2 [Pi]}, {v, 0, 2 [Pi]}, PlotStyle -> {LightBlue, Specularity[1, 20], Lighting -> Automatic}, Mesh -> None, Boxed -> False, Axes -> False]
Show[toruspic, Show[Table[u0 = lattice[[i, 1]]; v0 = lattice[[i, 2]]; ParametricPlot3D[ Module[{rad = c + Cos[1/(Sqrt[3]/2. yy/xx)*v3[r, [Theta], v0]]}, {rad Cos[ u3[r, [Theta], u0]], rad Sin[u3[r, [Theta], u0]], Sin[1/(Sqrt[3]/2. yy/xx)*v3[r, [Theta], v0]]}], {r, 0, max}, {[Theta], 0, 2 [Pi]}, PlotStyle -> {Specularity[White, 40], Blue, Opacity[.6]}, Mesh -> {2, 0}, MeshStyle -> Opacity[.4]], {i, Length[lattice]}]], PlotRange -> All, Boxed -> False, Axes -> False]
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@jc: Those are nice pictures! good ingenuity in finding a way to show a flat metric. Ideas: (a) Putting patterns on surfaces is called "texture mapping", implemented in hardware and well-suported in software on modern computers. It should be easy to draw an image of a photograph (say) wrapped around the torus. I don't know what tools do this conveniently. In Mma, check MeshShading and Image Processing. There are other options with separate software. (b) For conformal toruses: you can use stereographi projection from $S^3$ (or work out formula from derivatives). $\to$ next comment – Bill Thurston Sep 25 2010 at 10:40
@jc, cont'd: For hyperbolic surfaces: the universal covers of $(n m p)$ orbifold is a hypergeometric function (for the special case $22\infty$ it's $\cos$), and many finite-sheeted covers of these orbifolds have simple desriptions as algebraic curves. These are also related to congruence subgroups and modular forms. There are well-developed ways to describe minimal surfaces in $\mathbb R^3$ using complex analytic data of this sort. Willmore surfaces are nice. There are very large bodies of mathematics lurking nearby --- the problem is how to get just a little that you want. – Bill Thurston Sep 25 2010 at 10:59
@Bill Thurston: Thanks for the comments and directions for further exploration. I've confused myself a few times trying to think through a more basic question: is there a way of understanding what happens to small geodesic disks under such a texture mapping from just comparing the Gaussian curvature of the metric we'd like to impose to the principal curvatures of the embedding? The Jacobian of the mapping seems to enter as well. – jc Sep 25 2010 at 16:41
Here's an answer to an analogous question, not Bill's original question, but also a question about how to specify (in a simple way) a non-positively curved metric on a compact Riemann surface $C$ of genus $g>1$, in this instance, one that has been specified as an algebraic curve somewhere (as opposed to being given it as a surface in $3$-space).
The construction is easy: If the curve has been specified as an algebraic curve, then, more-or-less by algorithmic means, one can write down a basis for the holomorphic differentials on $C$ (which is a complex vector space of dimension $g$). Now select two of these differentials, say, $\omega$ and $\eta$, that have no common zeroes on $C$. (Again, this can be tested algebraically). Now consider the metric $g = \omega\circ\bar\omega + \eta\circ\bar\eta$. This $g$ will have non-positive curvature. In fact, the curvature will vanish at only a finite number of points and will otherwise be strictly negative. (Of course, you can add more terms. If you take a basis $\omega_1,\ldots,\omega_g$ of the holomorphic differentials on $C$, then the metric $g = \omega_1\circ\bar{\omega_1} +\cdots + \omega_g\circ\bar{\omega_g}$ will have strictly negative curvature except when $C$ is hyperelliptic, in which case, the curvature will vanish at the Weierstrass points of $C$.)
For example, if you take a hyperelliptic curve, say $y^2 = (x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{2g+2})$ (with the $\lambda_i$ being distinct and, say, nonzero), then a basis for the holomorphic differentials will be given by $\omega_i = x^{i-1}dx/y$ for $i = 1,\ldots, g$. Moreover, $\omega_1$ and $\omega_g$ (for example) have no common zeros. Thus, the smooth metric $g = (1 + |x|^{2(g-1)})|dx|^2/|y|^2$ has negative curvature on this curve except at a finite number of points.
- | {"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": 57, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.922257125377655, "perplexity_flag": "head"} |
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http://en.wikibooks.org/wiki/Numerical_Methods/Errors_Introduction | Numerical Methods/Errors Introduction
When using numerical methods or algorithms and computing with finite precision, errors of approximation or rounding and truncation are introduced. It is important to have a notion of their nature and their order. A newly developed method is worthless without an error analysis. Neither does it make sense to use methods which introduce errors with magnitudes larger than the effects to be measured or simulated. On the other hand, using a method with very high accuracy might be computationally too expensive to justify the gain in accuracy.
Accuracy and Precision
Measurements and calculations can be characterized with regard to their accuracy and precision. Accuracy refers to how closely a value agrees with the true value. Precision refers to how closely values agree with each other. The following figures illustrate the difference between accuracy and precision. In the first figure, the given values (black dots) are more accurate; whereas in the second figure, the given values are more precise. The term error represents the imprecision and inaccuracy of a numerical computation.
Accuracy
Precision
Absolute Error
The error between two values is defined as
$\epsilon_{abs}= \left \| \tilde{x} - x \right \| \quad ,$
where $x$ denotes the exact value and $\tilde{x}$ its approximation.
Relative Error
The relative error of $\tilde{x}$ is the absolute error relative to the exact value. Look at it this way: if your measurement has an error of ± 1 inch, this seems to be a huge error when you try to measure something which is 3 in. long. However, when measuring distances on the order of miles, this error is mostly negligible. The definition of the relative error is
$\epsilon_{rel} = \frac{\left \| \tilde{x} - x \right \|}{\left \| x \right \|} \quad .$
Sources of Error
In a numerical computation, error may arise because of the following reasons:
• Truncation error
• Roundoff error
Truncation Error
Truncation error refers to the error in a method, which occurs because some series (finite or infinite) is truncated to a fewer number of terms. Such errors are essentially algorithmic errors and we can predict the extent of the error that will occur in the method.
Roundoff Error
Roundoff error occurs because of the computing device's inability to deal with inexact numbers. Such numbers need to be rounded off to some near approximation which is dependent on the word size used to represent numbers of the device. | {"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": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9149149656295776, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/213762/question-about-linear-operators-on-hilbert-spaces | Question about linear operators on Hilbert spaces
I have the following question: "Let $V$ be a Hilbert space and let $T$ be a linear operator on $V$. If $S$ is any linear operator on $V$ that satisfies $\langle Tv,w \rangle = \langle Sv,w \rangle$ for all $v,w \in V$, then $S = T$."
My attempt at the problem went something like this: Notice that the equality $\langle (T-S)v,w \rangle = \langle Tv,w \rangle - \langle Sv,w \rangle = 0$ holds for all $v,w \in V$. If $S \neq T$, then $T-S$ is not the zero linear operator and there exists some $v_0 \in V$ such that $(T-S)v_0 = w_0 \neq 0$. But then $0 < \langle (T-S)v_0,w_0 \rangle$ by the properties of the inner product and this is a contradiction.
I have been informed that this argument is incorrect, but cannot seem to find the flaw, nor a correct proof. Any help would be appreciated.
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Your argument is not incorrect: you just have to write $\langle (T-S)v_0,w_0\rangle$ as $\lVert (T-S)v_0\rVert^2$. – Davide Giraudo Oct 14 '12 at 18:38
1 Answer
Fix $v\in V$. Taking $w:=(T-S)v$, we get that $\langle (T-S)v,(T-S)v\rangle=0$ hence $(T-S)v=0$ by properties of positive definiteness of an inner product.
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So it looks like my argument was correct after all, just not particularly elegant. Thanks! – James Miller Oct 14 '12 at 18:39 | {"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": 21, "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}, "config": {"markdown_headings": false, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9295424818992615, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/44113/whats-the-value-of-sum-limits-k-1-infty-frack2k/44120 | # What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?
For some series, it is easy to say whether it is convergent or not by the "convergence test", e.g., ratio test. However, it is nontrivial to calculate the value of the sum when the series converges. The question is motivated from the simple exercise to determining whether the series $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$ is convergent. One may immediately get that it is convergent by the ratio test. So here is my question:
What's the value of $$\sum_{k=1}^{\infty}\frac{k^2}{k!}?$$
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– Squirtle Sep 13 '12 at 4:12
## 4 Answers
The sum is equal to $2e$. First of all, the term $k^2/k!$ may be partly canceled as $k/(k-1)!$. Second, this can be written as $(k-1+1)/(k-1)!$. The term $k-1+1$ is divided to two terms. In the first term, the $k-1$ may be canceled again, giving us $e$. The second term leads to $e$ immediately. So the total sum is $2\exp(1)$.
In a similar way, one may easily calculate the sum even if $k^2$ is replaced by $k^n$, any positive integer power of $k$. The result is always a multiple of $e$.
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The value of $T_n := \displaystyle\sum_{k=1}^{\infty} \frac{k^n}{k!}$ is $B_n \cdot e$, where $B_n$ is the $n^{th}$ Bell number.
To see this, note that
$$\begin{align} T_{n+1} = \sum_{k=1}^{\infty} \frac{k^{n+1}}{k!} &= \sum_{k=0}^{\infty} \frac{(k+1)^n}{k!} \\ &= \sum_{k=0}^{\infty} \frac{1}{k!} \sum_{j=0}^n {n \choose j} k^j \\ &= \sum_{j=0}^n {n \choose j} \sum_{k=1}^{\infty} \frac{k^j}{k!} \\ &= \sum_{j=0}^{n} {n \choose j} T_j \end{align}$$
This is precisely the recursion formula that the Bell numbers follow, except that every term here is being multiplied by $e$.
Edit: I was unaware at the time I posted my answer, but the argument I gave goes by the name of Dobiński's formula.
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2
+1,I like this answer. – Eric♦ Jun 8 '11 at 16:28
2
+1, nice for providing the background. – Jack Jun 8 '11 at 16:41
@Eric/Jack Thanks! – JavaMan Jun 8 '11 at 16:49
Hint 1: $k^2=k(k-1)+k$. Hint 2: simplify the fractions $k(k-1)/k!$ and $k/k!$. Hint 3: write the series expansion around $x=0$ of the function $x\mapsto\exp(x)$.
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Wolfram Alpha says it is $2e$. Another derivation is to start with $e^x=\sum \limits_{n=0}^\infty \frac{x^n}{n!}$, apply $\frac{d}{dx}x\frac{d}{dx}$ to both sides, and evaluate at $x=1$.
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$\frac{d}{dx}x\frac{d}{dx}$, typo? – Jack Jun 8 '11 at 17:46
8
It means: differentiate, then multiply by $x$, then differentiate. – GEdgar Jun 8 '11 at 17:54
I would be curious to know why the downvote. – Ross Millikan Jul 7 '11 at 12:33 | {"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": 31, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9369695782661438, "perplexity_flag": "head"} |
http://mathoverflow.net/revisions/29039/list | ## Return to Answer
1 [made Community Wiki]
A polynomial, solvable in radicals, whose splitting field is not a radical extension (of $\bf Q$). Let $f(x)$ be any cyclic cubic, that is, any cubic with rational coefficients, irreducible over the rationals, with Galois group cyclic of order 3. Then $f(x)=0$ is solvable in radicals (every cubic is), so the splitting field $K$ of $f$ over $\bf Q$ is contained in a radical extension of $\bf Q$, but $K$ is not itself a radical extension of $\bf Q$. The degree of $K$ over $\bf Q$ is 3, so for $K$ to be radical over $\bf Q$ it would have to be an extension of $\bf Q$ by the cube root of some element of $\bf Q$, but such extensions are not normal. | {"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": 15, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9357467293739319, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/139108/explicit-example-of-a-toric-flip?answertab=oldest | # Explicit example of a toric flip
I am looking for a toy example of a flip between toric projective 3-folds. More precisely, I would like to see their defining fans (or polytopes).
Does anyone know where I can find something like this?
-
## 1 Answer
Consider the cube with vertices $(\pm1,\pm1,\pm1)$ and subdivide one of the faces by its diagonal. Consider the fan whose cones are spanned by the cells of the resulting polyhedral complex. Now, you can do this in two ways, so we get two fans $F_1$ and $F_2$.
Moreover, if we subdivide that face with the two diagonals, and build the cones on those cells, we get a third fan $F_3$ which refines both $F_1$ and $F_2$.
The swap $F_1\leftarrow F_3\rightarrow F_2$ is a flip, the corresponding construction on the associated toric varieties is a toric flip.
More generally, pick any polyhedron and subdivide one of its faces until you get triangles. Then —unless you picked a face which was a triangle!— you can always pick two triangles sharing a side and flip the cuadrilateral they form by removing that segment and adding back in the other diagonal of the cuadrilateral.
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Thanks for the answer. Why is that birational map a flip though (and not a flop for example). I looked at Cox-Little-Schenk and Matsuki's MMP book and and found some material for this. Do you know of any other good references? – hilbert May 6 '12 at 5:54
Wait, are these fans even simplicial? – hilbert May 8 '12 at 18:53
If you want them to be, then subdivide faces until it is. – Mariano Suárez-Alvarez♦ May 9 '12 at 16:33 | {"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": 7, "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}, "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}, "english_confidence": 0.9352231025695801, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/136856/definition-of-isomorphism-of-vector-spaces | # Definition of isomorphism of vector spaces
Suppose that $V_1$ is a vector space over the field $K_1$ and $V_2$ is a vector space over the field $K_2$. What is the definition of an isomorphism between these two vector spaces?
My best guess would be that $K_1$ and $K_2$ have to be isomorphic as fields. Say $\xi:K_1\to K_2$ is an isomorphism. And furthermore, we must have a bijective function $\sigma:V_1 \to V_2$ with
$$\sigma(\lambda v_1 + \mu v_2) = \xi(\lambda)\sigma(v_1) + \xi(\mu)\sigma(v_2)$$ for all $v_1,v_2 \in V_1$ and all $\lambda, \mu \in K_1$.
Is this correct, or have I missed or added extra conditions?
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It seems correct but silly to consider different but isomorphic scalar fields. – lhf Apr 25 '12 at 16:14
I have never encountered a notion of isomorphism of vector spaces over different fields. However, I agree that this is a good definition (and essentially the case you are considering does not use two different fields). Why would you want to use two different base fields? – Daan Michiels Apr 25 '12 at 16:14
Is it possible to have two vector spaces be isomorphic if their underlying fields are NOT isomorphic? – nullUser Apr 25 '12 at 16:16
No, unless you want to introduce this notion yourself. – Daan Michiels Apr 25 '12 at 16:17
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@nullUser It wouldn't make any sense. One of the best properties of vector spaces is that they are classified (up to isomorphism) by their dimension, making a given vector space isomorphic to a direct sum of copies of your base field. Different base fields would give you different decompositions of your vector space...? – M Turgeon Apr 25 '12 at 16:30
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## 1 Answer
I don't really know what it means for a definition to be correct. Certainly this is a sensible definition (e.g. it defines an "equivalence relation" on the "set" of vector spaces). It is the notion of isomorphism that one gets in the following category: objects are pairs $(k, V)$ of a field and a vector space over that field, and morphisms are pairs $(\phi, T)$ of field morphisms $\phi : k_1 \to k_2$ and maps $T : V_1 \to V_2$ such that $$T(av + bw) = \phi(a) T(v) + \phi(b) T(w).$$
However, when most people talk about isomorphism of vector spaces, they almost always work implicitly with a fixed base field $k$. This means more than that $k_1, k_2$ are isomorphic; it means that we have fixed an isomorphism between them.
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One reason to restrict to the case of a fixed field is that it makes $k$-vector spaces into algebras (in the sense of universal algebra), with one unary operation for every scalar in the field, and the category of $k$-vector spaces into a variety (with all that this entails). – Arturo Magidin Apr 25 '12 at 16:55
@Qiaochu: The category you describe is a reasonably interesting category. It can be obtained by applying the Grothendieck construction to the strict (!) 2-functor $\textbf{Vect}(-) : \textbf{Fld}^\textrm{op} \to \mathfrak{Cat}$, and is thus a fibred category over $\textbf{Fld}$. It's a sub-fibred-category of the stack of quasicoherent sheaves over the category of schemes... – Zhen Lin Apr 25 '12 at 17:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9439407587051392, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/25878/relevance-of-the-complex-structure-of-a-function-algebra-for-capturing-the-topolo | ## Relevance of the complex structure of a function algebra for capturing the topology on a space.
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This question is the outcome of a few naive thoughts, without reading the proof of Gelfand-Neumark theorem.
Given a compact Hausdorff space $X$, the algebra of complex continuous functions on it is enough to capture everything on its space. In fact, by the Gelfand-Neumark theorem, it is enough to consider the commutative C*-algebras instead of considering compact Hausdorff spaces.
The important thing here is that $C*$-algebras have a complex structure. The real structure is not enough. Given the algebra $C(X, \mathbb R) \oplus C(X, \mathbb R)$ of real continuous functions on $X$, the algebra $C(X, \mathbb{C})$ is simply the direct sum $C(X, \mathbb R) \oplus C(X, \mathbb R)$, as a Banach algebra(and this can be given a complex structure, (seeing it as the complexification...)). But to obtain a C*-algebra, we need an additional C*-algebra, and the obvious way, ie, defining $(f + ig)$* $= (f - ig)$ does not work out. More precisely, the C* identity does not hold.
So one cannot weaken (as it stands) the condition in the Gelfand-Neumark theorem that we need the algebra of complex continuous functions on the space $X$, since we do need the C* structure. Of course, this is without an explicit counterexample. Which brings us to:
Qn 1. Please given an example of two non-homeomorphic compact Hausdorff spaces $X$ and $Y$ such that the function algebras $C(X, \mathbb R)$ and $C(Y, \mathbb R)$ are isomorphic(as real Banach algebras)?
(Here I am hoping that such an example exists).
Then again,
Qn 2. From the above it appears that the structure of complex numbers is involved when the algebra of complex functions captures the topology on the space. So how exactly is this happening?
(The vague notions concerning this are something like: the complex plane minus a point contains nontrivial $1$-cycles, so perhaps the continuous maps to the complex plane might perhaps capture all the information in the first homology, etc..)..
Note : Edited in response to the answers. Fixed the concerns of Andrew Stacey, and changed Gelfand-Naimark to Gelfand-Neumark, as suggested by Dmitri Pavlov.
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If you know a little algebraic geometry, an exercise toward the end of chapter 1 in Atiyah-Macdonald walks you through an attractive proof that a CH space is determined by its ring of real valued functions. What the C* algebra formalism gets you in this context is the "image" of the functor $X \mapsto C(X, \mathbb{C})$ from CH spaces to $\mathbb{C}$-algebras. It is possible to formulate an analogous condition for $\mathbb{R}$-algebras, but the entire theory is more awkward (though more powerful in some ways!). – Paul Siegel May 25 2010 at 23:48
Let me try to justify my last remark. Aside from the awkwardness of the definition, the Bott periodicity theorem for the K-theory of real C* algebras has period 8 rather than period 2 in the complex case, and thus many arguments are much more arduous. But there is also room for more subtlety. For example, the complex structure of C* algebras is at some level responsible for the fact that the Atiyah Singer index theorem is most naturally formulated for even dimensional manifolds. One way to formulate an odd dimensional analogue is to use real C* algebras. This is not without applications. – Paul Siegel May 25 2010 at 23:57
@Paul: That is a nice exercise, but I am confused by "If you know a little algebraic geometry". I guess you mean because of the topology placed on the maximal ideal space? Also, thanks for addressing some differences between the real and complex case. @Akela: In spite of Paul's comments, I must say I'm a little confused by the noncommutative-geometry tag in light of the fact that there is nothing noncommutative in your question. – Jonas Meyer May 26 2010 at 2:40
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@Jonas: In AM, $X$ is recovered from $C(X)$ as the subspace of $Spec(C(X))$ (with the Zariski topology) consisting of maximal ideals. So the exercise requires a casual familiarity with $Spec$, which one usually accumulates via AG. For what it's worth, I support the NCG tag for this question. I don't always work directly with NC C*-algebras in a substantial way, but I usually tell people I work in NCG because I often use $C(X)$ as a proxy for $X$. Maybe I am "mistagging" myself. :) – Paul Siegel May 26 2010 at 14:03
Makes sense. This isn't the first time I've heard of someone working in commutative NCG, but I still appreciate the novelty. – Jonas Meyer May 26 2010 at 19:34
## 3 Answers
Here is a slightly different, perhaps simpler take on showing that $C(X,\mathbb{R})$ determines $X$ if $X$ is compact Hausdorff. For each closed subset $K$ of $X$, define $\mathcal{I}_K$ to be the set of elements of $C(X,\mathbb{R})$ that vanish on $K$. The map $K\mapsto\mathcal{I}_K$ is a bijection from the set of closed subsets of $X$ to the set of closed ideals of $C(X,\mathbb{R})$. Urysohn's lemma and partitions of unity are enough to see this, with no complexification, Gelfand-Neumark, or (explicitly) topologized ideal spaces required. I remember doing this as an exercise in Douglas's Banach algebra techniques in operator theory in the complex setting, but the same proof works in the real setting.
Here are some details in response to a prompt in the comments. (Added later: See Theorem 3.4.1 in Kadison and Ringrose for another proof. Again, the functions are assumed complex-valued there, but you can just ignore that, read $\overline z$ as $z$ and $|z|^2$ as $z^2$, to get the real case.)
I will take it for granted that each $\mathcal{I}_K$ is a closed ideal. This doesn't require that the space is Hausdorff (nor that $K$ is closed). Suppose that $K_1$ and $K_2$ are unequal closed subsets of $X$, and without loss of generality let $x\in K_2\setminus K_1$. Because $X$ is compact Hausdorff and thus normal, Urysohn's lemma yields an $f\in C(X,\mathbb{R})$ such that $f$ vanishes on $K_1$ but $f(x)=1.$ Thus, $f$ is in $\mathcal{I}_{K_1}\setminus\mathcal{I}_{K_2}$, and this shows that $K\mapsto \mathcal{I}_K$ is injective. The work is in showing that it is surjective.
Let $\mathcal{I}$ be a closed ideal in $C(X,\mathbb{R})$, and define $K_\mathcal{I}=\cap_{f\in\mathcal{I}}f^{-1}(0)$, so that $K_\mathcal{I}$ is a closed subset of $X$. Claim: `$\mathcal{I}=\mathcal{I}_{K_\mathcal{I}}$`.
It is immediate from the definition of $K_\mathcal{I}$ that each element of $\mathcal{I}$ vanishes on $K_\mathcal{I}$, so that `$\mathcal{I}\subseteq\mathcal{I}_{K_\mathcal{I}}.$` Let $f$ be an element of `$\mathcal{I}_{K_\mathcal{I}}$`. Because $\mathcal{I}$ is closed, to show that $f$ is in $\mathcal{I}$ it will suffice to find for each $\epsilon>0$ a $g\in\mathcal{I}$ with $\|f-g\|_\infty<3\epsilon$. Define $U_0=f^{-1}(-\epsilon,\epsilon)$, so $U_0$ is an open set containing $K_\mathcal{I}$. For each $y\in X\setminus U_0$, because $y\notin K_\mathcal{I}$ there is an $f_y\in \mathcal{I}$ such that $f_y(y)\neq0$. Define $$g_y=\frac{f(y)}{f_y(y)}f_y$$ and `$U_y=\{x\in X:|g_y(x)-f(x)|<\epsilon\}$`. Then $U_y$ is an open set containing $y$. The closed set $X\setminus U_0$ is compact, so there are finitely many points $y_1,\dots,y_n\in X\setminus U_0$ such that $U_{y_1},\ldots,U_{y_n}$ cover $X\setminus U_0$. Relabel: $U_k = U_{y_k}$ and $g_k=g_{y_k}$. Let $\varphi_0,\varphi_1,\ldots,\varphi_n$ be a partition of unity subordinate to the open cover $U_0,U_1,\ldots,U_n$. Finally, define $g=\varphi_1 g_1+\cdots+\varphi_n g_n$. That should do it.
In particular, a closed ideal is maximal if and only if the corresponding closed set is minimal, and because points are closed this means that maximal ideals correspond to points. (Maximal ideals are actually always closed in a Banach algebra, real or complex.)
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To wrap things up, I might as well purchase and wear a donkey costume.. – Akela May 25 2010 at 22:09
I'm not sure how to respond to that, but I hope you don't think I am disparaging your question. – Jonas Meyer May 25 2010 at 22:25
No I didn't mean that. I got thinking into all absurd directions with the wrong premises. I like your answer. This is a complete solution modulo the solution of the exercise. Would you please include a sketch of the proof? Also it may be a good idea to include explicitly that the maximal ideals in C(X,R) are precisely the points of the space. – Akela May 25 2010 at 22:30
I mean I accepted your answer. Since it's an accepted answer, it would be more helpful to any readers when it's complete with more details .. – Akela May 25 2010 at 22:30
Sure, I'll add some details. – Jonas Meyer May 25 2010 at 22:36
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Qn 1 is trivial because you said "topological spaces" rather than "compact Hausdorff spaces" (or "locally compact Hausdorff" would be okay, I guess). Simply $\lbrace 0,1\rbrace$ with the order topology and $\lbrace 0\rbrace$ will do.
If we refine to "compact Hausdorff spaces" then I take $C(X,\mathbb{R})$ and $C(Y,\mathbb{R})$, complexify, and apply GN to recover $X$ and $Y$, thus I claim that no counterexample exists.
I think that the issue stems from a confusion between the complexification of a real algebra and the underlying real algebra of a complex one. Since I can recover $C(X,\mathbb{C})$ from $C(X,\mathbb{R})$, all the information about the former is captured in the latter. However, since I can find several complex structures on the same real algebra, $C(X,\mathbb{C})_{\mathbb{R}}$ does not contain all the information that is contained in $C(X,\mathbb{C})$. There is a reason why they are called forgetful functors!
So $C(X,\mathbb{R})$ is not the underlying real algebra of $C(X,\mathbb{C})$, but $C(X,\mathbb{C})$ is the complexification of $C(X,\mathbb{R})$.
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Sorry, I unconsciously typed "topological" instead of "compact Hausdorff". How do you get the *-structure after complexification? – Akela May 25 2010 at 14:42
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The next-to-last R should be a C. – S. Carnahan♦ May 25 2010 at 15:01
@Scott: Thanks, fixed. @Akela: since C(X,R)oC = C(X,C), you get it from the same place as normally. – Andrew Stacey May 25 2010 at 20:14
I understand that C(X,R)oC is a complex Banach algebra. What I do not understand is how you gave it the structure of a C*-algebra, as "normal". As I understand Gelfand-Naimark theorem, you need commutative C*-algebras, not commutative Banach algebras. – Akela May 25 2010 at 20:21
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Dear Akela, $C(X,\mathbb R)\otimes_{\mathbb R} \mathbb C$ is isomorphic to $C(X,\mathbb C)$ as a $\mathbb C$-algebra. This is all that is needed to then recover $X$ from the usual Gelfand--Neumark theorem: one takes all maximal ideals, topologizes them via the weak topology, and this is $X$, by Gelfand--Neumark. – Emerton May 25 2010 at 20:55
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The noncommutative Gelfand-Neumark theorem can be stated and proved for real C*-algebras. See Corollary 4.10 in Johnstone's book “Stone Spaces”.
P.S. “Gelfand-Naimark” theorem is a misnomer. Take a look at the original paper and note how Gelfand and Neumark spell their names. In fact, they consistently use these spellings throughout all of their non-Russian papers.
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I don't think 'oxymoron' means what you think it means. – HW May 25 2010 at 22:04
"Misnomer" instead of "oxymoron" seems apt. However, it is confusing to us ignorant of Russian and the subtleties of its transliteration to Latin characters, because the name is written as "Naĭmark" in some of his other work. (For example, springerlink.com/content/v71158h17p227p39) I believe some choose "Gelfand-Naimark" for simplicity and consistency, but I appreciate your point. – Jonas Meyer May 25 2010 at 22:15
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@Henry: Inconceivable! – Yemon Choi May 25 2010 at 22:19
@Henry: By an oxymoron I meant “a combination of contradictory or incongruous words”. – Dmitri Pavlov May 26 2010 at 4:46
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@Jonas: It's either Gelfand-Neumark if you cite their non-Russian papers, or Gelʹfand-Naĭmark if you cite one of their Russian papers and use the AMS transliteration system. The spelling Naimark is incorrect and should never be used. As for the cited errata, note that the original paper springerlink.com/content/n3m5656p81712676 lists both spellings. – Dmitri Pavlov May 26 2010 at 5:23
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http://mathoverflow.net/questions/121018/intersections-of-anisotropic-tori-with-split-levi-subgroups | ## Intersections of anisotropic tori with split Levi subgroups
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Let G be a connected reductive group defined and split over a finite field k with Frobenius morphism F. Let T be an F-stable minisotropic maximal torus. Let P be an F-stable proper parabolic subgroup with F-stable Levi decomposition P=LN. Let Z be the center of $G^F$.
$T$ being minisotropic means that $T^F$ is contained in no proper (split) parabolic subgroup (ergo, in no proper Levi). My question is: when can $T^F$ can have some noncentral intersection with a Levi?
We know that $Z \subseteq T^F \cap L^F$. Equality holds if, for example:
(a) $P=B$, a Borel (since another characterization of minisotropic is that the maximal split torus contained in $T^F$ coincides with the maximal split torus contained in $Z(G^F)$)
(b) $G^F=GL(p,k)$, p prime (since an anisotropic torus is generated by an elliptic element, and there are no intermediate extension fields over which some element could split)
(c) $G=Sp(4,k)$ and $T$ the Coxeter torus (since the Coxeter torus has order $q^2+1$, whose gcd with the order of each torus in each proper Levi is 2)
In contrast, there exist choices of $T$ and $L$ for which the intersection is noncentral:
(a) $G=Sp(4,k)$ and $T$ the torus corresponding to $w=-1$, which has order $(q+1)^2$ and is two copies of the group of norm-1 elements of a quadratic extension field (and so meets the parabolic with Levi $SL(2)\times GL(1)$ in an anisotropic torus).
(b) In $GL(4,k)$, I believe one could find a $t \in T^F\setminus Z$ which splits over an intermediate extension field and has a conjugate lying in $L^F=GL(2,k)\times GL(2,k)$.
This question arises when one tries to compute the simple restriction of a Deligne-Lusztig cuspidal character to a parabolic subgroup; when equality holds, the result is just given by the central character and the Green function $Q_T^G$.
I think this is closely related to the following question on p-adic groups (relevant for the construction of supercuspidal representations): when does an anisotropic-mod-center torus lie in a proper twisted Levi subgroup?
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Can you elaborate on the line before the question? It's confusing at first sight to have a subgroup `$Z$` of the finite group `$G^F$` on the left of the inclusion, but on the right a closed algebraic subgroup of `$G$`. Is `$L$` here supposed to be `$F$`-stable? Also, is there a reference for the assertion before the question, and the details about your examples? (Minor suggestion: highight the question by replacing ""Q: When ..." with "> When ....") – Jim Humphreys Feb 6 at 23:38
Thank you very much for the corrections (sorry for the delay of this comment): indeed all groups are F-stable, T is maximal (ensuring, in particular, that it contains Z); and the groups in question are $T^F$ and $L^F$. – Monica Feb 8 at 16:12 | {"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": 25, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9159102439880371, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31431 | ## Can a mathematical definition be wrong?
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This question originates from a bit of history. In the first paper on quantum Turing machines, the authors left a key uniformity condition out of their definition. Three mathematicians subsequently published a paper proving that quantum Turing machines could compute uncomputable functions. In subsequent papers the definition of quantum Turing machine was revised to include the uniformity condition, fixing what was clearly a mathematical error the original authors made.
It seems to me that in the idealized prescription for doing mathematics, the original definition would have been fixed, and subsequent papers would have needed to use a different term (say uniform quantum Turing machine) for the class of objects under study. I can think of a number of cases where this has happened; even in cases where, in retrospect, the original definition should have been different.
My question is: are there other cases where a definition has been revised after it was realized that the first formulation was "wrong"?
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You only have to read e.g. James' History of Topology to learn of countless reformulations of the definition of manifold, fundamental group, homology... – Qiaochu Yuan Jul 11 2010 at 4:53
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I think whether or not 0 is a natural number and whether or not all "rings" have 1 are still not settled issues! – Greg Friedman Jul 11 2010 at 6:14
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@Greg: I will settle the latter controversy right now: Bourbaki says all rings have a multiplicative identity. This is a plain and simple argument from authority. – Harry Gindi Jul 11 2010 at 6:59
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The real question is why anyone thinks the phrase "natural numbers" is preferable to "nonnegative integers" or "positive integers." – JBL Jul 11 2010 at 19:34
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I want to praise the author of this question for the philosophical neutrality of the phrasing. – Alexander Woo Jul 11 2010 at 23:45
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## 16 Answers
Here's my favorite example. Imre Lakatos' book Proofs and Refutations contains a very long dialogue between a teacher and pupils who debate what are good definitions of polyhedra, with respect to a claimed proof that $V-E+F=2$ is true for polyhedra. It's common that a good definition (or reformulation) of a concept can help yield proofs of theorems, and this book promotes the "dual" view that a proof of a theorem can lead to a good definition in hindsight.
The footnotes of this dialogue show that Lakatos is actually tracing the history of the Euler characteristic in the mathematical literature. In short, both the definition(s) and the proof(s) went through substantial revisions over time.
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This issue is also masterfully discussed in Euler's Gem by Richeson: amazon.com/Eulers-Gem-Polyhedron-Formula-Topology/… – Qiaochu Yuan Jul 11 2010 at 5:45
The first appendix to Proofs and Refutations contains another example: definitions of continuity. – Seamus Jul 15 2010 at 14:50
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The problem is that Lakatos discusses a mathematical praxis before the "modern" revolution and the definition of polyhedron is not even stated. – Leo Alonso Dec 1 2010 at 10:07
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This is not quite an answer to the question you are asking, but the definition of a function is an obvious example of a concept that underwent considerable change, though perhaps one might argue that the eighteenth-century notion of a function was never formally defined. Amusingly, there are many textbooks that attempt to give a formal definition of function but get it wrong. (This is not my observation but something spotted by a colleague of mine who studies mathematical language.)
What the books do is say, "A function from A to B is a subset $F$ of AxB such that for every x in A there is a unique y in B such that $(x,y)\in F$." But if that is all you say, then two functions can be equal even if they have different codomains, which the authors of these same textbooks clearly don't intend if they ever mention surjections, bijections or inverse functions. There's an easy fix, which is to define a function to be an ordered triple consisting of A, B and the subset of AxB, but almost no books do this. (I'm talking here about introductions to undergraduate-level mathematics rather than books about axiomatic set theory.)
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There was a discussion of this point here very recently: mathoverflow.net/questions/30381/… – JBL Jul 12 2010 at 12:19
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@gowers: Many set theorists actually define a function this way. – Harry Gindi Jul 14 2010 at 6:22
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It seems inaccurate to portray the logician's standard definition of function as an amusing failed attempt at precision. Although other definitions are also common, this particular definition is used with consistency and precision throughout logic and set theory and other areas, including introductory undergraduate texts. (There are no problematic issues with surjections, bijections or inverse functions.) The link in JBL's comment has further explanation. – Joel David Hamkins Dec 2 2010 at 1:00
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@Tom Goodwillie: What is the problem with a function with empty domain? This is the empty function, which is 1-1, and its left inverse is the empty function itself (composition of the two functions is the identity on the epmty set, which is again the empty function). – Stefan Geschke Dec 2 2010 at 9:01
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@Stefan, there is no function from a non-empty set to the empty set. (Not even "the empty function".) – gowers Dec 2 2010 at 12:18
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I think there are many examples, spread out across a continuum of how "wrong" the definition really was. Of course, strictly speaking a definition cannot be "wrong", or can only be wrong in the logical sense of not umambiguously denoting a class of examples. E.g., Newton's and Leibniz's definition of the derivative was wrong -- or better, not well-defined! -- because it used infinitesimals in a way that was not formalized and could not be formalized in the context of known mathematics.
There are a lot of definitions that in retrospect look too limited or pedestrian: e.g., defining a manifold to be a certain kind of subset of Euclidean space. (Some people would say that the definition of a Riemann integrable function is "wrong" in this sense. I disagree -- the notion of Riemann integrability is a natural one that comes up e.g. in characterizations of uniform distribution of sequences.)
It seems like you are looking for examples of the following kind: the definition is given and then, in the same paper (or book, or whatever) a theorem is given using the definition. But contemporary mathematicians who look back at the theorem agree that the conclusion is not the desired one.
I can think of one instance of this, although it is of relatively minor importance. R.G. Bartle's 1955 paper Nets and Filters in Topology was one of the first to try to explicitly work out the folkloric "equivalence" between nets and filters when studying convergence on topological spaces. The way to do this is to, given a net on a topological space, associate a filter, and conversely, and then prove theorems about these associated nets and filters having the same convergence properties. But the definition Bartle gives of how to associate a net to a filter is "wrong", in the sense that certainly you want that when you in turn associate a filter to that net you get the filter that you started with, but his definition does not have this property (and the right definition does!). See for instance the last page of
http://math.uga.edu/~pete/convergence.pdf
for some more discussion of this.
In general, I would think that one has to be rather well-read in a subject area to come up with such examples, because -- thankfully! -- a truly "wrong" definition is usually swiftly drowned out by the correct definition.
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Wow,Pete-that's one of my favorite papers and I never noticed that!!! But it's a very subtle point and something even an analyst of Bartle's stature-and he was a VERY good one,apparently-missed. – Andrew L Jul 11 2010 at 6:47
Dear Pete L. Clark, there is a typo in the first reference in your above-mentioned paper. – Rasmus Jul 11 2010 at 13:18
@Rasmus: Thanks for pointing it out. I have reformatted the references. – Pete L. Clark Jul 11 2010 at 17:25
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"Of course, strictly speaking a definition cannot be "wrong", or can only be wrong in the logical sense of not umambiguously denoting a class of examples." I am not inclined to agree with this. One ought instead to say that the standard codified canons of logic that we are all taught do not tell us what right and wrong definitions are (beyond saying they must be unambiguous; they must not be circular, etc.). We should not rule out the possibility of future codification of such things based on understanding that has not yet been accomplished. – Michael Hardy Aug 7 2010 at 22:36
I was involved in such a case. My thesis advisor (S. Husseini) and his coauthor (E. Fadell) defined the "category weight" of a cohomology class in a way that allowed them to prove the theorem they wanted. But when I started looking at it, I noticed that it was not homotopy invariant, and all the applications were homotopy invariant. So I introduced the homotopy invariant version and called it "essential category weight." At roughly the same time Y. Rudyak made the same observation, and also defined a homotopy invariant version, calling it "strict category weight." After a few years of using competing terminology for the same concept, and not using the original "category weight" at all, everybody agreed to call the homotopy invariant version "category weight."
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Thanks. This is exactly the kind of case I was looking for when I asked the question. – Peter Shor Mar 16 at 15:51
In higher category theory, there have been examples of "wrong definitions". An example of such a definition is the definition of a strict 3-category. It is "wrong" because it doesn't include the fundamental 3-groupoid of the 2-sphere, see http://lanl.arxiv.org/abs/math/9810059.
Among mathematicians (unlike physicists!), it is common practice to accompany one's claims by proofs.
But we are not used to accompanying our definitions by proofs.
In higher category theory, things are different:
We should probably prove that our definitions are correct before being allowed to go on.
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You know,this is why I never feel comfortable with category theory no matter how hard I try to-insane things like THIS happen in it......... – Andrew L Aug 6 2010 at 16:27
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This is not insane. – Kevin Lin Aug 7 2010 at 23:37
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No, it's not insane. In fact I don't agree with Andre that this notion is "wrong" (although I understand why he says that) -- it's just a natural extension of defining a 2-category as a Cat-enriched category: a strict n-category is an (n-1)-Cat-enriched category. Strict n-categories are still a technically useful notion; for example, Batanin's notion of weak oo-category starts with the monad on globular sets whose algebras are strict oo-categories, and develops an associated notion of operad. Weak oo-categories are then algebras over certain contractible such operads. – Todd Trimble Sep 19 2010 at 12:53
An important historical example is the difficult evolution of the correct definition of "integer" in algebraic extensions, i.e. defining algebraic integers. It was only with great difficulty that Dedekind discovered the necessity of passing to integrally closed extensions in order to obtain nice factorization theories. Similar struggles were encountered while distilling the correct notion of integral elements for quaternion rings.
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This happened with model categories, where Quillen's original definition only required the existence of finite limits and colimits as well as a factorization of all maps that wasn't necessarily functorial. Almost all books today take model category to mean what Quillen called a closed model category (and in fact, they go one step further. They add functorial factorization as well), since it makes the proofs easier and the conclusions much more far-reaching.
This is at the cost of losing some categories of some kinds of finite chain complexes (I have never run into one of these in practice, but I suppose that some people do) as model categories, but this stronger definition includes almost every usage of model-category theory in homotopy theory.
I will note, however, that while model categories are almost always taken to be closed (outside of Quillen's original paper), functorial factorization is not nearly as standard (if I remember correctly, Jacob Lurie doesn't require functorial factorization in his definition in Higher Topos Theory).
For more, you can read the introductions to:
Model Categories by Mark Hovey
Model Categories and their Localizations by Phil Hirschhorn
Homotopy Limit Functors on Model Categories and Homotopical Categories by Dwyer, Hirschhorn, Kan, and Smith
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@Andrew L : I think you miss the point here. The opposite of what you described is true : Quillen's original definition was actually too general! – Andy Putman Jul 11 2010 at 4:57
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Andrew L, Quillen has even adopted the newer definition (he was one of the first to do so, in fact). (cf. Rational Homotopy Theory (1969) by Dan Quillen). – Harry Gindi Jul 11 2010 at 5:00
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I agree with Pete and Andrew -- changing the "default" level of generality of a definition (whether it's making it more general or less general) doesn't mean the original definition was "wrong." – Mike Shulman Jul 11 2010 at 5:26
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An example in the opposite direction is the change in terminology from "pre-scheme" to "scheme" in the second edition of EGA. – Victor Protsak Jul 11 2010 at 7:09
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For the record, I am interested in changes of definition even in the case where in cannot be argued that the original definition was "wrong" (whatever that means for a definition). – Peter Shor Jul 11 2010 at 23:58
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Here's a rather mundane example: a basis of a vector space. A basis is usually defined to be "a set of vectors such that...." The problem with this is the following: $$\begin{bmatrix}0 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix}$$Do the columns form a basis for $\mathbb{R}^2$? The answer is "yes" if a basis is a set of vectors...but this is obviously false.
The same applies to the usual definition of linear independence. Are the columns of the following matrix linearly independent? $$\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}$$The answer is apparently "yes," since in my experience linear [in]dependence is usually defined only for a set of vectors, and the set of the columns of that matrix consists of a single nonzero vector...again, obviously false: the columns of a square matrix shouldn't be linearly independent unless the matrix is invertible.
In my mind, the collections of vectors to which pretty much all linear algebra concepts apply are tuples of vectors, not sets. This stems from the fact that the fundamental operation on a vector space is the linear combination, which operates on a tuple of vectors and a tuple of scalars (indexed by the same set). If I want to figure out whether a collection of vectors spans a space or is linearly independent, the next thing I'm going to do is consider a linear combination of those vectors. Therefore, it only makes sense for that collection to be something to which "linear combination" readily applies, namely a tuple! Still, I've seen reputable textbooks define everything only in terms of sets.
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I agree that tuples are better than sets in this context, but I think indexed families are better yet. If the indexing set happens to be the integers from 1 to $n$, then you have tuples. But if you're dealing with (the underlying vector space of) the group algebra of a finite group $G$, then an obvious basis is indexed by $G$, and it seems unnecessary and unnatural to impose an (arbitrary) ordering on $G$ just to get tuples. (For similar reasons, I like to think of matrices as having their rows and columns labeled by some indices but not necessarily by natural numbers.) – Andreas Blass May 17 2011 at 13:56
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@Andreas - thanks for clarifying. That is in fact exactly what I meant! I understand "tuple" in the generalized sense, as in an "X-tuple" is nothing but a function with domain X. X doesn't need to be {1,...,n} or {0,...,n-1}; it could be any index set. (I'm not making this up: tuples are treated this way in Bergman's An Invitation to General Algebra and Universal Constructions, and the Wikipedia article mentions this perspective as well.) – Darsh Ranjan May 17 2011 at 17:31
If a definition can be tentative, it can also be wrong. Lakatos has been mentioned already. This is actually a fairly basic issue in understanding how "formal" mathematics advances. Something as fundamental as integration has seen inadequate definitions such as the Riemann integral pushed out by the Lebesgue integral. In this case the "wrong" definition has not simply been supplanted, though.
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Charles,this is a bad example since the Riemann integral is perfectly fine if the functions are defined on compact support and the number of discontinuities is at most countable. It wasn't that the Riemann integral was "wrong",it was simply discovered to be too restrictive for the purposes of modern analysis. – Andrew L Jul 11 2010 at 20:06
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No, it's not perfectly fine if you want a theory of Fourier transforms. – Charles Matthews Jul 11 2010 at 20:44
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@Charles If the same restrictions are imposed,I believe the theory of Fourier transforms can be established just fine with the Riemann integral. But of course,this is too restrictive for a general theory,which is why the Lebesgue integral supplanted it. – Andrew L Jul 11 2010 at 23:55
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In Rudin's Real and Complex Analysis he introduces the Lebesgue measure as the measure (say on the closed unit interval) such that integration with respect to this measure gives you the positive linear functional on C([0,1]) that is called "Riemann integral". So he actually defines the Lebesgue measure in terms of the Riemann integral. I agree that this approach is sort of odd, but very elegant from a certain point of view. My students didn't like this approach, though. I think they would have preferred one of the more common approaches to the Lebesgue integral. – Stefan Geschke Dec 2 2010 at 9:10
I'm not terribly familiar with this material, but since there's a good chance nobody will say something about it, I'll chip in with an example (I think?):
The Hardy spaces $H^p$ were originally defined in terms of complex functions on the unit disk. Namely, an element of $H^p$ is a holomorphic function $f$ on the unit disk such that $\sup_{r} \int_{0}^{2 \pi} |f(re^{i \theta})|^p d \theta$ is finite. This quantity (in analog with the $L^p$ norms) is used to define the norm on $H^p$, so this is a Banach space for $p \geq 1$ (and a Hilbert space for $p=2$). There is a very rich and interesting theory of these complex Hardy spaces. For instance, radial limits exist almost everywhere (though this is also true for the broader case of $f$ in the Nevanlinna class), and have vanishing Fourier coefficients at negative indices. The function $f$ can be reconstructed from the boundary data via a Poisson integral. More interestingly, the corona theorem is a statement about the spectrum of the Banach algebra $H^\infty$; it states that the ideals $M_z := {f: f(z)=0}$ for $z$ in the unit disk are dense. All this is based upon the complex-variable theory, which came first.
However, this original definition via complex-variable theory was "wrong" in the sense in that it had to be modified to allow for the real-variable theory in higher dimensions. The "real-variable" definition of an element of $H^p$ is defined in terms of distributions with "maximal functions" (defined with respect to a normalized Schwarz function) in $L^p$. Much of the modern theory of Hardy spaces (e.g. duality of $H^1$ and $BMO$, stability under singular integrals) was developed, I think, in this more general setting (which, according to Stein, took hold in the 1960s).
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This is a slightly different kind of example, namely, one where the original definition had to be revised when it later was realized that it was useless in certain contexts. Probably most readers have encountered the Dorroh extension [1] as a way to adjoin 1 to a rng (ring without unit). While various arguments can be made for the naturality of this construction, it turns out that this is the wrong definition in many contexts because it doesn't preserve crucial properties of the source rng and/or doesn't satisfy various minimality properties. For much further discussion see [2]. I mention it primarily as another perspective on the way that definitions may evolve. A word of warning: I've mentioned this many times over the years and almost always someone argues tooth-and-nail for the naturality of the Dorroh unital extension without first appreciating the issues that arise in contexts outside their expertise, e.g. see the AAA thread [3]. To avoid that here I highly recommend first perusing [2] before commenting.
Added later: remarks in the comments below lead me to believe that posting the the following excerpt from the introduction of [2] may help serve to alleviate any further confusion:
"It was observed long ago as 1932 (Dorroh's Theorem) that any non-unital ring $R$ may be embedded in a ring with unity. This is done by adjoining a copy of $\mathbb{Z}$, the ring of integers, to $R$. This does not preserve all the nice properties which $R$ might have, nor is it minimal in any of various senses; and so over the decades many embeddings have been invented to serve diverse purposes. For example, if $R$ is regular (or some generalization of regular such as $\pi$-regular) one would like to embed $R$ into a regular ring (or the generalization). There are other sorts of properties (semiprime, artinian domain, Ore domain) which one may wish to preserve in going from $R$ to some ring with 1, say $R^1$, all the while without adjoining anything more than necessary. It turns out that there is one construction [...] which will give all the main results as well as some new ones, although there is not yet one proof by which to do it. In the case of the generalized sorts of regularity, the ring formed [...] satisfies a universal property with respect to the adjunction of 1."
[1] http://en.wikipedia.org/wiki/Pseudo-ring
[2] W.D. Burgess; P.N. Stewart.
The characteristic ring and the "best" way to adjoin a one.
J. Austral. Math. Soc. 47 (1989) 483-496 http://anziamj.austms.org.au/JAMSA/V47/Part3/Burgess/p0483.html
[3] rings, ideals and correspondence theorem -- clarification requested. Ask an Algebraist, 8/4/2008
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I would like to request that Mr. Dubuque remove reference [3] from his response. It's not an example of mathematics but rather of several people squabbling in a rather unprofessional way (ostensibly) about mathematics. In the interest of full disclosure, let me add that one of those people is me, and this thread contains the things that I most regret having posted to the internet, ever. – Pete L. Clark Jul 11 2010 at 20:00
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I vote out of respect for Pete's wishes,Mr.Dubuque comply.That being said-and with the disclaimer I haven't carefully read it-that Pete shouldn't be afraid to act human in front of his collegues and they should be understanding if such incidents occur when passions flare. Scientists are human and there should be no shame in that. – Andrew L Jul 11 2010 at 20:03
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PS I just noticed that PLC revealed his anonymous participaton in the linked thread. Frankly I had forgotten about that since I have seen so many such discussions before that I didn't recall the particpants. I suggest instead of censoring my link that PLC simply remove his comment - thus alleviating any potential embarassment on his part. – Bill Dubuque Jul 11 2010 at 21:27
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I didn't downvote this because of what Pete said. Rather, I downvoted because I don't think this answers the question. Other mathematicians have not adopted the alternate definition despite the fact that it has been available for many years. Given that the notion of a "correct definition" is in many ways a question of aesthetics, the long-standing community consensus seems to trump your private campaign. – Andy Putman Jul 11 2010 at 22:57
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At the risk of commenting on an old discussion: this being a CW question, all answers are also CW, and so anyone, e.g. Pete, is doing nothing wrong by removing the reference themself. – Theo Johnson-Freyd Aug 6 2010 at 21:10
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An example from algebraic geometry :
At some point during the redaction of the EGA by Grothendieck and Dieudonne, Grothendieck discovered how to make parts of the theory work without finiteness (noetherian) hypotheses on schemes, by strenghtening finiteness for morphisms (finite presentation instead of finite type). The study of morphisms of finite presentation was carried out in EGAIV.
Unfortunately, some definitions of properties of morphisms were made before this discovery. In particular, the definition of a proper morphism in EGA only includes finite type and not finite presentation, mostly because it was first used for noetherian schemes. This can lead (and lead some fine french mathematicians) to spend entire classes repeating the words "morphismes propres de présentation finie"...
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As someone who regularly posts questions on MO asking things like "what is the correct definition for XYZ", I strongly believe that the answer to the question is "yes". For me, a definition is correct if it both matches and helps refine intuition.
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Yes, in the sense that a careless definition can actually introduce unsoundness into a formal framework. See Norm Megill's explanation [1] as well as Raph Levien's Ghilbert [2] which is (apparently) a solution to the problem.
[1] http://us.metamath.org/mpeuni/mmset.html#definitions
[2] http://wiki.planetmath.org/AsteroidMeta/Ghilbert
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The Euler Characteristic in the statement of Riemann-Roch was redefined by Grothendieck to be in the K-group, which he presumably(ie, I imagine so) constructed for this purpose.
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Here is a bad definition of integrability that my analysis instructor taught me about and explained why it was bad. Consider a real-valued function f with the interval [a, b] as its domain. Partition [a, b] evenly into n intervals of length (b-a)/n. If the upper Darboux sum and lower Darboux sum of f in these partitions converge and equal each other as n approaches infinity, then call f integrable on [a,b]. To see why this definition fails, consider the following function:
f(x) = n if x = 1/n for some natural number n
f(x) = 0 otherwise
On [0,1], the upper Darboux sums of f on partitions from this definition must be greater than or equal to 1 and the lower Darboux sums converge to 0. So f is not integrable with respect to this definition. But f is Riemann integrable.
Edit: Okay, I don't remember the exact example he used.
Riemann-integrable functions should be bounded by the usual definition, so f is not Riemann-integrable. But nonetheless the limit $R-\int_a^1 f(x)\,dx$ for $a\to 0$ exists. – Johannes Hahn Dec 1 2010 at 16:12
Agreed, this function $f$ is not Riemann integrable according to the usual definition. But is the above is a bad definition? Can we "simplify" the usual definition (allowing arbitrary finite partitions) by requiring only "equally spaced" partitions? Isn't this an exercise in "advanced" presentations such as Rudin? – Gerald Edgar Dec 1 2010 at 16:31 | {"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": 37, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9431182146072388, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/65233/lifting-varieties-from-char-p-to-char-0-after-alterations | ## Lifting varieties from char. $p$ to char. 0 after alterations
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The question is related to this MO question:
http://mathoverflow.net/questions/25337/lifting-varieties-to-characteristic-zero
Let $X$ be a projective smooth variety over $k$ alg. closed field of char. $p.$ Does there always exist an alteration $Y\to X,$ with $Y$ also projective smooth, such that $Y$ lifts to char. 0?
Side remark: over $k=\overline{\mathbb F}_p,$ modulo Tate conjecture, abelian varieties "generate" the motives of all proj. smooth varieties. Since abelian varieties are liftable, one can say that the (irred. components of) motives of any proj. smooth varieties is liftable in some sense. And I wonder if this can be realized geometrically. "Alteration" in my question is just a try; replace it with any reasonable geometric construction if you want. For instance, "a proper surjection" would be fine.
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I know Bhargav Bhatt in his thesis asked whether "any variety can be dominated by a smooth one that lifts $W_2(k)$?" – Karl Schwede May 17 2011 at 15:39
Remark 5.5.5 by the way. – Karl Schwede May 17 2011 at 15:40
I'm not sure if this is also part of your motivation, but I'll just add that Torsten Ekedahl points out that after a sequence of curve contractions and a deformation even Hirokado's example of a non-liftable CY threefold lifts to char 0. – Matt May 17 2011 at 16:45
@Karl and Matt: These are certainly good motivations, which I didn't know before. Thank you for letting me know. Karl, do you know where I can find Bhatt's thesis? – shenghao May 17 2011 at 18:55
OK, it seems that this question is still open. – shenghao Jun 27 2011 at 15:51 | {"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": 9, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9160706400871277, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/36167/does-juggling-balls-reduce-the-total-weight-of-the-juggler-and-balls/36212 | # Does juggling balls reduce the total weight of the juggler and balls?
A friend offered me a brain teaser to which the solution involves a $195$ pound man juggling two $3$-pound balls to traverse a bridge having a maximum capacity of only $200$ pounds. He explained that since the man only ever holds one $3$-pound object at a time, the maximum combined weight at any given moment is only $195 + 3=198$ pounds, and the bridge would hold.
I corrected him by explaining that the acts of throwing up and catching the ball temporarily make you 'heavier' (an additional force is exerted by the ball to me and by me onto the bridge due to the change in momentum when throwing up or catching the ball), but admitted that gentle tosses/catches (less acceleration) might offer a situation in which the force on the bridge never reaches the combined weight of the man and both balls.
Can the bridge withstand the man and his balls?
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did anyone else chuckle when they read that last sentence? – David Peterman Sep 11 '12 at 20:59
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...must ask question about a bridge that can withstand a truck but not the flightful birds inside it. – aitchnyu Sep 12 '12 at 5:56
Strange, I just read this in a book called something like "Fantastic Science Puzzles" (I don't have it on me), and was going to ask this very question tomorrow, as I am almost certain it is wrong. – BlueRaja - Danny Pflughoeft Sep 12 '12 at 8:11
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If the man juggles vigorously for long enough, he'll lose weight because of the exercise; there will also be some wear on the balls. – Keith Thompson Sep 12 '12 at 18:58
Although... if you toss a single ball in to the air really high, step on to the bridge, run across really fast, and catch the ball on the other side, you could make it! – NeuroFuzzy Sep 13 '12 at 5:27
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## 7 Answers
Suppose you throw the ball upwards at some speed $v$. Then the time it spends in the air is simply:
$$t_{\text{air}} = 2 \frac{v}{g}$$
where $g$ is the acceleration due to gravity. When you catch the ball you have it in your hand for a time $t_{\text{hand}}$ and during this time you have to apply enough acceleration to it to slow the ball from it's descent velocity of $v$ downwards and throw it back up with a velocity $v$ upwards:
$$t_{\text{hand}} = 2 \frac{v}{a - g}$$
Note that I've written the acceleration as $a - g$ because you have to apply at least an acceleration of $g$ to stop the ball accelerating downwards. The acceleration $a$ you have to apply is $g$ plus the extra acceleration to accelerate the ball upwards.
You want the time in the hand to be as long as possible so you can use as little acceleration as possible. However $t_{\text{hand}}$ can't be greater than $t_{\text{air}}$ otherwise there would be some time during which you were holding both balls. If you want to make sure you are only ever holding one ball at a time the best you can do is make $t_{\text{hand}}$ = $t_{\text{air}}$. If we substitute the expressions for $t_{\text{hand}}$ and $t_{\text{air}}$ from above and set them equal we get:
$$2 \frac{v}{g} = 2 \frac{v}{a - g}$$
which simplifies to:
$$a = 2g$$
So while you are holding one 3kg ball you are applying an acceleration of $2g$ to it, and therefore the force you're applying to the ball is $2 \times 3 = 6$ kg.
In other words the force on the bridge when you're juggling the two balls (with the minimum possible force) is exactly the same as if you just walked across the bridge holding the two balls, and you're likely to get wet!
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Am I the only person who noticed you converted the OPs pounds to kg using a 1:1 ratio? Were you assuming that g~=22m/s/s? – Dan Neely Sep 12 '12 at 20:40
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I didn't want to put the force in Newtons because I suspected that might confuse the OP. I also didn't want to put the units in as kgf for the same reason. I used kg as the unit because outside of us physics geeks your normal public doesn't make a distinction between mass and weight/force. – John Rennie Sep 13 '12 at 6:15
After thinking about this, I don't buy the $t_{\text{hand}}$ = $t_{\text{air}}$ thing. If I launched the balls a mile into the air with a cannon, acceleration is much greater, but the time in the air could still equal the time in the cannon... – adamdport Sep 16 '12 at 1:22
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@adamdport: when you fire the cannon the force on the bridge will be much higher than the weight of the ball. The lowest acceleration, so the lowest force on the bridge, is got by accelerating the ball uniformly for as long as possible i.e. you spread the acceleration out. – John Rennie Sep 16 '12 at 7:46
Nice neat simple answer! Like it alot. – Killercam Nov 29 '12 at 15:54
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I love this class of problem as a fantastic physical example of the mean value theorem. Allow me to describe a specific case that fits the following conditions:
• The man plus the balls has a total weight of $m$
• The entire system (man+balls) starts at rest and ends at rest
From these relatively simple assumptions, I will claim that the average normal force (the force the ground exerts upward) is equal to the weight of the the system. In other words, for a given period of time of length $T$ we have this:
$$m g = \frac{1}{T} \int_0^T \vec{F}(t) \cdot \vec{n} dt$$
This is a spectacular claim actually. To simplify the notation, consider that $\vec{F}(t) \cdot \vec{n}$ is just equal to the weight a scale would read (this isn't a bad assumption, depending on the scale). Imagine the man is juggling, standing on a scale, and the scale reads a value that depends on time, $w(t)$. The average value the scale reads will be equal to gravity times his mass, including everything he's holding or wearing.
In the story of the man walking across the bridge juggling balls, the total weight is $201 lb$. For every second he weighs $200 lb$, he spends one second weighing $202 lb$ or something similar. The point is that the average value is the same.
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I just realized - the dot product with the normal vector isn't actually needed, provided you write $g$ as a vector too. – AlanSE Sep 11 '12 at 21:50
"The entire system (man+balls) starts at rest and ends at rest" - this assumption is vital for this solution, otherwise it should be possible to reduce the weight on the bridge by either catching off the bridge or throwing when off the bridge – Casebash Sep 12 '12 at 16:05
– Casebash Sep 12 '12 at 16:27
But if you're allowed to do your first throw and last catch off either end of the bridge, aren't you better off doing a spectacular throw of one of the balls and then dashing across with the other? – Emilio Pisanty Sep 12 '12 at 16:52
@EmilioPisanty That's right, either way it violates the assumption of starting and ending at rest. The average weight may be less if the balls are moving up upon entering the bridge, or moving down upon exiting. This applies for the cases that you catch them in the middle or don't. There is another answer where there is significant discussion on this point, although I don't think it's been mathematically addressed yet. – AlanSE Sep 12 '12 at 17:29
Put one ball down. Walk the other across. Go back, get the second ball.
Or, roll the two balls across, then run after them.
Or, the juggler takes off his shoes and walks across barefoot.
This is solved as a "nonlinear thinking" problem, not with "juggling is anti-gravity". The ball-man system must be accelerated downwards with an average of 1 lb of force or the bridge will break. Otherwise you could build a perpetual motion machine from two jugglers on a see-saw who take turns juggling.
(Also, running is like juggling in that the weight is up in the air much of the time--if this could work, you could also just hold the balls and run.)
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+1 good point about the jugglers on a seesaw making a perpetual motion machine. – John Rennie Sep 12 '12 at 8:59
The original riddle actually stated that you had to do it in one pass (no second trip), a 3-pound cell phone and its 3-pound battery (no rolling), and that you were naked (no stripping). – adamdport Sep 12 '12 at 12:43
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@adamdport - Better start thinking about making a rope out of your hair, then. Or...have you gone to the bathroom lately? – Rex Kerr Sep 12 '12 at 15:34
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My allegedly "incorrect" answers had been "take a 1-pound plank off the bridge before crossing", "run leaning forward so your body acts like an airfoil" and "wait for high-tide and exploit lunar gravitational influences" ^_^ – adamdport Sep 12 '12 at 16:22
Imagine for simplicity that the juggler at some instant repeats himself, i.e. that the juggler and balls (with masses $M$ and $2m$, respectively) are in the precise same kinematic state at times $t_1$ and $t_2$.
Consider man + 2 balls as the system, and bridge, etc., as the environment.
Let $p(t)$ be (the vertical component of) the total momentum of the system.
Newton's second law applied to the system yields:
$$\tag{1} \dot{p}(t) ~=~ F_n(t) - F_g,$$
where
$$\tag{2} F_g~=~(M+2m)g,$$
and where $F_n(t)$ is the normal force from the bridge, which may vary in time $t$ as the juggler does his routine.$^1$
Because of our simplifying assumption of repeating states, we have
$$\tag{3} 0~=~p(t_2)-p(t_1)~=~ \int_{t_1}^{t_2} F_n(t)dt - (t_2-t_1)F_g,$$
or
$$\tag{4} F_g ~=~ \frac{1}{t_2-t_1} \int_{t_1}^{t_2} F_n(t)dt ~=~\langle F_n \rangle.$$
But if the average $\langle F_n \rangle$ is $F_g$, then clearly at at least one instance $t_3\in [t_1,t_2]$, one must have$^2$
$$\tag{5} F_n(t_3)\geq F_g.$$
In other words, the bridge collapses.
$^1$ The juggler is allowed to do whatever motion he thinks would benefit his case. Whether he wants to jump with both feet leaving the bridge, or lower his center-of-mass, or fall down, is up to him. It seems physically reasonable to assume that the normal force $F_n(t)$ is a piecewise continuous function of time $t\in [t_1,t_2]$, with only finitely many discontinuity points. In that case the integral $\int_{t_1}^{t_2} F_n(t)dt$ can be defined using the Riemann integral without involving the technically more complicated Lebesgue integral. (Also note that the mean value theorem does not apply for discontinuous functions, and from a mathematical purist point of view, the mean value theorem is not needed, i.e., the crucial ineq.(5) may be established with considerations that are even more elementary.)
$^2$ Indirect proof of eq.(5): Assume
$$\tag{6} \forall t\in [t_1,t_2]:~ F_n(t)~<~ F_g.$$
Then
$$\tag{7} \int_{t_1}^{t_2} F_n(t)dt ~<~ (t_2-t_1)F_g,$$
if we assume piecewise continuity $t\mapsto F_n(t)$. But eq.(7) is inconsistent with eq.(3). QED.
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... unless the period is greater than the time it takes for him to cross the bridge, i.e. he could throw the balls in the air, cross the bridge, then catch them on the other side ;) – UncleZeiv Sep 12 '12 at 15:10
I'm only interested in an idealized version of the puzzle(v1) that deals with whether the man can stay on the bridge juggling rather than whether he can make it across the bridge. – Qmechanic♦ Sep 12 '12 at 18:19
It depends on how long his arms are!! (and how long the bridge is) If he starts in first position, arms held high, and imparts -0.17G to his balls while crossing, he will make it. Oops. I did the math wrong in my comment.
Besides, he can do a juggler's trick, and !gradually lower his center of gravity! as he walks across the bridge. The juggling is optional, a distraction from what they are really doing. He only has to accelerate at G*(1/201) to have the bridge bear, not 201 lbs (195+6), but 200 lbs. If he can crouch down to 2 ft, I get 5 seconds to cross the bridge.
````1/2 ( 0.16 ft / s^2 ) t^2 = 2 ft
t = sqrt[ 4ft/(0.16ft) sec^2 ]
````
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+1 if you can work out how long his arms would have to be hahaha – adamdport Sep 12 '12 at 13:13
1/2 A t^2 = drop (he goes from arms up to arms down, so l=0.25At^2), but we do not know how fast the juggler walks. A is 32 ft/sec^2 * (1-1/6), apx 27 ft/sec^2. In one second, he needs 7 ft arms. – Bobbi Bennett Sep 12 '12 at 14:51
I think it might be possible if the man first throws one of the balls into the air before he steps on the bridge. In that case the man could apply 4 pounds of force upwards on one ball initially, then step on the bridge. At that point the bridge would be holding 198 pounds. The man can then accelerate the other ball upwards with 4 pounds of force before the other ball lands. This would mean that the bridge would be holding 199 pounds at that point. When both balls are in the air the bridge would be holding 195 pounds. Then the first ball would land in the man's hand, and the man would need to apply 4 pounds of force to decelerate it to rest. During deceleration the bridge would be holding 199 pounds. After deceleration the bridge would be holding 198 pounds. Then the man can repeat accelerating the ball upwards with 4 pounds of force as he crosses the bridge.
It may also be possible to do this if the balls had a large volume and you counted air resistance, in which case the air would help decelerate the balls as they fell down, but the man would still have to throw one of the balls into the air before he stepped on the bridge.
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No he can't, unless he throws the ball across the bridge, see Alan SE's answer. No downvote for now, but please delete this. – Ron Maimon Sep 11 '12 at 20:49
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AlanSE's answer assumes that the whole system starts at rest, which is not an assumption my answer makes. – Thomas Sep 11 '12 at 21:04
It makes no difference, if the average motion is not parabolic, conservation laws ensure that the average force supports the weight. This is found by drawing a big sphere around the man and the balls, and no matter what is happening in the sphere, the net momentum flux through the bridge must balance gravity. – Ron Maimon Sep 11 '12 at 21:31
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@RonMaimon The upward force on your control volume won't balance gravity by the initial and/or final upward/downward momentum. If the ball is thrown up right before entering the bridge, the man-ball system has upward momentum. – AlanSE Sep 11 '12 at 21:37
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@RonMaimon Yes, the momentum would be decreasing. If the man-ball starts with an upward momentum at the moment he enters the bridge, there will be a negative average rate of vertical momentum change over time. The proposition is to make it a non-steady state problem. – AlanSE Sep 12 '12 at 12:56
show 5 more comments
It thinks that its reasonable to assume: "The entire system (man+balls) starts at rest and ends at rest". Then we can completely avoid integrals and dealing with time. For the moment, let's just consider the speeds of the balls and pretend his arms have unlimited length. We can only provide 5 pounds of force per second => an acceleration of `5/3 g`, although this can be divided between two balls. The balls experience a downward acceleration of `g` each or `2g` overall. Therefore, the total acceleration downwards (possibly divided between the two balls) is g/3 and we can't end up with them both at rest. The only way we could end up with them both at rest, is if we we were allowed 6 pounds of weight instead of 5 pounds (ie. same as carrying)
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http://math.stackexchange.com/questions/139929/how-to-apply-the-poincare-formula-to-a-regular-n-gon | # How to apply the Poincaré formula to a regular n-gon?
I've been trying to solve the following home task:
Choose $n$ points ($n\ge 2$) on the circle's circumference and connect them all with each other using chords. In result, the circle is divided into $R$ number of regions. Let $P$ be the number of intersection points and $S$ the number of (chord) segments.
Prove that $P-S+R=1$.
A hint is provided that it's not advised nor needed to try and find the exact counts for $P$, $S$ or $R$.
My first insight is that we can always create a regular polygon with $n$ sides if we put the points on circle equally spaced. So for example a pentagon has $P=5$, $S=20$ and $R=16$.
But here I'm stuck badly. Maybe we could apply Euler's polyhedron formula but I don't see how since it's for polyhedrons. The hint makes me think about an inductive approach. But if we start with $n$ points and add one more (that is, go from $n$-gon to ($n+1$)-gon), I don't see how we can make use of the induction assumption to proof the $(n+1)$ case.
Any insight is welcome!
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You may use Euler's formula: Imagine that your drawing is placed on the northern half of a sphere and the southern half is an "empty" $n$-gon. – Christian Blatter May 2 '12 at 15:40
## 1 Answer
Here is an idea of a proof, similar to a proof for the Euler's formula:
Pick to adjacent points in the circumference, remove the chord corresponding to these points. Think about how R, S and P change, it is "obvious" that S decreases in 1, R decreases in 1 and P does not change.
This is the base for an inductive proof, you should work out what is the base case for the induction and you should write the "obvious" part above carefully.
EDIT: This does not work as pointed by randomguy in the comment below, but: You can cut the chords at every intersection point, this adds one point and one segment, so the invariant does not change. At the end you have a planar graph and you can apply the Euler's formula. But the comment of Christian Blatter gives a better idea.
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Thank you. I do see the similarity with the proof of Euler's formula (in regards how the change in counts is observed), but fail to see how the results of removal of outer chords can be helpful to understand what goes on with the removal of points, diagonals or "inner" segments (the changes there seem to be much more complicated and unpredictable). – randomguy May 2 '12 at 18:11
To make it more clear what I meant to say: AFAIK Euler's formula is for planar graphs. Removal of the non-intersecting chords is trivial as you point out, but I fail to see how we can systematically remove an intersecting chord, since that has a variable effect on the $R$, $S$ and $P$. – randomguy May 2 '12 at 23:25
Ok, you have a point. I don't know how to fix the idea without copying the proof of the Euler's formula. – Quimey May 2 '12 at 23:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429633617401123, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/83149?sort=votes | ## When is a reflective subcategory of a topos a topos?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Suppose $\mathcal{E}$ is a topos and $\mathcal{F}\subseteq \mathcal{E}$ is a reflective subcategory with reflector $L$, say. Under what conditions is $\mathcal{F}$ a topos?
A well-known sufficient condition for this is that $L$ be left exact. But this is certainly not necessary. For instance, let $f\colon C\to D$ be a functor such that $f^*\colon Set^D \to Set^C$ is fully faithful. A sufficient condition for this is given in C3.3.8 of Sketches of an Elephant — for every $d\in D$ the category of $c\in C$ with $d$ exhibited as a retract of $f(c)$ must be connected, and every morphism of $C$ must be a retract of the $f$-image of some morphism of $C$. These conditions do not imply that $\mathrm{Lan}_f$ is left exact, but nevertheless they allow us to identify $Set^D$ with a reflective subcategory of $Set^C$, and of course both are toposes.
Is there any general sufficient condition for a reflective subcategory of a topos to be a topos which includes this case?
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from "Topos Theory" (p.Johnstone), Exercise1 p. 43, a general condiction is that the (reflexive) subcategory has power objects, and is ereditary for subobjects of (each) power object. – Buschi Sergio Dec 11 2011 at 9:57
@Buschi Sergio - sure, but that condition is almost identical to saying "the subcategory is a topos". I'd like some condition that is easier to verify! – Mike Shulman Dec 11 2011 at 23:06
What is the role of the reflector? – David Carchedi Dec 12 2011 at 14:31
I find this Street article, may be you just know it "Notions of topos" R. Street journals.cambridge.org/… – Buschi Sergio Jun 3 at 20:30
@Buschi Sergio - Yes, I know it. I don't think it answers the question. – Mike Shulman Jun 5 at 3:16
## 1 Answer
Let me pay no attention to size issues:
Denote the adjunction by $R$ right adjoint to $L$. Equip $\mathcal{E}$ with the canonical topology $J$ (so generated by jointly surjective epimorphisms), so that we have $Sh_J(\mathcal{E}) \simeq \mathcal{E}.$ Denote the induced sheafication functor $a:Set^{\mathcal{E}^{op}} \to \mathcal{E}.$ Now consider the Yoneda embedding $y:\mathcal{F} \hookrightarrow Set^{\mathcal{F}^{op}}.$ Since $\mathcal{F}$ is reflective in $\mathcal{E}$, it is cocomplete, so we can left-Kan extend the identify functor of $\mathcal{F}$ along Yoneda, to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which, by construction is left-adjoint to the Yoneda embedding.. EDIT: Since $\mathcal{E}$ is a topos, and therefore total, so is the reflective subcategory $\mathcal{F}.$ So we get to get a functor $a_\mathcal{F}:Set^{\mathcal{F}^{op}} \to \mathcal{F}$ which is left-adjoint to this Yoneda embedding. However, it is also canonically equivalent to $L \circ a \circ R_{!}$ since this composite is colimit preserving and along representables is the identity. So, it follows that $\mathcal{F}$ is a (Grothendieck) topos if and only if the composite $L \circ a \circ R_{!}$ is left-exact.
Note: By one of the comments I made below, $a \circ R_!$ is left-exact, so $\mathcal{F}$ is a topos if and only if $L$ preserves those finite limits of the form $\rho:\Delta_{aR_!(C)} \Rightarrow a \circ R_! \circ G,$ with $G:D \to Set^{\mathcal{F}^{op}}$ a finite diagram.
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Interesting approach! I think your ignoring of size issues is in danger of being wrong when you construct $a_{\mathcal{F}}$ by "cocompleteness" of $\mathcal{F}$, since $\mathcal{F}$ will almost never have colimits of the same size as itself. However, I think you're in luck: any reflective subcategory of a total category (nlab.mathforge.org/nlab/show/total+category) is total, and that suffices to show the existence of $a_{\mathcal{F}}$. Why is $L \circ a \circ R_!$ the identity on representables? – Mike Shulman Dec 12 2011 at 18:10
Ah, good point. I guess I should be careful about sizes. Anyhow, $R_!$ is the left-Kan extension along Yoneda of the functor $\mathcal{F} \to Set^{\mathcal{E}^{op}},$ which sends $X \in \mathcal{F}$ to $Hom_{\mathcal{E}}(blank,X),$ which is representable. And, $a$ is the left-Kan extension along Yoneda of the identity functor, so it follows that for $X \in \mathcal{F},$ $a\circ R_!{X}=X$ which is secretly in $\mathcal{F},$ so the reflector does nothing, yielding $L \circ a \circ R_!(X)=X.$ – David Carchedi Dec 12 2011 at 18:27
Just an additional cute remark: Notice that $\mathcal{F}$ has finite limits. Now, $R_!$ is the left-Kan extension along Yoneda of $y \circ R$, which is left-exact, hence flat, because of the finite limits in $\mathcal{F}$, hence $R_!$ is left-exact. So is $a$. In particular, this tells us something we already knew, that if $L$ is left-exact, then $\mathcal{F}$ is a topos. – David Carchedi Dec 12 2011 at 19:12
That's actually quite cute! Note that it suffices to restrict even further to ask only that $L$ preserve pullbacks in the image of $a \circ R_!$, since any reflector preserves the terminal object. I wonder whether there is a way to refine this into something more manageable that doesn't refer to arbitrary pullbacks of presheaves on the large category $\mathcal{F}$. – Mike Shulman Dec 13 2011 at 6:38
In particular, you've shown that there is a theorem of the sort I was hoping for: for any reflective subcategory $\mathcal{F}$ of a topos $\mathcal{E}$, there is a class of pullbacks in $\mathcal{E}$ such that $\mathcal{F}$ is a topos if and only if its reflector preserves those pullbacks. Even if it would be nice to have a more explicit description of those pullbacks. – Mike Shulman Dec 13 2011 at 6:41
show 1 more comment | {"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": 68, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9429276585578918, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/126610/problem-on-linear-diophantine-equation-over-3-variables | # Problem on Linear Diophantine Equation over 3 variables
How to solve $ax+by+cz=d$ over integers where $a,b,c,d$ are integers?
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what is given in your equation? – Victor Mar 31 '12 at 17:14
@Victor: $a,b,c,d$, obviously – t.b. Mar 31 '12 at 17:16
Some details depend on relationships between greatest common divisors. For example, if $\gcd(a,b,c)$ does not divide $d$, there are no solutions. But generally there is a two-parameter infinite family of solutions. Individual solutions can be found using the Extended Euclidean algorithm. Situation is simplest if $\gcd(a,b)=\gcd(b,c)=\gcd(c,a)=1$. Then from two suitably chosen solutions, can find a formula for all. Is your problem really about specific numbers, or is it general? – André Nicolas Mar 31 '12 at 17:16
So the answer is in terms of a,b,c? – Victor Mar 31 '12 at 17:17
## 1 Answer
You can solve it using the same "row reduction" techniques used in the bivariate form of the extended Euclidean algorithm, viz. start with the given linear combinations
$$\rm a \:=\: 1\cdot a + 0\cdot b + 0\cdot c$$
$$\rm b\: =\: 0\cdot a + 1\cdot b + 0\cdot c$$
$$\rm c\: =\: 0\cdot a + 0\cdot b + 1\cdot c$$
then used the division algorithm to find smaller linear combinations of $\rm\:a,b,c\:$ till you reach $\rm\: g = gcd(a,b,c)\:$ expressed as a linear combination of $\rm\:a,b,c.$
The given equation has a solution iff $\rm\: g\ |\ d.\:$ If so, then a solution arises by scaling the the representation of $\rm\:g\:$ by $\rm\:d/g\:$ to yield a representation of $\rm\:d\:$ as a linear combination of $\rm\:a,b,c.$
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Thank you very much for your help. – user12290 Apr 1 '12 at 12:34 | {"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": 14, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8840529918670654, "perplexity_flag": "head"} |
http://regularize.wordpress.com/2012/04/04/the-augemented-lagrangian-with-variational-inequalities-and-on-necessary-conditions-for-variational-regularization/ | # regularize
Trying to keep track of what I stumble upon
April 4, 2012
## The Augmented Lagrangian with variational inequalities and on necessary conditions for variational regularization
Posted by Dirk under Math, Regularization | Tags: ill-posed problems, regularization |
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Today I’d like to blog about two papers which appeared on the arxiv.
1. Regularization with the Augmented Lagrangian Method – Convergence Rates from Variational Inequalities
The first one is the paper “Regularization of Linear Ill-posed Problems by the Augmented Lagrangian Method and Variational Inequalities” by Klaus Frick and Markus Grasmair.
Well, the title basically describes the content quite accurate. However, recall that the Augmented Lagrangian Method (ALM) is a method to calculate solutions to certain convex optimization problems. For a convex, proper and lower-semicontinuous function ${J}$ on a Banach space ${X}$, a linear and bounded operator ${K:X\rightarrow H}$ from ${X}$ into a Hilbert space ${H}$ and an element ${g\in H}$ consider the problem
$\displaystyle \inf_{u} J(u)\quad\text{s.t.}\quad Ku=g. \ \ \ \ \ (1)$
The ALM goes as follows: Start with an initial dual variable ${p_0}$, choose step-sizes ${\tau_k>0}$ and iterate
$\displaystyle u_k \in \text{argmin}\Big(\frac{\tau_k}{2}\|Ku-g\|^2 + J(u) + \langle p_{k-1},Ku-g\rangle\Big)$
$\displaystyle p_k = p_{k-1}+\tau_k(g-Ku_k).$
(These days one should note that this iteration is also known under the name Bregman iteration…). Indeed, it is known that the ALM converges to a solution of (1) if there exists one. Klaus and Markus consider the ill-posed case, i.e. the range of ${K}$ is not closed and ${g}$ is replaced by some ${g^\delta}$ which fulfills ${\|g-g^\delta\|\leq\delta}$ (and hence, ${g^\delta}$ is generally not in the range of ${K}$). Then, the ALM does not converge but diverges. However, one observes “semi-convergence” in practice, i.e. the iterates approach an approximate “solution to ${Ku=g^\delta}$” (or even a true solution to ${Ku=g}$) first but then start to diverge from some point on. Then it is natural to ask, if the ALM with ${g}$ replaced by ${g^\delta}$ can be used for regularization, i.e. can one choose a stopping index ${k^*}$ (depending on ${\delta}$ and ${g^\delta}$) such that the iterates ${u_{k^*}^\delta}$ approach the solution of (1) if ${\delta}$ vanishes? The question has been answered in the affirmative in previous work by Klaus (here and here) and also estimates on the error and convergence rates have been derived under an additional assumption on the solution of (1). This assumption used to be what is called “source condition” and says that there should exist some ${p^\dag\in H}$ such that for a solution ${u^\dagger}$ of (1) it holds that
$\displaystyle K^* p^\dagger \in\partial J(u^\dagger).$
Under this assumption it has been shown that the Bregman distance ${D(u_{k^*}^\delta,u^\dag)}$ goes to zero linearly in ${\delta}$ under appropriate stopping rules. What Klaus and Markus investigate in this paper are different conditions which ensure slower convergence rates than linear. These conditions come in the form of “variational inequalities” which gained some popularity lately. As usual, these variational inequalities look some kind of messy at first sight. Klaus and Markus use
$\displaystyle D(u,u^\dag)\leq J(u) - J(u^\dag) + \Phi(\|Ku-g\|^2)$
for some positive functional ${D}$ with ${D(u,u)=0}$ and some non-negative, strictly increasing and concave function ${\Phi}$. Under this assumption (and special ${D}$) they derive convergence rates which again look quite complicated but can be reduced to simpler and more transparent cases which resemble the situation one knows for other regularization methods (like ordinary Tikhonov regularization).
In the last section Klaus and Markus also treat sparse regularization (i.e. with ${J(u) = \|u\|_1}$) and derive that a weak condition (like ${(K^*K)^\nu p^\dag\in\partial J(u^\dag)}$ for some ${0<\nu<1/2}$ already imply the stronger one (1) (with a different ${p^\dag}$). Hence, interestingly, it seems that for sparse regularization one either gets a linear rate or nothing (in this framework).
2. On necessary conditions for variational regularization
The second paper is “Necessary conditions for variational regularization schemes” by Nadja Worliczek and myself. I have discussed some parts of this paper alread on this blog here and here. In this paper we tried to formalize the notion of “a variational method” for regularization with the goal to obtain necessary conditions for a variational scheme to be regularizing. As expected, this goal is quite ambitions and we can not claim that we came up with ultimate necessary condition which describe what kind of variational methods are not possible. However, we could first relate the three kinds of variational methods (which I called Tikhonov, Morozov and Ivanov regularization here) and moreover investigated the conditions on the data space a little closer. In recent years it turned out that one should not always use a term like ${\|Ku-g^\delta\|^2}$ to measure the noise or to penalize the deviation from ${Ku}$ to ${g^\delta}$. For several noise models (like Poisson noise or multiplicative noise) other functionals are better suited. However, these functionals raise several issues: They are often not defined on a linear space but on a convex set, sometimes with the nasty property that their interior is empty. They often do not have convenient algebraic properties (e.g. scaling invariance, triangle inequalities or the like). Finally they are not necessarily (lower semi-)continuous with respect to the usual topologies. Hence, we approached the data space from quite abstract way: The data space ${(Y,\tau_Y)}$ is topological space which comes with an additional sequential convergence structure ${\mathcal{S}}$ (see e.g. here) and on (a subset of) which there is a discrepancy functional ${\rho:Y\times Y\rightarrow [0,\infty]}$. Then we analyzed the interplay of these three things ${\tau_Y}$, ${\mathcal{S}}$ and ${\rho}$. If you wonder why we use the additional sequential convergence structure, remember that in the (by now classical) setting for Tikhonov regularization in Banach spaces with a functional like
$\displaystyle \|Ku-g^\delta\|_Y^q + \alpha\|u\|_X^p$
with some Banach space norms ${\|\cdot\|_Y}$ and ${\|\cdot\|_X}$ there are also two kinds of convergence on ${Y}$: The weak convergence (which is replaced by ${\tau_Y}$ in our setting) which is, e.g., used to describe convenient (lower semi-)continuity properties of ${K}$ and the norm ${\|\cdot\|_Y}$ and the norm convergence which is used to describe that ${g^\delta\rightarrow g^\dag}$ for ${\delta\rightarrow 0}$. And since we do not have a normed space ${Y}$ in our setting and one does not use any topological properties of the norm convergence in all the proofs of regularizing properties, Nadja suggested to use a sequential convergence structure instead.
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http://math.stackexchange.com/questions/46092/computing-singular-value-question | # Computing Singular Value question
Good afternoon.
I'm studying for my finals of this year, currently studying for the exam "Numerical Linear Algebra". I'm trying to solve some of the questions the teacher asked the past years (for preparation), but I can't seem to figure out a few, namely this one. We've got this setup:
• $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},{-}\dfrac{1}{\sqrt{2}},0\Bigr]$
$Ax=[2,0,0]$
• $x=\Bigl[{-}\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}},0\Bigr]$
$Ax=[0,-3,0]$
• $x=[0,0,1]$
$Ax=[0,0,6]$
The questions were as follows: 1. Determine the SVD of A (WITHOUT COMPUTING THE ACTUAL MATRIX) 2. Determine the condition number of A 3. Determine the rank and the determinant
Questions 2 and 3 are quite logical (condition number = largest SVD divided by the smallest, rank = number of non zero SV's and the determinant is the product of the SV's, if I'm not mistaken), but I can't seem to find a logical answer or explanation to question 1. So I'm hoping anyone here has a pointer in the right direction? I'm not asking for the asnwer (since I have to be able to replicate it on the exam, most likely), but some hints would be appreciated.
Thanks.
-
## 1 Answer
Since the three given vectors are orthonormal and their images are orthogonal, these are the right singular vectors. Thus, the singular value decomposition of $A$ is $$A \;=\; \begin{bmatrix}1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 6\end{bmatrix}\begin{bmatrix}-1/\sqrt{2} & -1/\sqrt{2} & 0 \\ -1/\sqrt{2} & 1/\sqrt{2} & 0 \\ 0 & 0 & 1\end{bmatrix}.$$ The rows of the last matrix are the given vectors, the diagonal entries of the middle matrix are the norms of the images of these vectors, and the columns of the first matrix are the normalizations of these images.
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That makes insanely good sense, thanks. Only question remaining is: As far as I understand, the SV's have to be ordered from highest to lowest on the diagonal, or doesn't that matter? – Timmy Nelen Jun 18 '11 at 17:49
There are different possible conventions. If you want to switch the order of the diagonal entries, all you have to do is switch the columns of the first matrix and the rows of the last matrix in the same way. – Jim Belk Jun 18 '11 at 18:05
Alright, I've got it :) If I swap columns 1 & 3 in the first matrix, I swap the 2 and the 6 in matrix 2 (Thus making the diagonal [2,3,6]), and swap the first and third rows in the third matrix I have an SVD where the elements are in decreasing order (as per the standard that got utilized in our lectures), and thus I have a correct solution for it, thank you :) – Timmy Nelen Jun 18 '11 at 18:13 | {"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": 7, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9410584568977356, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/39431/the-residue-class-functor-from-a-frobenius-category-to-its-stable-category-induce | ## The residue class functor from a Frobenius category to its stable category induces a functor on cube-shaped diagrams - is it dense?
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Let $\mathcal{E}$ be a Frobenius category, i.e. an exact category with sufficiently many bijective objects. (Such as e.g. the category of complexes over an additive category.)
Let $\underline{\mathcal{E}}$ be its stable category, defined as the factor category of $\mathcal{E}$ modulo its full subcategory of bijective objects. We have the residue class functor $R:\mathcal{E}\to\underline{\mathcal{E}}$. (For complexes, this is the canonical functor from the category of complexes to its homotopy category.)
Consider the category $D := \Delta_1\times\Delta_1\times\Delta_1$, i.e. a cube. Denote by $\mathcal{E}(D)$ the category of functors from $D$ to $\mathcal{E}$, i.e. the category of $\mathcal{E}$-valued diagrams of shape $D$. Etc.
Consider the functor $R(D):\mathcal{E}(D)\to\underline{\mathcal{E}}(D)$, obtained by pointwise application of $R$. Is this functor $R(D)$ dense, i.e. surjective on isoclasses? In other words, given a stably commutative cube-shaped diagram, does there exist a strictly commutative cube-shaped diagram isomorphic to it in $\underline{\mathcal{E}}(D)$?
My guess would be: no. But I was unable to construct a counterexample. I was searching in $\mathcal{E} = {\bf Z}/p^a\text{-mod}$.
The paper [1] deals with the question in greater generality, but it doesn't seem to give an explicit answer to my question. (This question can be generalised: one does not have to stick to the Frobenius case, and one can ask for obstructions depending on $D$ against the density of $R(D)$.)
[1] Dwyer, W. G.; Kan, D. M.; Smith, J. H., Homotopy commutative diagrams and their realizations, J. Pure Appl. Alg. 57, p. 5-24, 1989.
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http://www.physicsforums.com/showthread.php?p=4257779 | Physics Forums
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## How can evaporation occur?
to confirm my understanding the overall temperature will be the same even when some particles have more KE than others.
So long as their kinetic energies average to get the specified temperature, yes.
However, usually we would say that the temperature of the liquid would drop after evaporation. Would the explanation be something like this: after they transfer some KE to the surface molecules they have a smaller KE, hence once those surface molecules escape to form a gas the average KE of the liquid decreases and hence the temperature drops?
Exactly. The molecules left behind have a lesser kinetic energy than those that vaporized.
Quote by AbsoluteZer0 So long as their kinetic energies average to get the specified temperature, yes. Exactly.
Hi thanks for the help
What about melting could it be possible for those lucky bumps to occur as well allowing it to turn into a liquid?
Quote by sgstudent Hi thanks for the help What about melting could it be possible for those lucky bumps to occur as well allowing it to turn into a liquid?
A solid can evaporate as well, at temperatures well below the melting point.
See "sublimation".
Quote by nasu A solid can evaporate as well, at temperatures well below the melting point. See "sublimation".
But what about turning into a liquid before its melting point? Like how the liquid turns into a gas before it reaches its boiling point.
Thanks for the help :)
I'm no expert in this subject, but wanted to put my 2 cents into it. When a solid or liquid turns into a gas, the molecules escape (along with their energy) and are lost forever. I'm not sure about them turning into liquid (can one molecule even ever be considered a liquid?) but even if it does happen, it will, by definition, not "evaporate" away. It will instead remain on the surface, where it will turn back into solid almsot immediately.
Quote by sgstudent But what about turning into a liquid before its melting point? Like how the liquid turns into a gas before it reaches its boiling point.
Consider the surface of water in a covered pot simmering on the stove. At boiiling temperature the rate of evaporation of water into the saturated vapor is equal to the rate of condensation of the saturated vapor back into the water. There is an equilibrium.
In the case of your clothing drying on the line in sub-zero temperatures the partial pressure of water vapor in the air is very low. So low that the boiling temperature of water at that pressure would be even further below zero. There is no equililibrium. The clothes dry out.
The point here is that the "boiling point" of a liquid is a function of pressure. In some sense, a liquid does not turn into a gas before it reaches its boiling point.
In the case of the transition between water and ice there is some pressure dependency. It is possible to melt ice by applying pressure rather than by increasing temperature. It is also possible to melt ice by applying salt. Also, some solids soften as their melting temperature is approached.
Quote by Lsos I'm no expert in this subject, but wanted to put my 2 cents into it. When a solid or liquid turns into a gas, the molecules escape (along with their energy) and are lost forever. I'm not sure about them turning into liquid (can one molecule even ever be considered a liquid?) but even if it does happen, it will, by definition, not "evaporate" away. It will instead remain on the surface, where it will turn back into solid almsot immediately.
Oh would the reason it turns into a solid immediately be that since it is at a liquid state where it does not have enough energy to escape like a gas. So it's own kinetic energy would be transferred to the other particles causing it to solidify again?
Quote by AbsoluteZer0 Yes. $E_k = \frac{1}{2} mv^2$ What this equation states is that Kinetic energy is equivalent to half of the mass times the square of the velocity. When molecules collide with each other, they can either increase in velocity or decrease, changing their kinetic energy. Molecules are always in motion and are always colliding with each other. There will always be a difference in kinetic energy between most molecules in a substance. One molecule in a substance may have the same temperature as another, but this commonality would be very short lived.
Hi AbsoluteZer0, I was thinking about the evaporation and when those few molecules gain enough kinetic energy and escape as a gas, what would their temperatures be? When they were still in a liquid state despite having more KE, the temperature still remained the same. Once they leave the liquid to becomes a gas, how will their temperatures considered to be?
Thanks for the help :)
Quote by sgstudent Oh would the reason it turns into a solid immediately be that since it is at a liquid state where it does not have enough energy to escape like a gas. So it's own kinetic energy would be transferred to the other particles causing it to solidify again?
That's what I'm thinking. I didn't read this anywhere though, just arriving to conclusons...
When they were still in a liquid state despite having more KE, the temperature still remained the same
Are you talking about the temperature of the whole system or the temperature of the individual molecules?
Quote by AbsoluteZer0 Are you talking about the temperature of the whole system or the temperature of the individual molecules?
I guess it would be those individual atoms? Since they have a greater kinetic energy so will they be considered to be at 100 degrees?
Quote by sgstudent I guess it would be those individual atoms? Since they have a greater kinetic energy so will they be considered to be at 100 degrees?
They wouldn't necessarily be at 100°C. Evaporation can take place at temperatures lower than the boiling point.
http://en.wikipedia.org/wiki/Tempera...to_temperature
Classical thermodynamics concerns mostly the macroscopic (systems as a whole.)
As a consequence, it's inefficient to discuss molecules in terms of their thermal temperature. It's easier to discuss their statistical temperature. Statistical Thermodynamics deals with large populations of particles.
http://en.wikipedia.org/wiki/Statistical_mechanics
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http://mathoverflow.net/questions/99590/the-properties-of-functions-from-w-21 | ## The properties of functions from $W_{2}^{1}$ [closed]
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Is it true that the product of the two functions from $W_{2}^{1}(0,1)$ belongs to $W_{2}^{1}(0,1)$?
-
2
yes, because they are in $L^\infty$ with derivative in $L^2$, and $(uv)'=uv'+u'v$. – Pietro Majer Jun 14 at 10:45 | {"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": 6, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9107285141944885, "perplexity_flag": "middle"} |
http://euler.slu.edu/escher/index.php/Angles_of_Polygons_and_Regular_Tessellations_Exploration | # Angles of Polygons and Regular Tessellations Exploration
### From EscherMath
Objective: Calculate the interior angles of polygons and classify the regular tessellations of the plane.
## Exploration
### Interior Angles of Polygons
1. Check that the sum of the angles in a triangle is 180° as follows: Cut out a triangle. Tear off the corners and put them together so that their vertices are touching. What do you see?
2. Draw some quadrilaterals. For each one, show how to cut it into two triangles. Since the angle sum of each triangle is 180°, explain how you know the angle sum of each quadrilateral. What is the angle sum of a quadrilateral?
3. Any polygon can be cut into triangles by connecting its vertices with additional lines. How many triangles make up a 4-gon? How many triangles make up a 5-gon? How many triangles make up a 6-gon? How many triangles make up an n-gon?
4.
Cutting polygons into triangles. A bad way to cut into triangles.
5. Using the information from question 3 argue that:
The sum of the interior angles of an n-gon is $(n-2)\times 180^\circ$
6. Why does the "bad way to cut into triangles" fail to find the sum of the interior angles?
### Regular Polygons
A regular polygon is a polygon with all sides the same length and all angles having the same angle measure.
1. Explain the following formula:
Each angle of a regular n-gon is $\frac{(n-2)180^\circ}{n}$.
Would this formula work for just any n-gon? Why or why not?
2. Complete the following table:
| | | | | | | | | | | | | Number of Sides
| Corner angle = $\frac{(n-2)180^\circ}{n}$
|
|-----|-----|----|----|----|----|----|----|----|----|----|----|---------------------|-----------------------------------------------------------------------------------------------|
| 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 15 | 20 | 50 | 100 |
| 60° | 90° | | | | | | | | | | | | |
3. If regular polygons are going to fit around a vertex, then their angle measures have to divide evenly into 360°. Explain. Which of the angle measures in the table divide evenly into 360˚?
4. The table doesn't list every possible number of sides. How do you know that there are no other regular polygons with angles that divide evenly into 360˚, besides the ones mentioned on the list?
5. Which regular n-gons are the only ones that can tessellate the plane using just one type of tile?
6. Three equilateral triangles and two squares can fit together, since 60+60+60+90+90 = 360°. What other combinations of corner angles in the table can be combined to make 360°?
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http://mathoverflow.net/questions/9936/kummerian-fields/9937 | ## “Kummerian” fields?
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This is sort of a random, spur of the moment question, but here goes:
We define [with apologies to Conan the Barbarian] a field K to be $\textbf{Kummerian}$ if there exists an index set I, and functions $x: I \rightarrow K, n: I \rightarrow \mathbb{Z}^+$ such that the algebraic closure of K is equal to $K[(x(i)^{\frac{1}{n(i)}})_{i \in I}]$. More plainly, the algebraic closure is obtained by adjoining roots of elements of the ground field, not iteratively, but all at once.
Questions:
QI) Is there a classification of Kummerian fields?
QII) What about a classification of "Kummerian (topological) groups", i.e., the absolute Galois groups of Kummerian fields?
Here are some easy observations:
1) An algebraically closed or real-closed field is Kummerian. In particular, the groups of order 1 and 2 are Galois groups of Kummerian fields. By Artin-Schreier, these are the only finite absolute Galois groups, Kummerian or otherwise.
2) A finite field is Kummerian: the algebraic closure is obtained by adjoining roots of unity. Thus $\hat{\mathbb{Z}}$ is a Kummerian group.
3) An algebraic extension of a Kummerian field is Kummerian. Thus the class of Kummerian groups is closed under passage to closed subgroup. Combining with 2), this shows that any torsionfree procyclic group is Kummerian. On the other hand, the class of Kummerian groups is certainly not closed under passage to the quotient, since $\mathbb{Z}/3\mathbb{Z}$ is not a Kummerian group.
4) A Kummerian group is metabelian: i.e., is an extension of one abelian group by another. This follows from Kummer theory, using the tower $\overline{K} \supset K^{\operatorname{cyc}} \supset K$, where $K^{\operatorname{cyc}}$ is the extension obtained by adjoining all roots of unity.
In particular no local or global field (except $\mathbb{R}$ and $\mathbb{C}$) is Kummerian.
5) The field $\mathbb{R}((t))$ is Kummerian. Its absolute Galois group is the profinite completion of the infinite dihedral group $\langle x,y \ | \ x^2 = 1, \ xyx^{-1} = y^{-1} \rangle$. In particular a Kummerian group need not be abelian.
Can anyone give a more interesting example?
ADDENDUM: In particular, it would be interesting to see a Kummerian group that does not have a finite index abelian subgroup or know that no such exists.
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I was under the impression that K((t)) is Kummerian for any algebraically closed field K of characteristic zero. Am I misinformed? – Qiaochu Yuan Dec 28 2009 at 5:20
I feel like the term "kummerian" may not be ideal because kummer extensions "are a thing". – Harry Gindi Dec 28 2009 at 5:22
@QY: You are absolutely correct. In my haste to get to the somewhat more interesting case in which K is real-closed, I guess I forgot to mention this. The Galois group is again Z-hat. – Pete L. Clark Dec 28 2009 at 5:25
@Harry: The term "absolutely Kummerian" would probably be more appropriate. But let's talk about the mathematics, not the nomenclature, shall we? – Pete L. Clark Dec 28 2009 at 5:38
Let me just say that I find this question interesting, in connection with my more general interest in the maximal radical extension of any given field (the extension obtained by adjoining all roots of all elements of the basic field, not iteratively but all at once). – Leonid Positselski Dec 28 2009 at 12:47
## 2 Answers
Recall that a field $K$ is pseudo finite if (1) $K$ is perfect (2) the absolute Galois group of $K$ is $\hat{\mathbb{Z}}$, and (3) every absolutely irreducible non-void variety defined over $K$ has a $K$-rational point. Ax proved that large finite fields have the same elementary theory as a pseudo finite field.
I guess that the statement: the unique extension of degree $n$ is generated by a root of an element of the field is elementary, hence from since finite fields are Kummerian, so are pseudo finite fields are Kummerian.
If I got it right so far, this gives an abundance of examples using Jarden's theorem (non of them explicit): Choose a Galois automorphism $\sigma$ in the absolute Galois group of $\mathbb{Q}$. Then with probability one, the fixed field of $\sigma$ in the algebraic closure is pseudo finite. (Recall that Galois groups are profinite, hence compact, hence equipped with Haar probability measure.)
Jarden's theorem holds in more generality, namely when we replace $\mathbb{Q}$ with any countable Hilbertian field (for some examples see here).
Anyway, I guess it doesn't answer any of your questions, in particular, it doesn't give any new absolute Galois group. But I thought some more examples might be interesting...
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I had actually thought about pseudo-finite fields myself, and convinced myself that at least some of them are Kummerian. Here was my train of thought: (i) by Kummer theory, a field of characteristic 0 containing all roots of unity and with abelian Galois group is Kummerian. (ii) One can arrange for a pseudofinite field to contain all roots of unity, by making sure that for any given N, all but finitely many finite fields in an ultraproduct have N roots of unity. (Similarly, one can construct pseudofinite fields which do not contain all roots of unity.) ... – Pete L. Clark Dec 28 2009 at 21:06
...Anyway, your approach is interesting: I agree that the statement that the unique field extension of degree n is radical is first order. This raises the following question: is the property of being Kummerian itself elementary? – Pete L. Clark Dec 28 2009 at 21:09
I'm always confused with that elementary stuff :P There are infinitely many statements, one for each $n$, right? So are you aiming that somehow we can see it via merely finitely many statements. – Lior Bary-Soroker Dec 28 2009 at 21:32
No, that's not what I mean (and yes, the terminology here is confusing). I just mean that the Kummerian fields form an elementary class, meaning that if K is Kummerian and K' is a field such that Th(K) = Th(K') -- i.e., a statement in the language of fields is true in K iff it is truein K' -- then K' is Kummerian. I think this is probably true, and I think it is probably not true that Kummerianity (yikes!) is finitely axiomatizable. – Pete L. Clark Dec 28 2009 at 22:54
It is true if the absolute Galois group of a Kummerian field is f.g. as in all the examples. However I'm not sure it's true: what about $\mathbb{C}((t_1))((t_2))\cdots$? (the limit field of the field you suggested) I just throw it to, I'm not sure is Kummerian, but one the first thought it is – Lior Bary-Soroker Dec 29 2009 at 6:49
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Here is one more easy construction: let $K_n = \mathbb{C}((t_1))\ldots((t_n))$, an $n$ times iterated Laurent series field over the complex numbers. Then the absolute Galois group of $K_n$ is $\hat{\mathbb{Z}}^n$, obtained by adjoining all roots of the $t_i$'s. Thus $\hat{\mathbb{Z}}^n$ is a Kummerian group. Combined with 3) above, I believe this shows that any topologically finitely generated torsionfree abelian profinite group is Kummerian.
ADDENDUM: I now believe that an infinite abelian profinite group is Kummerian iff it is torsionfree. I'll sleep on this and see if someone else can say more...
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http://mathoverflow.net/questions/45448?sort=votes | ## Aleph 0 as a large cardinal
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The first infinite cardinal, $\aleph_0$, has many large cardinal properties (or would have many large cardinal properties if not deliberately excluded). For example, if you do not impose uncountability as part of the definition, then $\aleph_0$ would be the first inaccessible cardinal, the first weakly compact cardinal, the first measureable cardinal, and the first strongly compact cardinal. This is not universally true ($\aleph_0$ is not a Mahlo cardinal), so I am wondering how widespread of a phenomenon is this. Which large cardinal properties are satisfied by $\aleph_0$, and which are not?
There is a philosophical position I have seen argued, that the set-theoretic universe should be uniform, in that if something happens at $\aleph_0$, then it should happen again. I have seen it specifically used to argue for the existence of an inaccessible cardinal, for example. The same argument can be made to work for weakly compact, measurable, and strongly compact cardinals. Are these the only large cardinal notions where it can be made to work? (Trivially, the same argument shows that there's a second inaccessible, a second measurable, etc., but when does the argument lead to more substantial jump?)
EDIT: Amit Kumar Gupta has given a terrific summary of what holds for individual large cardinals. Taking the philosophical argument seriously, this means that there's a kind of break in the large cardinal hierarchy. If you believe this argument for large cardinals, then it will lead you to believe in stuff like Ramsey cardinals, ineffable cardinals, etc. (since measurable cardinals have all those properties), but this argument seems to peter out after a countable number of strongly compact cardinals. This doesn't seem to be of interest in current set-theoretical research, but I still find it pretty interesting.
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Even measurability and strong compactness are out if defined in terms of elementary embeddings. The point is that definitions that are equivalent for uncountable cardinals may be inequivalent for `$\aleph_0$`. – Andreas Blass Nov 9 2010 at 16:17
But if I knew the answer, I wouldn't be asking the question, would I? Many of the large cardinal axioms have alternate characterizations -- honestly I thought that was part of the case for their mathematical interest. The language characterization of weakly compact and strongly compact cardinals applies to $\aleph_0$, as does the ultrafilter characterization of measurable and strongly compact. I don't know enough about large cardinals to know if the cardinals I mentioned are unusual that regard, or typical. – arsmath Nov 9 2010 at 16:45
So for example, supercompactness has a characterization in terms of normal measures. Being a Vopenka cardinal has a characterization in terms of families of binary relations. – arsmath Nov 9 2010 at 16:52
arsmath: The current direction of research regards the embeddings template as the basic format for large cardinal notions. Sometimes rather than the embeddings themselves we focus on the strong reflection principles they provide. There are equivalent formulations for very small large cardinals, but the search for such formulations at the higher infinite is not pursued. That being said, one could regard statements of the form the second order theory of arithmetic is $\Omega$-provable'' as large cardinal statements (roughly at the level of a proper class of Woodin cardinals). – Andres Caicedo Nov 9 2010 at 16:54
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It's slightly amusing that 0, 1, and 2 have some large cardinal properties that 3, 4, 5, ... lack. 2 is not a sum of fewer than 2 numbers, each of which is smaller than 2. Similarly for 0 and 1. Maybe the paradox dissolves when you realize that "large" means in effect large by comparison to smaller cardinals. – Michael Hardy Nov 9 2010 at 17:25
show 5 more comments
## 1 Answer
This probably isn't the kind of thing anyone just knows off hand, so anyone who's going to answer the question is just going to look at a list of large cardinal axioms and their definitions, and try to see which ones are satisfied by $\aleph _0$ and which aren't. You could've probably done this just as well as I could have, but I decided I'd do it just for the heck of it.
First of all, this doesn't cover all large cardinal axioms. Second, many large cardinal axioms have different formulations which end up being equivalent for uncountable cardinals, or perhaps inaccessible cardinals, but may end up inequivalent in the context of $\aleph _0$. So even for the large cardinals that I'll look at, I might not look at all possible formulations.
1. weakly inaccessible - yes, obviously
2. inaccessible - yes, obviously
3. Mahlo - no, since the only finite "inaccessibles" are 0, 1, and 2 as noted by Michael Hardy in the comments, and the only stationary subsets of $\omega$ are the cofinite ones
4. weakly compact:
• in the sense of the Weak Compactness Theorem - yes, by the Compactness Theorem
• in the sense of being inaccessible and having the tree property - yes, by Konig's Lemma
• in the sense of `$\Pi ^1 _1$`-indescribability - no, it's not even `$\Pi ^0 _2$`-indescribable as witnessed by the sentence $\forall x \exists y (x \in y)$
5. indescribable - no, since it's not even `$\Pi ^0 _2$`-indescribable
6. Jonsson - no, the algebra $(\omega, n \mapsto n \dot{-} 1)$ has no proper infinite subalgebra
7. Ramsey - no, the function $F : [\omega ]^{< \omega} \to \omega$ defined by $F(x) = 1$ if $|x| \in x$ and $0$ otherwise has no infinite homogeneous set
8. measurable:
• in the sense of ultrafilters - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition, i.e. closed under finite intersections
• in the sense of elementary embeddings - no, obviously
9. strong - no, obviously (taking the elementary embedding definition)
10. Woodin - ditto
11. strongly compact:
• in the sense of the Compactness Theorem - yes, by the Compactness Theorem
• in the sense of complete ultrafilters - yes, as in the case of measurables
• in the sense of fine measures - yes, by Zorn's Lemma, and because filters are $\omega$-complete by definition
12. supercompact:
• in the sense of normal measures - no, if $x \subset \lambda$ is finite and $X$ is the collection of all finite subsets of $\lambda$ which contain $x$, then the function $f : X \to \lambda$ defined by $f(y) = \max (y)$ is regressive, but for any $Y \subset X$, if $f$ is constant on $Y$ with value $\alpha$, then Y avoids the collection of finite subsets of $\lambda$ which contain `$\{ \alpha + 1\}$` and hence Y cannot belong to any normal measure on $P_{\omega }(\lambda)$
• in the sense of elementary embeddings - no, obviously
13. Vopenka - no, take models of the empty language of different (finite) sizes
14. huge - no, obviously (taking the elementary embedding definition)
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Thanks. I half-expected that there was a succinct characterization in terms of some general property somewhere in the literature. "Everything that happens for $\aleph_0$ happens again" has the flavor of a reflection principle, and there's a big literature on reflection principles, so I found it plausible that it had a tidy answer. – arsmath Nov 10 2010 at 8:13 | {"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": 31, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9495764970779419, "perplexity_flag": "head"} |
http://nrich.maths.org/1955/clue | ### Logosquares
Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares.
### Shape and Territory
If for any triangle ABC tan(A - B) + tan(B - C) + tan(C - A) = 0 what can you say about the triangle?
### So Big
One side of a triangle is divided into segments of length a and b by the inscribed circle, with radius r. Prove that the area is: abr(a+b)/ab-r^2
# Three by One
##### Stage: 5 Challenge Level:
Here is a hint to help on you on the way :
Let $\alpha + \beta = \gamma$ and $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$
where $\alpha = \tan^{-1}{1\over3}$, $\beta = \tan^{-1}{1\over2}$, $\gamma = \tan^{-1}1$.
The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here. | {"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": 5, "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}, "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}, "english_confidence": 0.8939582705497742, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/939/counting-primes?answertab=active | # Counting primes
Let $\pi(x)$ be the number of primes not greater than $x$.
Wikipedia article says that $\pi(10^{23}) = 1,925,320,391,606,803,968,923$.
The question is how to calculate $\pi(x)$ for large $x$ in a reasonable time? What algorithms do exist for that?
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3
What's wrong with the ones given on the page? – Noldorin Jul 28 '10 at 11:34
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There is only one algorithm there. And I don't think it's possible to calculate $pi(10^{23})$ using it. Also there are no bounds on it's complexity. – falagar Jul 28 '10 at 11:48
## 3 Answers
The most efficient prime counting algorithms currently known are all essentially optimizations of the method developed by Meissel in 1870, e.g. see the discussion here http://primes.utm.edu/howmany.shtml
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The Sieve of Atkin is one of the fastest algorithm used to calculate $pi(x)$. The Wikipedia page says that its complexity is O(N/ log log N).
(edit)
I found a distributed computation project which was able to calculate $pi(4\times 10^{22})$, maybe it could be useful.
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There is no way that $\pi(10^{23})$ could be calculated using the sieve of atkin (in any reasonable amount of time) – BlueRaja - Danny Pflughoeft Jul 28 '10 at 14:51
Right. This pages could be useful: ams.org/mcom/1996-65-213/S0025-5718-96-00674-6/… ieeta.pt/~tos/primes.html – zar Jul 28 '10 at 15:56
This is one of the papers linked to on Bill's page; they were able to use it calculate $\pi(10^{18})$, but more advanced refinements were needed to calculate higher than that – BlueRaja - Danny Pflughoeft Jul 28 '10 at 16:22
You can use inclusion exclusion principle to get a boost over the Eratosthenes sieve
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I'm not sure what do you mean, but Eratosthenes sieve is way too slow for numbers like 10^23. Could you provide an asymptotic complexity of your approach? – falagar Jul 28 '10 at 11:23 | {"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": 10, "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}, "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}, "english_confidence": 0.9213634133338928, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/155796-integral-square-root-print.html | # integral (square root)
Printable View
• September 10th 2010, 07:23 PM
mybrohshi5
integral (square root)
I am having trouble solving this integral
$\int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$
I think i can reduce the equation under the square root to an expression that is squared which would get rid of the square root (then it would be easy to evaluate from there), but i cannot seem to figure it out (by completing the square or any other way)
Thanks for any help :)
• September 10th 2010, 07:41 PM
TheEmptySet
Quote:
Originally Posted by mybrohshi5
I am having trouble solving this integral
$\int ^2_1 \sqrt(4x^2+4+\frac{1}{x^2}) dx$
I think i can reduce the equation under the square root to an expression that is squared which would get rid of the square root (then it would be easy to evaluate from there), but i cannot seem to figure it out (by completing the square or any other way)
Thanks for any help :)
$4x^2+4+\frac{1}{x^2}=\left(2x+\frac{1}{x} \right)\left(2x+\frac{1}{x} \right)$
• September 10th 2010, 07:43 PM
Soroban
Hello, mybrohshi5!
Quote:
$\displaystyle \int ^2_1 \sqrt{4x^2+4+\frac{1}{x^2}}\;dx$
Look at what's under the square root . . .
. . $4x^2 + 4 + \dfrac{1}{x^2} \;=\;\dfrac{4x^4 + 4x^2 + 1}{x^2} \;=\;\dfrac{(2x^2+1)^2}{x^2}$
Or you can factor it like this: . $4x^2 + 4 + \dfrac{1}{x^2} \;=\;\left(2x + \dfrac{1}{x}\right)^2$
Got it?
• September 10th 2010, 07:49 PM
mybrohshi5
Oh boy i feel stupid that i missed that haha.
Thanks for both of the fast replies :D
All times are GMT -8. The time now is 03:27 AM. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9484699964523315, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/810/what-is-the-difference-between-a-stream-cipher-and-a-one-time-pad/814 | # What is the difference between a stream cipher and a one-time-pad?
A (synchronous) stream cipher is an algorithm which maps some fixed-length key to an arbitrary-length key-stream (i.e. a sequence of bits): $C : \{0,1\}^k \to \{0,1\}^{\infty}$.
This key-stream is then XOR-ed with the plain text stream, giving the ciphertext stream. For decrypting, the same key-stream (generated from the key at the receiver side) will be XOR-ed with the ciphertext stream, giving again the key stream.
A One-time pad is an algorithm which takes a key of large size (at least message size), and XORs its start with the plaintext to get the ciphertext. For decryption, we XOR the start of the key with the ciphertext to get back the plaintext.
These look quite similar – could one say that a stream-cipher is a (special way to create/use a) one-time pad, or that the one-time pad is a kind of stream cipher?
Are there any important differences between these two classes of algorithms?
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## 4 Answers
There is no universally accepted definition of the expression "stream cipher"; but the one I most often encounter is the following: a stream cipher is a symmetric encryption algorithm which accepts as inputs arbitrary sequences of bits (or bytes) such that:
• the length of the output is equal to the length of the input (no padding);
• for any $n$ (possibly any $n$ which is a multiple of 8 if we restrict ourself to bytes), the first $n$ output bits depend only on the key and the first $n$ input bits, regardless of the value of the subsequent input bits.
In that sense, the One-Time Pad is a stream cipher. A block cipher used in CTR mode or CFB mode is also a stream cipher. Note that the latter is not of the kind "XOR with a stream generated from the key independently of the input data". The Wikipedia page you link to talks about "synchronous stream ciphers" and "self-synchronzing stream ciphers".
However, the ultimate security of the One-Time Pad comes from the key size: it is unbreakable because it assumes that the key is as long as the message and was generated by an unpredictable mechanism. If you generate the pad with a more conventional stream cipher, working over a small fixed-size key, then it is no longer a One-Time Pad, just a "regular" stream cipher. The expression "One-Time Pad" refers to, exclusively, the mythical scheme which uses truly random long keys. So while One-Time Pad is a stream ciphers, stream ciphers are not One-Time Pads.
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Also note that a computationally unbounded adversary could break a stream cipher (having enough known plaintext) but not a one-time pad. – Henno Brandsma Sep 27 '11 at 19:32
Since the one-time-pad was actually used, it is not mythical. For example, the original Moscow–Washington hotline starting in 1967 used a one-time pad. – David Cary Oct 17 '11 at 12:11
@David: I have no detail on how the pads were generated. The "red phone" had the operational constraints of a one-time pad (namely requiring weekly distribution of large keys on tapes) but unless the pads were generated with a 100% physical RNG, it was not a "true" one-time pad. – Thomas Pornin Oct 17 '11 at 12:50
The one-time pad and Venona project articles say that the USSR and a few other organizations heavily used physical pads of random-looking letters called one-time pads. I think it's pretty likely that practically all of those pads were, in fact, generated with a 100% physical RNG, although I don't really have any evidence one way or another. I suspect that we may never know exactly how those pads were generated. – David Cary Jul 30 '12 at 17:25
One important difference between the one-time pad and a stream cipher is the proof of security of the one-time pad. Shannon proved that the one-time pad provides perfect secrecy. He also provided another proof that is interesting to this dicussion. His proof was that no cipher can provide perfect secrecy unless the key is at least a long as the message. Therefore, we know that no stream cipher can provide perfect secrecy unless it meets that requirement. But, just because it meets that requirement does not mean that it automatically provides perfect secrecy.
Now, as to whether or not they are the same, the answer is no. The one-time pad uses a fixed length key (where the length is at least a long as the message among other requirements) and the xor operation, period (i.e., there is no key expanion/stream generation).
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Mathematically speaking. The entropy of a stream cipher is upper-bounded by the key size.
The entropy of a one time pad, on the other hand is upper-bounded by the plaintext size.
For true one-time pads and good stream ciphers, this bound is tight.
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the stream cipher is similar to the one-time pad but exact difference is that
`one-time pad uses a genuine random number stream, whereas a stream cipher uses a pseudorandom number stream`
pseudorandom numbers calculated by a computer through a deterministic process, cannot, by definition, be random. Given knowledge of the algorithm used to create the numbers and its internal state, you can predict all the numbers returned by subsequent calls to the algorithm,
whereas with genuinely random numbers, knowledge of one number or an arbitrarily long sequence of numbers is of no use whatsoever in predicting the next number to be generated.
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http://mathoverflow.net/revisions/73019/list | ## Return to Question
2 added 1110 characters in body
Let $c(r)$ be a nice, continuous function with compact support. For example, $c(r) = \tfrac 1 5 (1-r)^{11} \big( 5 + 55r + 239 r^2 + 429 r^3 \big)$ for $r \in [0,1]$, and $c(r) = 0$ otherwise.
On a rectangular domain $D$ in $\mathbb R^2$, we define $c(x,y) = c(|x-y|)$, which we interpret as the kernel of an integral operator. Consider the integral equation $$\iint_D c(x,y) f(y) \ \mathrm{d}y = \lambda f(x).$$ By Mercer's theorem, we can find a sequence of non-negative eigenvalues $\lambda_k$ and eigenfunctions $f_k(x)$ such that:
• The sequence $f_k$ is an orthonormal basis for $L^2(D)$,
• The eigenfunctions $f_k$ corresponding to non-negative eigenvalues are continuous on $D$, and
• We have the representation $c(x,y) = \sum \lambda_k f_k(x) f_k(y)$.
Question: How can we numerically approximate the first $K$ eigenfunctions $f_1, \cdots, f_K$ and keep the total error within a desired error? (Polynomial or Fourier approximations would suffice)
Edit:A number of commentators have described the natural technique: discretize space, turn the covariance into a matrix, easily solve for the eigenvectors, then reconstruct the functions $f_k(t)$ by interpolation.
Our motivation is to generate a Gaussian random field $\xi$ using the Karhunen-Loève transformation: let $Z_1, \cdots, Z_K$ be independent Gaussian RVs, and define the random field $$\xi(t) = \sum_{k=1}^K Z_k f_k(t).$$ Since the covariance is so smooth, the eigenfunctions will be smooth, and the Gaussian field will be smooth. I'm sure that a competent numerical analyst could use the discretization technique to reconstruct the eigenfunctions up to arbitrary derivatives. Unfortunately, I am an incompetent numerical analyst, and I would surely introduce systematic errors along the way. Yet the problem is so common that I am still hoping somebody could point us to a simple reference which does this well.
To make the question more mathematical precise: the numerical approximations $f_k$ should be within an arbitrarily small $C^3$-error of the actual eigenfunctions.
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# Numerically finding a Mercer expansion for a given covariance kernel
Let $c(r)$ be a nice, continuous function with compact support. For example, $c(r) = \tfrac 1 5 (1-r)^{11} \big( 5 + 55r + 239 r^2 + 429 r^3 \big)$ for $r \in [0,1]$, and $c(r) = 0$ otherwise.
On a rectangular domain $D$ in $\mathbb R^2$, we define $c(x,y) = c(|x-y|)$, which we interpret as the kernel of an integral operator. Consider the integral equation $$\iint_D c(x,y) f(y) \ \mathrm{d}y = \lambda f(x).$$ By Mercer's theorem, we can find a sequence of non-negative eigenvalues $\lambda_k$ and eigenfunctions $f_k(x)$ such that:
• The sequence $f_k$ is an orthonormal basis for $L^2(D)$,
• The eigenfunctions $f_k$ corresponding to non-negative eigenvalues are continuous on $D$, and
• We have the representation $c(x,y) = \sum \lambda_k f_k(x) f_k(y)$.
Question: How can we numerically approximate the first $K$ eigenfunctions $f_1, \cdots, f_K$ and keep the total error within a desired error? (Polynomial or Fourier approximations would suffice) | {"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": 37, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8570418357849121, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/37356?sort=votes | ## Realizing groups as automorphism groups of graphs.
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
Frucht showed that every finite group is the automorphism group of a finite graph. The paper is here.
The argument basically is that a group is the automorphism group of its (colored) Cayley graph and that the colors of edge in the Cayley graph can be coded into an uncolored graph that has the same automorphism group.
This argument seems to carry over to the countably infinite case.
Does anybody know a reference for this?
In the uncountable, is it true that every group is the automorphism group of a graph? (Reference?) It seems like one has to code ordinals into rigid graphs in order to code the uncountably many colors of the Cayley graph.
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## 2 Answers
According to the wikipedia page, every group is indeed the automorphism group of some graph. This was proven independently in
de Groot, J. (1959), Groups represented by homeomorphism groups, Mathematische Annalen 138
and
Sabidussi, Gert (1960), Graphs with given infinite group, Monatshefte für Mathematik 64: 64–67.
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Thanks. I somehow missed that wikipedia page. I only found the one with the Frucht graph and related paper. – Stefan Geschke Sep 1 2010 at 8:37
From de Groot's paper you can see that this argument you described goes pretty far, for graphs, topological spaces etc. Coding arbitrary number of colors is done through some rigid spaces with no automorphisms. – Gjergji Zaimi Sep 1 2010 at 8:53
Yes, Stefan certainly had the right idea in mind. I'll just remark that Sabidussi's paper meets my requirements for recreational math reading; it's only three pages long. – Tony Huynh Sep 1 2010 at 9:10
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In the topological setting or if you want to relate the size of the graph to the size of the group, there are two relevant results:
(1) Any closed subgroup of $S_\infty$, i.e., of the group of all (not just finitary) permutations of $\mathbb N$, is topologically isomorphic to the automorphism group of a countable graph.
(2) The abstract group of increasing homeomorphisms of $\mathbb R$, ${\rm Homeo}_+(\mathbb R)$, has no non-trivial actions on a set of size $<2^{\aleph_0}$. So in particular, it cannot be represented as the automorphism group of a graph with less than continuum many vertices.
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Hi Christian! Nice to see you here. – Andres Caicedo Sep 1 2010 at 17:04
Hi Christian! Nice to see you! I fixed a small typo in your post. – François G. Dorais♦ Sep 1 2010 at 17:47
This is interesting, thank you. – Stefan Geschke Sep 2 2010 at 8:52 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9160827994346619, "perplexity_flag": "middle"} |
http://mathhelpforum.com/calculus/169645-differentiate-respect-x.html | # Thread:
1. ## Differentiate with respect to x
Y = $\displaystyle \sqrt{(x^3+5)}$
2. Use the chain rule: $\dfrac{d}{dx} f[g(x)] = f'[g(x)] \cdot g'(x)$
In your case you have $f(x) = \sqrt{g(x)}$ and $g(x) = x^3+5$
3. $\dfrac{d}{dx}\left(\sqrt{u(x)}\right)=\dfrac{1}{2\ sqrt{u(x)}}\cdot u'(x)$
Fernando Revilla
Edited: Sorry, I didn't see e^(i*pi)'s post.
4. You can also write this as $(x^3+5)^{1/2}$ to use the "power law" and the chain rule.
5. Is this the final answer?
$=\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}$
6. How do I expand on this?
7. Yes that's fine, I wouldn't bother expanding that though.
8. Originally Posted by HNCMATHS
Is this the final answer?
$=\dfrac{3 x^2}{2\sqrt{(x^3 + 5)}}$
Let's see!
$\displaystyle\ u=x^3+5\Rightarrow\frac{du}{dx}=3x^2$
From the Chain Rule
$\displaystyle\frac{dy}{dx}=\frac{dy}{du}\;\frac{du }{dx}=\frac{d}{du}u^{\frac{1}{2}}\;3x^2=\frac{1}{2 }u^{-\frac{1}{2}}\;3x^2=\frac{1}{2u^{\frac{1}{2}}}3x^2= \frac{3x^2}{2\sqrt{x^3+5}}$
So you see, those fast methods can be quite useful as an alternative to messing with fractional indices!
9. Originally Posted by HallsofIvy
You can also write this as $(x^3+5)^{1/2}$ to use the "power law" and the chain rule.
HI Halls of Ivy it is this method I would like you to expand on please, Thanks
10. Originally Posted by HNCMATHS
HI Halls of Ivy it is this method I would like you to expand on please, Thanks
Have you been taught the chain rule? Your class notes and textbook will have many examples for you to follow and relate to the question you posted. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9622585773468018, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/9335?sort=newest | ## Why is every symplectomorphism of the unit disk Hamiltonian isotopic to the identity?
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That is, for any symplectomorphism $\psi: D^2 \to D^2$, there should be a time-dependent Hamiltonian Ht on D2 such that the corresponding flow at time 1 is equal to $\psi$.
I found this in claim a paper, and I think it should be easy, but nothing comes to mind. I'd be happy with a reference to a page in McDuff-Salomon, but I couldn't find this there immediately.
Thanks!
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## 2 Answers
It is a theorem of Smale that the group of orientation-preserving diffeomorphisms of $D^2$, rel boundary, is contractible. If the diffeomorphisms can move the boundary, you can establish a homotopy equivalence between that and the circle. The diffeomorphisms do not have to preserve area. Then, a theorem of Moser establishes a deformation retract from diffeomorphisms to volume-preserving diffeomorphisms. Moser's result is easier to see if you have a closed manifold, but it extends to manifolds with boundary with the doubling trick. Together, this indirectly gives you a curve of symplectomorphisms connecting the identity to $\psi$, since in two dimensions the symplectic structure is just a volume structure. Finally if you have a smooth curve of area-preserving diffeomorphisms of a disk, I think there is a time-dependent Hamiltonian obtained by integrating the corresponding vector field.
I shared the same concern that Ilya expresses in the comments, but after considering it, here is why I think that it works. To have a clean view of the boundary conditions, let's double the disk to the sphere and let everything be equivariant with respect to reflection across the equator.
Moser's theorem truly is a deformation retraction. Let $M$ be a Riemannian manifold, let $\mu$ be a volume form on $M$ (not necessarily Riemannian volume), and let $\phi_\alpha:M \to M$ be a family of diffeomorphisms of $M$ that may or may not preserve $\mu$. Then `$\mu_\alpha = (\phi_\alpha)_*(\mu)$` is "wrong". Let $\mu_{\alpha,t}$ be a family of volume forms defined as the weighted geometric mean of $\mu_\alpha$ and $\mu$: $$\mu_{\alpha,t} = \mu^t \mu_\alpha^{1-t}.$$ Then there is a corresponding Moser flow $\phi_{\alpha,t}$ such that $\phi_{\alpha,0} = \phi_\alpha$ and $\phi_{\alpha,1}$ is volume-preserving. Moreover, $\phi_{\alpha,t} = \phi_\alpha$ for all $t$ if $\phi_\alpha$ is already volume-preserving for some fixed $\alpha$.
In particular, if $\phi_t$ is a curve of diffeomorphisms as produced by Smale's theorem with $\phi_0$ the identity, then Moser gives you an improvement $\phi_{t,s}$ such that $\phi_{t,1}$ is then what you want. What worried us is whether $\phi_{1,1} = \phi_1$; if $\phi_1$ is area-preserving, then it is true.
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Thank you for the answer! There's one thing I don't understand however: the Moser's theorem I know does not seem strong enough to give a deformation retraction from diffeo. to volume-preserving diffeo. All it would do is, given a time dependent volume form $\omega_t \in H^2(D^2)$, it'd give use a flow $\gamma_t: D^2 \to D^2$ such that $\gamma_t^*(\omega_t)=\omega_0$ and $\gamma_0 = id$. In particular, if $\omega_1 =\omega_0$, there is no guarantee that $\gamma_1 = id$. In other words, in our situation, if we have a path of diffeo $\psi_t$, apply Moser to $\psi_t^*(vol)$ to get $\gamma_t$, – Ilya Grigoriev Dec 19 2009 at 18:36
then the composition $\phi_t \circ \psi_t^-1$ will be a path of volume-preserving diffeomorphism, but the volume preserving diffeo. at time 1 will not be $\phi_1$. Is there a way to strengthen Moser to fix this and get a deformation retraction from diffeo. to volume-preserving diffeo.? Am I missing something? – Ilya Grigoriev Dec 19 2009 at 18:38
Thank you very much, this now works very well. It's a neat idea! – Ilya Grigoriev Dec 19 2009 at 23:01
You're very welcome; I enjoyed the question. – Greg Kuperberg Dec 19 2009 at 23:21
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I just wanted to expand two points of Greg's answer. Both are rather trivial additions, but took me a short while to understand, so I'm putting them here for completeness's sake and my own future reference.
First, here's a picture of Greg's idea of fixing the problem with Moser's trick. In his notation, we have volume forms $\mu_{\alpha,t} = \mu^t \mu_\alpha^{1-t}$. They are defined because the two volume form must have the same sign everywhere (since everything is orientation-preserving; note that because of this I doubt that this argument can be generalized to symplectic forms in higher dimensions).
The obvious, but wrong, solution (described in my comments to his answer) would be to apply Moser's theorem to $\mu_{\alpha,0}$ in this notation. This would correspond to flowing along the horizontal axes of the picture below. However, Greg's idea is to flow in a different direction: we fix $\alpha$ and vary $t$. Strictly speaking, we should also show that the resulting flow $\phi_{\alpha,1}$ will be smooth, but this should follow from the proof of Moser's theorem quite easily.
```` Moser flow,
t defines phi_alpha,t
(mu_{alpha,1} = mu) 1 ^ ^
| .
| .
| .
| .
(mu_{alpha,0} = mu_alpha) 0 ----------------> alpha
(phi_{alpha,0} = phi_alpha)
````
(Note that, for all t, $\mu_{0,t}=\mu_{1,t}=\mu$)
Secondly, Greg's answer implies that there is an isotopy $\psi_t: D^2 \to D^2$ such that each $\psi_t$ is volume preserving (or, equivalently, a symplectomorphism), $\psi_0$ is the identity, and $\psi_1=\psi$. Here's how we find the time-dependent Hamiltonian such that this is the Hamiltonian flow. Let Xt be the time-dependent vector field that is the derivative of the flow $\psi_t$. The fact our flow is volume-preserving is equivalent to the fact that the 1-form $\iota_{X_t} \omega$ is be closed for all $t$. Since we are on a disk, this form will also be exact. So, let $H_t$ be the function such that $dH_t = \iota_{X_t} \omega$. The Hamiltonian flow of the function $H_t$ is precisely $\psi_t$; this is immediate from the definitions.
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Two points on your points: (1) What you graciously call my idea is really an elaboration of two general precepts that are not due to me. One is that deformation retracts, or even more generally homotopy equivalences, are very powerful. The other is that Moser's trick is very powerful. (2) There is a symplectic version of Moser's trick, but it indeed does not give you a homotopy equivalence between diffeomorphisms and symplectomorphisms. The volume case is nicer because the set of volume structures is convex and therefore contractible. – Greg Kuperberg Dec 20 2009 at 0:15
Also I didn't have to be so fancy as to use a weighted geometric mean of the measures. You could also use an arithmetic mean. – Greg Kuperberg Dec 20 2009 at 5:48 | {"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": 54, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9533225297927856, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/206575-difference-between-gradient-vector-surface-normal-vector-s.html | 1Thanks
• 1 Post By HallsofIvy
# Thread:
1. ## The difference between a gradient vector to a surface and the normal vector to that s
...urface. Plz help I don't understand this difference. Thanks
2. ## Re: The difference between a gradient vector to a surface and the normal vector to th
Hey nicksbyman.
If you have a definition, you should post it (if its from lectures notes or a book) but a gradient vector can refer to the tangential vector with respect to a particular variable (i.e. one based on the partial derivative of the surface at a particular point).
I think your problem is where you refer to the "gradient vector" of a surface. There is no such thing. Rather, the gradient vector is the gradient of a function. If we have a function f(x,y,z) then the "gradient of f", also written $\nabla f$, is the "vector" $\frac{\partial f}{\partial x}\vec{i}+ \frac{\partial f}{\partial y}\vec{j}+ \frac{\partial f}{\partial z}\vec{k}$. Given such a function, the equation f(x,y,z)= constant, could, theoretically, be "solved" for one of the variables in terms of the other two. Since we can then have z= g(x,y), say, that equation defines a surface. Given the equation f(x,y,z)= C, $\nabla f$, the gradient vector of the function is a normal vector to the surface at every point. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9351446628570557, "perplexity_flag": "head"} |
http://mathhelpforum.com/calculus/70468-college-calculus.html | Thread:
1. college calculus
Let p(x) be a polynomial function of degree n. Determine if each of the following statements is TRUE or FALSE and justify your answer in writing.
NOTE: If you can find a counterexample, the counterexample is enough to justify saying the statement is FALSE.
(a) The domain of p(x) is all real numbers.
(b) The range of p(x) is all real numbers.
(c) p(x) has exactly n x-intercepts (i.e. exactly n roots)
My answers: Just to let you know we just started this class and the professor has taught us nothing so i have no clue what I am doing.
(a) n has all real numbers
(b) n has all real numbers
(c) x=n and x=p
2. Originally Posted by asweet1
Let p(x) be a polynomial function of degree n. Determine if each of the following statements is TRUE or FALSE and justify your answer in writing.
NOTE: If you can find a counterexample, the counterexample is enough to justify saying the statement is FALSE.
(a) The domain of p(x) is all real numbers.
(b) The range of p(x) is all real numbers.
(c) p(x) has exactly n x-intercepts (i.e. exactly n roots)
My answers: Just to let you know we just started this class and the professor has taught us nothing so i have no clue what I am doing.
(a) n has all real numbers
(b) n has all real numbers
(c) x=n and x=p
I believe the question is asking you to have "true" or "false" as an answer for each part. (True if you believe the statement is true, and otherwise)
For example,
(b) False. Here is an example: $y=x^2+2$ is a polynomial of degree 2. However, the range of such function is $y \geq 2$. Thus the statement is false since you can find a counterexample. | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9103504419326782, "perplexity_flag": "middle"} |
http://mathhelpforum.com/differential-geometry/156579-2-simple-outer-measure-problems.html | # Thread:
1. ## 2 simple outer measure problems
Hello all.
These are homework problems, so partial solutions / advice would be preferred.
These come from the book Real Analysis (4th ed) by Royden and Fitzpatrick
Chapter 2 problems 15 and 20
15. Show that if E has finite measure and $\epsilon>0$, then E is the disjoint union of a finite number of measurable sets, each of which has measure at most $\epsilon$.
[For this one I don't really know how to begin. At this point we know
Theorem 9: The collection M of measurable sets is a $\sigma$-algebra that contains the $\sigma$-algebra B of Borel sets. Each interval, each open set, each closed set, each $G_\delta$ and each $F_\sigma$ set is measurable.
If I'm not mistaken this means that once we prove problem 15, we will have also shown that any set with finite outer measure is measurable (since it is a finite union of measurable sets)
Thoughts? ]
20.(Lebesgue) Let E have finite outer measure. Show that if E is measurable if and only if for each open, bounded interval (a,b),
b-a=m*( $(a,b) \cap E$)+m*((a,b)~E)
[Here one direction is very easy. If E is measurable, just let (a,b) be a testing set and we're done. As for the other direction...?
In this section we learned the excision property m*(B~A)=m*(B)-m*(A) and
Theorem 11: The following are equivalent
0)E measurable
1) $\exists$ Outer Approximation by Open Sets and $G_\delta$ Sets
2) $\exists$ Inner Approximation by Closed Sets and $F_\sigma$ Sets
and
Theorem 12: Let E be a measurable set of finite outer measure. Then for each $\epsilong>0$, there is a disjoint collection of open intervals ${{\{I_k\}}_{k=1}}^n$ for which if O=union of $I_k$, then m*(E~O)+m*(O~E) $<\epsilon$
I don't really know how to go from making a statement about all open intervals to a statement about all sets of real numbers.
Thoughts?]
Thank you all. | {"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": 15, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9418390989303589, "perplexity_flag": "head"} |
http://physics.stackexchange.com/questions/32011/what-is-a-tensor | # What is a tensor?
I have a pretty good knowledge of physics but couldn't understand what a tensor is. I just couldn't understand it, and the wiki page is very hard to understand as well. Can someone refer me to a good tutorial about it?
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1
– Qmechanic♦ Jul 14 '12 at 7:45
I wrote a short one. The main problem is that the mathematical definitions are absurdly abstract, and don't match the domain of discourse well. The abstract definition is that a tensor is a linear map from vectors and covectors to numbers. The concrete definition is below. – Ron Maimon Jul 14 '12 at 8:52
2
Really short answer: A $[p, q]$ valent tensor is a thing that has $p$ upper and $q$ lower indices. Or it is something, that has to be multiplied by $q$ vectors and $p$ covectors to become a scalar. – queueoverflow Nov 2 '12 at 17:12
@queueoverflow can you explain what covector is ?thanks – 0x90 Nov 2 '12 at 17:33
A covector is a row vector ($1 \times n$ matrix), or transposed (i. e. rotated) vector as opposed to a vector, which is a column vector ($n \times 1$ matrix). So a regular matrix $M$ is a $[1, 1]$ valent tensor because you have to multiply it with a vector $v$ and a covector $c$ to get a scalar $s$: $s = c M v$. – queueoverflow Nov 3 '12 at 9:04
## 6 Answers
A (rank 2 contravariant) tensor is a vector of vectors. If you have a vector, it's 3 numbers which point in a certain direction. What that means is that they rotate into each other when you do a rotation of coordinates. So that the 3 vector components $V^i$ transform into
$$V'^i = A^i_j V^j$$
under a linear transformation of coordinates.
A tensor is a vector of 3 vectors that rotate into each other under rotation (and also rotate as vectors--- the order of the two rotation operations is irrelevant). If a vector is $V^i$ where i runs from 1-3 (or 1-4, or from whatever to whatever), the tensor is $T^{ij}$, where the first index labels the vector, and the second index labels the vector component (or vice versa). When you rotate coordinates T transforms as
$$T'^{ij} = A^i_k A^j_l T^{kl} = \sum_{kl} A^i_k A^j_l T^{kl}$$
Where I use the Einstein summation convention that a repeated index is summed over, so that the middle expression really means the sum on the far right.
A rank 3 tensor is a vector of rank 2 tensors, a rank four tensor is a vector of rank 3 tensors, so on to arbitrary rank. The notation is $T^{ijkl}$ and so on with as many upper indices as you have a rank. The transformation law is one A for each index, meaning each index transforms separately as a vector.
A covariant vector, or covector, is a linear function from vectors to numbers. This is described completely by the coefficients, $U_i$, and the linear function is
$$U_i V^i = \sum_i U_i V^i = U_1 V^1 + U_2 V^2 + U_3 V^3$$
where the Einstein convention is employed in the first expression, which just means that if the same index name occurs twice, once lower and once upper, you understand that you are supposed to sum over the index, and you say the index is contracted. The most general linear function is some linear combination of the three components with some coefficients, so this is the general covector.
The transformation law for a covector must be by the inverse matrix
$$U'_i = \bar{A}_i^j U_j$$
Matrix multiplication is simple in the Einstein convention:
$$M^i_j N^j_k = (MN)^i_k$$
And the definition of $\bar{A}$ (the inverse matrix) makes it that the inner product $U_i V^i$ stays the same under a coordinate transformation (you should check this).
A rank-2 covariant tensor is a covector of covectors, and so on to arbitrarily high rank.
You can also make a rank m,n tensor $T^{i_1 i_2 ... i_m}_{j_1j_2 ... j_n}$, with m upper and n lower indices. Each index transforms separately as a vector or covector according to whether it is up or down. Any lower index may be contracted with any upper index in a tensor product, since this is an invariant operation. This means that the rank m,n tensors can be viewed in many ways:
• As the most general linear function from m covectors and n vectors into numbers
• As the most general linear function from a rank m covariant tensor into a rank n contravariant tensor
• As the most general linear function from a rank n contravariant tensor into a rank m covariant tensor.
And so on for a number of interpretations that grows exponentially with the rank. This is the mathemtician's preferred definition, which does not emphasize the transformation properties, rather it emphasizes the linear maps involved. The two definitions are identical, but I am happy I learned the physicist definition first.
In ordinary Euclidean space in rectangular coordinates, you don't need to distinguish between vectors and covectors, because rotation matrices have an inverse which is their transpose, which means that covectors and vectors transform the same under rotations. This means that you can have only up indices, or only down, it doesn't matter. You can replace an upper index with a lower index keeping the components unchanged.
In a more general situation, the map between vectors and covectors is called a metric tensor $g_{ij}$. This tensor takes a vector V and produces a covector (traditionally written with the same name but with a lower index)
$$V_i = g_{ij} V^i$$
And this allows you to define a notion of length
$$|V|^2 = V_i V^i = g_{ij}V^i V^j$$
this is also a notion of dot-product, which can be extracted from the notion of length as follows:
$$2 V\cdot U = |V+U|^2 - |V|^2 - |U|^2 = 2 g_{\mu\nu} V^\mu U^\nu$$
In Euclidean space, the metric tensor $g_{ij}= \delta_{ij}$ which is the Kronecker delta. It's like the identity matrix, except it's a tensor, not a matrix (a matrix takes vectors to vectors, so it has one upper and one lower index--- note that this means it automatically takes covectors to covectors, this is multiplication of the covector by the transpose matrix in matrix notation, but Einstein notation subsumes and extends matrix notation, so it is best to think of all matrix operations as shorthand for some index contractions).
The calculus of tensors is important, because many quantities are naturally vectors of vectors.
• The stress tensor: If you have a scalar conserved quantity, the current density of the charge is a vector. If you have a vector conserved quantity (like momentum), the current density of momentum is a tensor, called the stress tensor
• The tensor of inertia: For rotational motion of rigid object, the angular velocity is a vector and the angular momentum is a vector which is a linear function of the angular velocity. The linear map between them is called the tensor of inertia. Only for highly symmetric bodies is the tensor proportional to $\delta^i_j$, so that the two always point in the same direction. This is omitted from elementary mechanics courses, because tensors are considered too abstract.
• Axial vectors: every axial vector in a parity preserving theory can be thought of as a rank 2 antisymmetric tensor, by mapping with the tensor $\epsilon_{ijk}$
• High spin represnetations: The theory of group representations is incomprehensible without tensors, and is relatively intuitive if you use them.
• Curvature: the curvature of a manifold is the linear change in a vector when you take it around a closed loop formed by two vectors. It is a linear function of three vectors which produces a vector, and is naturally a rank 1,3 tensor.
• metric tensor: this was discussed before. This is the main ingredient of general relativity
• Differential forms: these are antisymmetric tensors of rank n, meaning tensors which have the property that $A_{ij} =-A_{ji}$ and the analogous thing for higher rank, where you get a minus sign for each transposition.
In general, tensors are the founding tool for group representations, and you need them for all aspects of physics, since symmetry is so central to physics.
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A tensor is a generalization of the notion of scalars and vectors. A tensor of rank 0 is a scalar (it has $3^0$ compenent), while a tensor of rank 1 is a vector (which has $3^1$ components). In general, a tensor of rank $n$ has $3^n$ components.
See http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf for a nice introduction.
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The above link sometimes doesn't work. If so, google "tensor" and "nasa", which will bring up the article by JC Kolecki. You can then use Google QuickView. – Joebevo Jul 14 '12 at 8:04
I've edited the link so that it points directly to the pdf, rather than going through a google search result. I hope that's ok... – Nathaniel Jul 14 '12 at 9:34
This was an excellent tutorial! The example on page 10 on why not all scalars are rank-0 tensors was very clear. – recipriversexclusion Jul 27 '12 at 22:27
Tensors are objects with usually multiple indices, a generalization of vectors and matrices, with definite transformation properties under a change of basis. They are introduced differently in different traditions, with different notations.
You may find the entry ''How are matrices and tensors related?'' from Chapter B8 of my theoretical physics FAQ relevant to disentangle some of the associated problems.
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My very simple answer is really just one of many situations where a tensor is handy when describing the forces on a body...they are used almost everywhere in physics however...this just one SIMPLE example.
A cubic body is moving through air and is feeling the resistance to motion orthogonal to it's trajectory. This normal force could occur on any SIDE of of the cube. OR, If the cube sits still, it experiences pressure from the atmosphere, the pressure can be decomposed into normal forces on each side.
Now there's the SHEAR force, of the viscose air that clings to the top of the cube and the drag deforms the top of the cube. This shear force is on the sides parallel to the motion of the moving cube. This can occur FOR EVERY parallel surface.
Tensors are handy when ALL the possibilities are actually possible and do occur. Then there are tricks for summing the forces. That is what all the fancy tenor math above is about.
I was told by a great fluid mechanics professor, that tensors should only be used when we understand the forces and/or system well. Typically, when learning something new, we start with each dimension separately and tediously work out all the math....then when we know what's going on, tensors can be used.
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There are some good answers already, and they show how varied the equivalent definitions can be. For a succinct answer, the Wikipedia page has a good one, depending on how familiar you are with the notion of a dual space.
For application to physics, though, know that many different subfields use tensors, and all in different ways. We might be able to provide more pertinent references if you specify the context you're interested in. For example, Carroll's Spacetime and Geometry gives a physically intuitive introduction to tensors in general relativity without any loss of rigor.
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There are a lot of answers already hope I can make it even more clear.
Tensors are the generalization of the linear transformations.
Tensor is something that takes $m$ vectors and makes $n$ vectors from it.
The $n+m$ is the order (or rank) of the tensor.
Their type is denoted by $(n,m)$ (n: output vectors, m: input vectors)
When a tensor takes 0 vectors it means it calculates something from a scalar (or is a constant), if a tensor makes 0 vectors, it produces a scalar.
Some examples of tensors by type:
• (0,0): scalar, just a number.
• (1,0): single vector.
• (2,0): a bivector
• (1,1): Linear transformation.
• (0,2): dot product of two vectors.
• (1,2): cross product of two vectors in 3D.
• (1,3): Riemann curcature tensor (if you are interested in general relativity, you will need this.)
Tensors can be described using an $n+m$ dimensional array of numbers. So the tensor's elements can be accessed using $n+m$ indexes.
For example, linear transformation is a 2nd order tensor.
The elements of the multidimensional tensor can be accessed by index, a matrix has obviously 2 indexes.
Now something about the notation. Tensor elements usually has multiple indexes, some upper indexes and some lower ones. Lower indexes going for the input vectors, upper indexes are for the output vectors. Note: upper indexes has nothing to do with exponents!
So a linear transformation tensor would look like this: $L_j^i$.
You do a linear transformation (aka calculating the elements of the resulting vector) like this:
$b^i = \displaystyle\sum_j L_j^i a^j$
So assume you are in 3D and multiply a 3×3 matrix with a column vector. In this case the upper index is for the lines, and the lower is for the columns of the matrix. $i$ and $j$ runs from 1 to the dimension you are in (usually 3).
You can chain these linear transformations like this:
$c^k = \displaystyle\sum_i M_i^k \displaystyle\sum_j L_j^i a^j$
Einstein noted, that in these summation formulas the index below the summation sign appears exactly twice. So it can be removed. So the previous two expressions will look like this:
$b^i = L_j^i a^j$
$c^k = M_i^k L_j^i a^j$
Which is very analogous with the matrix formulas you use in linear algebra. The upper index kills the lower index during calculation, while the lone indexes remain intact.
So you can multiply the two matrixes as tensors like this:
$T_j^k = M_i^k L_j^i = \displaystyle\sum_i M_i^k L_j^i$
And finally a cross-product with tensors would look like this:
$r^k = C_{ij}^k a^i b^j$
The $C$ is a 3×3×3 array of numbers multiplied by a vector will produce and ordinary matrix, which multiplied by another vector will produce the final vector.
A dot product in the language of tensors would look like this:
$r = D_{ij} a^i b^j = \displaystyle\sum_{i,j} D_{ij} a^i b^j$
Where $D_{ij}$ is an identity matrix.
Now the wiki article on Tensors should be more comprehensible.
Hope this will give an aha moment someone.
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http://leifjohnson.net/post/on-database-wrangling | # Leif Johnson » On database wrangling
Wasn't I just sitting here writing about time passing quickly ? Well, as the end of the semester approaches and I find myself increasingly stressed, time just zips right along !
## Kohonen maps
This week I wrote up a little JavaScript visualization of a bimodal Kohonen map that I think is pretty ok. The dataset stored in the map is a sequence of $(\rho, \theta, x, y)$ tuples. The $x$ and $y$ components of each tuple identify a single point in the 2-dimensional cartesian plane, and the $\rho$ and $\theta$ components describe the normal vector to a line in the $2$-dimensional cartesian plane. The point $(x, y)$ is constrained to lie on the line that's normal to the $\rho$-$theta$ vector, so any three of the four dimensions are sufficient to calculate the fourth. (This is actually not entirely true, since $\theta$ can take on two values for any particular $\rho$, $x$, and $y$. But the other dimensions are ok.)
A training dataset consists of randomly sampled points from this 4-dimensional space, such that the constraints are met. The sampling process draws $x$ and $y$ from Normal(0, 1), and $\theta$ from Uniform(0, $2\pi$). Then $\rho$ is calculated using the three sampled values.
The bimodal map actually consists of three separate Kohonen maps, each of the same topology—in this dataset, I used 2-dimensional rectangular grid maps with 50 rows and 50 columns. One map stores the “external” cues (the $\rho$ values), one map stores the “internal” responses (the $(\theta, x, y)$ values), and the third stores the Jacobian that maps changes in the internal values to changes in the external value.
The external map is presented with the $\rho$ components of each data point, and the coordinates of the winner are used to identify both the internal vector and Jacobian matrix for that external cue. These values are mapped onto a series of (hopefully increasingly accurate) $\rho$ values. This process is described by [Ritter, Martinetz and Schulten][] in their work on controlling a robot arm.
The maps have actually been working fairly well. There are some issues remaining with initialization, since the map tends to explode into NaN territory relatively often, but in general I've been happy with the results. The next step is to generate a more biologically relevant dataset by simulating the motions of a human shoulder/elbow(/wrist eventually).
I'm also working on putting my Python Kohonen map code up on bitbucket.
## MySQL boo hoos
On a completely separate front, I'm finding myself once again butting heads with MySQL. As hokey as it may be to complain about such things, I thought I'd gotten away from battling with this relational database system once I left my post in the lawless territory of corporate software. But MySQL is just too useful for my task : I'm implementing the algorithm described in "Finding community structure in very large networks", but because the graph in question has about 3MM vertices and 70MM edges, I just can't hold the thing in memory very well. So it's off to the races with the MySQL documentation, setting variables like key_buffer_size and such nonsense. Mostly I'm just whining ; it's really the graph's fault that these operations take so much time. But waiting for a database to finish is worse than watching paint dry—you can't even blow on it to get it to hurry up already.
## The softer side
The garden is going fairly well, but I'm a bit worried that the peas are drying out—their bottom leaves are all shriveled and yellow, but the top leaves seem fine. I think it might just be a growth strategy, so I'm leaving it alone for now. I have a couple of photos to put up, whenever I get photos set up on this web site (soon !).
I've been refreshing my hands on the piano more often lately, which has been quite pleasant. It's difficult to play so badly right now, knowing that the only thing lying between where I am now and where I want to get is a lot of practice time. Right now I'm still struggling with reading the music ; I think a lot of my practice time will improve once I've learned the mapping between black dots on paper and motions of my hands.
Finally, I think I've just managed to squeeze one game of disc golf in since the last update, but this time I did remember my stroke counts on each hole, so I put them on a neato graph :
It'll be fun to track this over time. It's too bad the labels in the legend aren't sorted, but whatevs.
© Copyright 2009 Leif Johnson. Licensed for reuse under CC-BY-SA. Originally published at 15:04 on 23 Apr 2009. | {"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": 22, "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}, "config": {"markdown_headings": true, "markdown_code": false, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9472241401672363, "perplexity_flag": "middle"} |
http://unapologetic.wordpress.com/2009/09/09/ | # The Unapologetic Mathematician
## Shears Generate the Special Linear Group
We established that if we restrict to upper shears we can generate all upper-unipotent matrices. On the other hand if we use all shears and scalings we can generate any invertible matrix we want (since swaps can be built from shears and scalings). We clearly can’t build any matrix whatsoever from shears alone, since every shear has determinant ${1}$ and so must any product of shears. But it turns out that we can use shears to generate any matrix of determinant ${1}$ — those in the special linear group.
First of all, let’s consider the following matrix equations, which should be easy to verify
$\displaystyle\begin{aligned}\begin{pmatrix}1&0&0\\{0}&1&x\\{0}&0&1\end{pmatrix}\begin{pmatrix}y&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}&=\begin{pmatrix}y&0&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&1&x\\{0}&0&1\end{pmatrix}\\\begin{pmatrix}1&x&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&y\end{pmatrix}&=\begin{pmatrix}1&0&0\\{0}&1&0\\{0}&0&y\end{pmatrix}\begin{pmatrix}1&x&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\\\begin{pmatrix}1&x\\{0}&1\end{pmatrix}\begin{pmatrix}y&0\\{0}&1\end{pmatrix}&=\begin{pmatrix}y&0\\{0}&1\end{pmatrix}\begin{pmatrix}1&\frac{x}{y}\\{0}&1\end{pmatrix}\\\begin{pmatrix}1&x\\{0}&1\end{pmatrix}\begin{pmatrix}1&0\\{0}&y\end{pmatrix}&=\begin{pmatrix}1&0\\{0}&y\end{pmatrix}\begin{pmatrix}1&xy\\{0}&1\end{pmatrix}\end{aligned}$
These show that we can always pull a scaling to the left past a shear. In the first two cases, the scaling and the shear commute if the row and column the scaling acts on are uninvolved in the shear. In the last two cases, we have to modify the shear in the process, but we end up with the scaling written to the left of a shear instead of to the right. We can use these toy examples to see that we can always pull a scaling from the right to the left of a shear, possibly changing the shear in the process.
What does this mean? When we take a matrix and write it out in terms of elementary matrices, we can always modify this expression so that all the scalings are to the left of all the shears. Then we have a diagonal matrix to the left of a long product of shears, since the product of a bunch of scalings is a diagonal matrix. But now the determinant of each shear is ${1}$, and the determinant of the diagonal matrix must be the product of the diagonal entries, which are the scaling factors. And so the product of the scaling factors is the determinant of our original matrix.
We’re specifically concerned with matrices of determinant ${1}$, meaning the product of all the diagonal entries must come out to be ${1}$. I’m going to use this fact to write the diagonal matrix as a product of scalings in a very particular way. Let’s say the diagonal entry in row $n$ is $\lambda_n$. Then I’m going to start by writing down
$\displaystyle C_{1,\lambda_1}C_{2,\lambda_1^{-1}}$
I’ve scaled the first row by the right amount, and then scaled the second row by the inverse amount so the product of the two scaling factors is ${1}$. Then I write down
$\displaystyle C_{2,\lambda_1\lambda_2}C_{3,\lambda_1^{-1}\lambda_2^{-1}}$
The product of the two scalings of the second row ends up scaling it by $\lambda_2$, and we scale the third row to compensate. We continue this way, scaling each row to the right amount, and the next one by the inverse factor. Once we scale the next-to-last row we’re done, since the scaling factor for the last row must be exactly what we need to make the total product of all the scaling factors come out to ${1}$. That is, as long as the total scaling factor is ${1}$, we can write the diagonal matrix as the product of these pairs of scalings with inverse scaling factors.
Now let’s take four shears, alternating upper and lower, since two upper shears in a row are the same as a single upper shear, and similarly for lower shears. We want it to come out to one of these pairs of scalings.
$\displaystyle\begin{aligned}\begin{pmatrix}1&a\\{0}&1\end{pmatrix}\begin{pmatrix}1&0\\b&1\end{pmatrix}\begin{pmatrix}1&c\\{0}&1\end{pmatrix}\begin{pmatrix}1&0\\d&1\end{pmatrix}=\begin{pmatrix}1+ab&a\\b&1\end{pmatrix}\begin{pmatrix}1+cd&c\\d&1\end{pmatrix}&\\=\begin{pmatrix}1+ab+ad+cd+abcd&a+c+abc\\b+d+bcd&1+bc\end{pmatrix}&\\=\begin{pmatrix}x&0\\{0}&\frac{1}{x}\end{pmatrix}&\end{aligned}$
This gives us four equations to solve
$\displaystyle\begin{aligned}1+ab+ad+cd+abcd&=x\\a+c+abc&=0\\b+d+bcd&=0\\1+bc&=\frac{1}{x}\end{aligned}$
These quickly simplify to
$\displaystyle\begin{aligned}1+ab&=x\\\frac{a}{x}+c&=0\\b+\frac{d}{x}&=0\\1+bc&=\frac{1}{x}\end{aligned}$
Which can be solved to find
$\displaystyle\begin{aligned}a&=\frac{x-x^2}{d}\\b&=-\frac{d}{x}\\c&=\frac{x-1}{d}\\d&=d\end{aligned}$
So we could pick $d=-x$ and for any scaling factor $x$ write
$\displaystyle\begin{pmatrix}x&0\\{0}&\frac{1}{x}\end{pmatrix}=\begin{pmatrix}1&x-1\\{0}&1\end{pmatrix}\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&\frac{1-x}{x}\\{0}&1\end{pmatrix}\begin{pmatrix}1&0\\-x&1\end{pmatrix}$
And so we can write such a pair of scalings with inverse scaling factors as a product of four shears. Since in the case at hand we can write the diagonal part of our elementary matrix decomposition with such pairs of scalings, we can translate them all into shears. And at the end of the day, we can write any special linear transformation as a product of a bunch of shears.
Posted by John Armstrong | Algebra, Linear Algebra | 3 Comments
## About this weblog
This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).
I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. | {"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": 21, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254040122032166, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/41701/how-does-one-compute-cos-pi-4k-1 | # How does one compute $\cos((\pi/4)(k-1 ))$?
How to compute $\cos \left( \frac{\pi}{4}(k-1) \right)$ ?
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## 2 Answers
Think about it; $\pi/4$ is $45°$. With each different value of $k$ you're summing up $45°$. Since $\pi/4$ divides $2\pi$, you'll eventually get a cycle of values for $\cos(k(\pi/4))$, and then you can use this to derive an answer.
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It's probably worthwhile pointing out that we're assuming that $k$ is an integer. This wasn't stated in the original post, but it isn't a very meaningful question otherwise. – cch May 27 '11 at 20:04
As a more difficult extension of fmartin's answer, we might consider by using the angle difference formula for cos: $\cos(\alpha \pm \beta) = \cos (\alpha ) \cos (\beta ) \mp \sin (\alpha ) \sin (\beta )$, which can be derived using complex numbers or matrices very quickly.
Then we consider $\cos (k \pi /4 - \pi /4) = \cos (k \pi/4) /\sqrt2 + \sin (k \pi /4)/\sqrt2$. Of course, we would then proceed by fmartin's answer, noting the cyclic nature for different integer values of k.
And if k is not an integer, then this question is just sort of asking how to evaluate cos(x)? With a Taylor series, I suppose.
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http://nanoexplanations.wordpress.com/2011/12/09/polygon-rectangulation-part-2-minimum-number-of-fat-rectangles/ | the blog of Aaron Sterling
# Polygon rectangulation, part 2: Minimum number of fat rectangles
Posted on December 9, 2011
This post is the second in a series on polygon rectangulation. In my previous post, I discussed methods to decompose an orthogonal polygon into a minimum number of rectangles. (See that post for definitions and motivation.) In my next post, I will consider finding a rectangulation of a minimum length — a topic very important in VLSI. In this post, I will consider a modification to the minimum-number-of-rectangles problem; the modification was motivated by concerns in VLSI, but, as yet, only a theoretical algorithm exists, with a running time of $O(n^{42})$. (That is not a typo, and it is obtained through a “natural” dynamic programming solution to the problem I am about to state.)
Printed circuits are created through a process called photolithography, in which electron beams etch a design onto a substrate. While these electron beams are, in one sense, extremely narrow, as the Wikipedia article on VLSI states, current photolithography techniques “tend closer to the fundamental laws of optics.” Among other things, this means that the fixed minimum width of an electron beam is suddenly important. In principle, it implies a “fat fingers” problem. Suppose our substrate is in the shape of orthogonal polygon $P$, and we use a rectangulation technique from the previous post to rectangulate $P$. We may not be able to apply the rectangulation in real life, because we have no guarantee that all of the rectangles are wider than our electron beam. Therefore, we would like to constrain the space of rectangulations we consider to ones that are feasible to etch — informally, ones that contain only “fat” rectangles. We formalize this optimization problem as follows.
Fat Rectangle Optimization Problem: Given an orthogonal polygon $P$, maximize the shortest side $\delta$ over all rectangulations of $P$. Among the partitions with the same $\delta$, choose the partition with the fewest number of rectangles.
This optimization problem has been studied by O’Rourke and co-authors in at least three papers. In this blog post, I will focus on consideration of The Structure of Optimal Partitions of Orthogonal Polygons into Fat Rectangles, by O’Rourke and Tewari (2004).
Rectangulations limited to vertex cuts
We saw in the last post that to obtain a minimum-rectangle rectangulation, we could limit ourselves to considering rectangulations whose line segments were boundary edges of $P$, or rooted at convex vertices of $P$. Given $P$, we can define its vertex grid, by drawing horizontal and vertical lines through each vertex of $P$, and a point $q$ is a member of the vertex grid iff it is located on the intersection of two of those lines. We will see in the next post that, to obtain a minimum-length rectangulation, we can limit consideration to line segments between points that lie on the vertex grid. Neither of these restrictions is sufficient for the Fat Rectangle Optimization Problem, as the figure below demonstrates.
This is Figure 4 in the O’Rourke and Tewari paper. If we try to move (or redraw) the segments so they are rooted at vertices, or at points on the vertex grid, we necessarily produce a rectangle that is skinnier than any in the figure. Moreover, there is a segment that is floating, i.e., not anchored to the boundary at either endpoint.
For the rest of this post, we will call a non-boundary segment of a rectangulation a cut. The terminology is motivated by photolithography: those are the segments we would need to cut into the substrate. One way to simplify this problem is to limit the types of cuts we permit to appear in rectangulations. Let us call a vertex cut a cut with at least one endpoint a polygon vertex.
As long ago as 2001, O’Rourke, Pashchenko and Tewari designed an $O(n^5)$ algorithm to solve Fat Rectangle Optimization limited to vertex cuts, and they implemented their algorithm (postscript file of CCCG abstract). They noted empirically that the algorithm appeared to perform at $O(n^2)$ with inputs of random orthogonal polygons. Their algorithm was inspired by the Liou et al. $O(n \log \log n)$ algorithm for polygon rectangulation that I outlined in my previous post.
As an aside, the notion of a “random orthogonal polygon” is surprisingly nontrivial. One paper that provides methods for this is Generating Random Orthogonal Polygons, by Tomás and Bajuelos. No polynomial time algorithm is known for the problem of uniformly randomly generating polygons with a given set of vertices, so researchers must use heuristics, or limit the types of polygons they generate. Further, Tomás and Bajuelos base (some of) their work on a computer program supplied by O’Rourke that is originally based on methods from a 1991 technical report by O’Rourke and Virmani in which the authors state, “A precise definition of what constitutes a random polygon seems elusive, and no attempt will be made here to clarify this notion.” I have no additional wisdom to impart, but I thought you might want to know that there is something important (or at least ill-defined) hiding just under the surface here.
Rectangulations limited to anchored cuts
An anchored cut is a cut that is not a floating cut; in other words, at least one endpoint of an anchored cut is on the boundary of the polygon. The figure above shows that there exist polygons for which anchored cuts do not suffice to find the optimal solution of the Fat Rectangle Optimization Problem. There is a nice structure on solutions obtainable from anchored cuts, but O’Rourke and Tewari initially found it surprising. Their motivating example for their proof was this figure (which is Figure 8 in the paper).
The surprising aspect of this is that the vertical anchored cuts do not lie midway between any pair of vertices, but rather lie 1/3 and 2/3 of the distance between vertices 3 and 11. It turns out that this phenomenon can be captured in a theorem, which is Theorem 6 of O’Rourke and Tewari.
Theorem: There exists an optimal anchored partition of a polygon $P$ such that every movable anchored cut has coordinate $\frac{1}{2} a + \frac{1}{2} b$ or $\frac{1}{3} a + \frac{2}{3} b$, where $a$ and $b$ are (either x or y) coordinates of two vertices of $P$.
This means that one can define a grid of possible anchor points from the vertices of $P$, and then evaluate rectangulations based on “vertex cuts” anchored at that grid of possible anchor points, using the already-existing $O(n^5)$ vertex-cut algorithm mentioned in the previous section. Since there are $O(n^2)$ possible anchor points, this then gives a $O(n^{10})$ algorithm for the Fat Rectangle Optimization Problem limited to anchored cuts.
Rectangulations with no restriction on types of cut
Analysis of rectangulations based on unrestricted cuts makes up about half of O’Rourke and Tewari’s paper. They prove many interesting combinatorial and geometric facts about fat rectangulations, but, as they put it at the end of one lemma, “There is an overestimation at several points in the above argument; we do not believe the quoted upper bounds can be achieved. It would be of interest to establish tight bounds.” So they end up proving that a solution to the Fat Rectangle Optimization Problem lies, in principle, in polynomial time — but the exponent is one of the largest ones I have seen to come out of a “natural” argument. (For other algorithms that “naturally” produce large exponents and/or large constants, see this CSTheory question.)
Where does the exponent 42 come from? This is just mentioned briefly in the “Structure” paper, but is explained in detail in Partitioning Orthogonal Polynomials into Fat Rectangles in Polynomial Time by O’Rourke and Tewari (2002). The authors obtain combinatorial bounds on the maximum number of rectangles in an optimal partition ($<18n$), on the total number of cuts required ($<12n$), and on the amount of nesting of floating cuts inside floating rectangles that might occur in worst case (depth $\leq 2$). Bounds like this allow them to prove the following theorem.
Theorem: There is an optimal partition whose cuts fall on a subset of $O(n^4)$ gridlines.
This can be considered an analogue to the fact that finding a minimum-length rectangulation can be done by considering only cuts anchored to the polygon’s vertex grid, as I mentioned at the beginning of the post. Unfortunately, lines instead of points means that a dynamic programming algorithm needs a much larger table of subproblems to consider.
There are, of course $O(n^2)$ pairs of vertices of the input polygon $P$. One combinatorial bound the authors prove is that any set of cuts between two vertices which is “good” — that is, which may be considered an optimal solution to a subproblem, and available for solution to a larger problem — is at most distance 4 away from a chain of cuts that has chain length at most 10. It then requires time and space $O(n^{42})$ to build this table of subproblem solutions, for use in a dynamic programming algorithm to solve the Fat Rectangle Optimization Problem without restrictions.
O’Rourke, J., & Tewari, G. (2004). The structure of optimal partitions of orthogonal polygons into fat rectangles Computational Geometry, 28 (1), 49-71 DOI: 10.1016/j.comgeo.2004.01.007
J. O’Rourke, & G. Tewari (2002). Partitioning orthogonal polygons into fat rectangles in polynomial time Canadian Conference on Computational Geometry, 97-100
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• ### About me
Occasionally streetwise researcher of DNA self-assembly and other nonstandard applications of theoretical computer science. | {"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": 32, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9026318192481995, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/105933/associated-graded-of-a-filtration-of-a-tensor-product | ## Associated graded of a filtration of a tensor product
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
I'm trying to understand a part of the PhD thesis of Kenji Lefèvre-Hasegawa (e.g. available here). My question is about the proof of Lemma 1.3.2.3b stating:
Remarquons que nous avons un isomorphisme de complexes $\bigoplus_{i\geq 1}Gr_i(\Omega BA)\to \Omega \overline{T^c}V$ qui identifie à la composante $Gr_i(\Omega BA), i\geq 1$, à la somme des $S^{-1} V^{\otimes i_1}\otimes\dots\otimes S^{-1} V^{\otimes i_k}\subset (S^{-1} \overline{T^c}V)^{\otimes k}$, où $k\geq 1$ et où $i_1+\dots+i_k=i$.
Here $V=SA$, where $S$ is the shift functor. $A$ is a dg algebra and $B$ and $\Omega$ are the bar and cobar resolution, respectively, i.e. $BA=\bigoplus_{i\geq 1} V^{\otimes i}$ (with the coproduct $\Delta: V^{\otimes i}\to \bigoplus_{p+q=i} V^{\otimes p}\otimes V^{\otimes q}$ and the filtration $BA_{[i]}=ker\Delta^{(i)}$) and $\Omega BA=\bigoplus_{i\geq 1}S^{-1}BA^{\otimes i}$ with the filtration induced by that of $BA$.
Can someone give more details?
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## 1 Answer
OK let me try a naive answer, and then maybe you will elaborate a bit on what it is that worries you?
The key idea is very simple - the bar differential has a part coming from the differential on $A$, and the remaining part, encoding the product of $A$. The first part preserves the number of tensor factors, the remaining part consists of terms with fewer tensor factors. Therefore, for the filtration by the number of tensor factors, the corresponding graded object will just remember the differential on $A$, that is the structure of a chain complex, not the product. It remains to notice that for obvious reasons the number of tensor factors is precisely what is counted by the filtration by kernels of iterated deconcatenations... Does it help?
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If I understand you correctly, your answer is on the associated graded of $BA$, whose parts are $(SA)^{\otimes i}$. That I did understand, what I am missing is how this leads to the associated graded parts of $\Omega BA$. – Julian Kuelshammer Aug 30 at 15:56
1
Well, the induced filtration on the cobar-bar construction is naturally by the total number of tensor factors, $i_1+\ldots+i_k$ in the notation of your post, so I am puzzled by what worries you. $BA$ is a cofree coalgebra, so the differential of its cobar complex has a part coming from the differential of $BA$, and you seem to agree that after passing to the graded object this becomes the differential on $\overline{T^c(sA)}$ obtained by extending the differential of $A$, and a part coming from the (deconcatenation) coproduct on $BA$, it preserves the number of tensor factors and is unchanged. – Vladimir Dotsenko Aug 30 at 16:10
The problem was more elementary, I got the definition of induced filtration wrong. Thanks for the help. – Julian Kuelshammer Aug 30 at 20:38
Glad you sorted it out! – Vladimir Dotsenko Aug 30 at 20:41 | {"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": 27, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9274764060974121, "perplexity_flag": "head"} |
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Let $(X,\tau)$ be compact and $(X,\tau^*)$ be a Hausdorff space. How can we show that $\tau=\tau^*$ if $\tau^*\subset \tau$? | {"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": 171, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9327822923660278, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/45419/interpreting-correlation-coefficient-of-determination | # Interpreting correlation & coefficient of determination
Correlation (r) is a measure of linear association between two variables. Coefficient of determination (r^2), is a measure of how much of the variability in one variable can be "explained by" variation in the other.
For example, if r 0.8 is the correlation between two variables, then, r^2 = 0.64. Hence, 64% of the variability in one can be explained by differences in the other. Right?
My question is, for the example stated, is any one of the following statements correct?
1. 64% of values fall along the regression line
2. 80% of values fall along the regression line
I am quite confused. I would highly appreciate any help. Thanks!
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## 2 Answers
The first part of this is basically correct - but it's 64% of the variation is explained by the model. In a simple linear regression: Y ~ X, if $R^2$ is .64 it means that 64% of the variation in Y is determined by the linear relationship between Y and X. It is possible to have a strong relationship with very low $R^2$, if the relationship is strongly non-linear.
Regarding your two numbered questions, neither is correct. Indeed, it is possible that none of the points may lie exactly on the regression line. That's not what's being measured. Rather, it is a question of how close the average point is to the line. If all or nearly all points are close (even if none are exactly on the line) then $R^2$ will be high. If most points are far from the line, $R^2$ will be low. If most points are close but a few are far, then the regression is incorrect (problem of outliers). Other things can go wrong, too.
In addition, I've left the notion of "far" rather vague. This will depend on how spread out the X's are. Making these notions precise is part of what you learn in a course on regression; I won't get into it here.
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Well that cleared up a lot for me! Thank you Mimshot and Peter Flom! Much grateful to you both! :) – Bradex Dec 8 '12 at 15:50
1
+1, good answer, would you mind adding something like "Indeed, [it is possible that] none of the points may lie... ". Also, it might be worth discussing that the notion of how far the points are from the line is also relative to how spread out the X's are. – gung Dec 8 '12 at 15:56
Niether 1 nor 2 is correct.
Let's say you are trying to predict a set of values $\pmb{y}$ from a set of values $\pmb{x}$ using a linear regression. Your model is
$$y_i = b + mx_i + \epsilon_i$$
Where $\epsilon_i \sim \mathcal{N(0,\sigma^2)}$ is some noise. $R^2=.64$ means that 64% of the variance of $y$ can be explained by variability in $x$ under your model. The residual variance (i.e., the variance unexplained) is 0.36. That is, if:
$$\hat{y}_i = b + mx_i$$
Then
$$1-0.64 = 0.36 = \frac{\mathrm{var}(\pmb{y}-\pmb{\hat{y}})}{\mathrm{var}(\pmb{y})}$$
- | {"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": 10, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9379416704177856, "perplexity_flag": "head"} |
http://stats.stackexchange.com/questions/30560/how-to-give-entire-word-a-quality-score-to-account-for-the-phred-score-of-each-l | # How to give entire word a quality score to account for the phred score of each letter?
I have a word representing a DNA sequence and each letter of that word is a nucleotide base (A,T,G,C). Each letter has a phred quality score. The formula used for the phred quality score is $Q=-10\log_{10} P$ where $Q$ is the phred score and $P$ the probability of that letter to be incorrect( for example, there was a problem in DNA sequencing).
My question is:
what is the best way to give the entire word a quality score, so that it takes into account the phred score of each letter?
One thing I though about was to take the average and another was to add the phred scores.
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I think a good tag for this post would be phred, but I cannot create a tag. – Julio Diaz Jun 18 '12 at 18:55 | {"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": 3, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9504567980766296, "perplexity_flag": "middle"} |
http://physics.stackexchange.com/questions/44773/vector-nature-of-angular-velocity/44795 | # Vector Nature Of Angular Velocity
I am currently reading about angular position, angular velocity, and angular acceleration. I came across this paragraph that was particularly confusing, and was wondering if someone could perhaps help me. Here is the paragraph:
"If a particle rotates in the $xy$ plane, the direction of $\vec{\omega}$ for the particle is out of the plane of the diagram when the rotation is counterclockwise and into the plane of the diagram when the rotation is clock-wise. To illustrate this convention, it is convenient to use the right-hand rule demonstrated in Figure 10.3. When the four fingers of the right hand are wrapped in the direction of rotation, the extended right thumb points in the direction of $\vec{\omega}$The direction of $\vec{\alpha}$ follows from its definition $\vec{\alpha}=d\vec{\omega}/dt$. It is in the same direction as $\vec{\omega}$ if the angular speed is increasing in time, and it is antiparallel to $\vec{\omega}$ if the angular speed is decreasing in time."
The above text in bold is what I am having most difficulty with. What do they mean? Also, how do I color words?
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1
If you imagine your computer screen as a 2D plane, then if you look at the analog clock app your computer has it is rotating around clockwise, then the angular momentum vector points into your computer screen (like out the back). If something rotates the other way (counter-clockwise) there is an arrow pointing at you out of the screen which represents the angular momentum vector, and is orthogonal to the screen. I hope this helps. PS you should quick search the forum and see if you can find this question before asking, it's a pretty typical confusion :) – kηives Nov 21 '12 at 17:53
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I'd recommend you leave that for now and concentrate on the right-hand rule, which is more intuitive and easier to apply. Once you have a feeling for that, go back to the plane and the normal. – Emilio Pisanty Nov 21 '12 at 18:17
## 2 Answers
Say you're looking at the piece of paper on your desk; that is the $xy$ plane. You place a dot in the center of the paper; that's your origin.
Your angular momentum is $\vec{L}=\vec{r}{\times}m\vec{v}$. For this example, $\vec{\omega}$ points in the same direction as your angular momentum, because $\vec{L}=mr^2\vec{\omega}$.
The way I remember the directions for the right-hand rule are as follows:
• Thumb points up (this is $\vec{L}$, the cross product)
• Index finger points forward (this is $\vec{r}$, the left vector being crossed)
• Middle finger points to the left (this is $\vec{v}$, the right vector being crossed)
Let's say the particle is on the paper, directly above the origin, and moving counterclockwise. Then, $\vec{v}$ points to the left edge of the page. The direction of $\vec{r}$ goes from the origin to the point, so it points to the top of the page.
My thumb points up, so this is the direction of $\vec{L}$, and hence the direction of $\vec{\omega}$.
Crane your hand around so that your middle finger points right and your index finger continues to point to the top of the page, and your thumb points down, which is the direction of $\vec{\omega}$ for clockwise motion.
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The least ambiguous way to describe it would be using cartesian vector notation. so we have $\hat x$, $\hat y$ and $\hat z$. If the particle is moving in a circle from the $\hat x$ axis to the $\hat y$ axis, it's moving counter clockwise, like moving from 3 to 12 on an analog clock. We say it's angular velocity is in the $+ \hat z$ direction, or up out of the clock.
If it's moving from the $\hat x$ axis to the $-\hat y$ axis, it's moving clockwise, like moving form 3 to 6 on an analog clock. It's convention to say it's angular velocity is in the $- \hat z$ direction, or down into the clock. These relations fall out of the right hand rule and the cross-product in vector arithmetic, since angular velocity is $\hat r \times \hat V=\hat \omega$.
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I've replaced $x$ by $\times$. – Luboš Motl Nov 21 '12 at 18:48 | {"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": 31, "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}, "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}, "english_confidence": 0.9353974461555481, "perplexity_flag": "head"} |
http://www.physicsforums.com/showthread.php?p=1057320 | Physics Forums
## Prime Number Gaps
Hello everyone,
I'd first like to say that I am uninformed on this subject and that I have a question to the mathematicians on these forums who know about the subject.
In the set of all prime numbers, has the integer gaps between two prime numbers been studied? I mean, do mathematicans know what the largest difference is between two prime numbers?
Im interested in this subject and I would like to know..
Thanks in advance.
--rad
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Recognitions: Homework Help Science Advisor The gap can be arbitrarily large. Just consider n!+2, n!+3,...n!+n. Lots of work has been done, you might want to take a look at Caldwell's prime pages: http://primes.utm.edu/notes/gaps.html
Recognitions: Homework Help Science Advisor Since the primes have measure 0, prime gaps must be unbounded in length. Think about it -- if every k integers had a prime number for some fixed k, then some primes would have common (nontrivial) factors.
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## Prime Number Gaps
i refer you to the work of helmut maier.
Helmut Maier
Primes in short intervals.
Source: Michigan Math. J. 32, iss. 2 (1985), 221
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Quote by CRGreathouse Since the primes have measure 0
the integers have measure zero, but the gaps between consecutive integers is not unbounded.
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Quote by matt grime the integers have measure zero, but the gaps between consecutive integers is not unbounded.
What I mean is that for a set $$X\in\mathbb{N}$$ with
$$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0$$
$$\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}$$ such that there is no $$x\in X$$ with $$m\le x\le m+n$$. (The set of primes is of course such a set by the PNT.) I'm sorry if I was ambiguous.
Recognitions: Homework Help Science Advisor You want: $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0$$ or equivalently here pi(n)/n->0 as n->infinity. In otherwords, the asymptotic density of the primes is zero.
According to Bertrand's postulate, there is at least one prime between n and 2n-2, for any n>3. I wonder if there is a theorem about the number of primes between n and 2n exclusive (see http://www.research.att.com/~njas/sequences/A060715 ), because that number seems to be steadily increasing over a sufficiently large period of the sequence (sorry if this is not precise enough); what I mean is that, for n=5, for instance, the number of primes between n and 2n is 1, but, it seems, for any n > 5 the number of primes between n and 2n is greater than 1; similarly, for n= 8, the number of primes between n and 2n is 2, but for any n>8, the number of primes between n and 2n is greater than 2 (?), and so on. If it is true that for any m >= 1 there is an n for which the number of primes between k and 2k, k>=n, is greater than m (is it?) then there is a limit on prime gaps as well, depending on m (or n), I think (although Bertrand's postulate itself puts a limit on prime gaps, depending on n). What I mean by Bertrand's postulate putting a limit on prime gaps is that for any prime p, there is another prime between p+1 and 2p.
Recognitions: Homework Help Science Advisor pi(2n)-pi(n)~n/log(n) by the prime number theorem, so the number of primes in [n,2n] tends to infinity as n does. The bound Bertrands puts on the gap to the next prime is pretty far from what's known to be true (though correspondingly simpler to prove!). For example, if n is large enough, we can guarantee a prime in [n,n+n^0.525].
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Quote by CRGreathouse What I mean is that for a set $$X\in\mathbb{N}$$ with $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}x=0$$ $$\forall n\in\mathbb{N}\;\;\exists m\in\mathbb{N}$$ such that there is no $$x\in X$$ with $$m\le x\le m+n$$. (The set of primes is of course such a set by the PNT.) I'm sorry if I was ambiguous.
density, not measure.
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Quote by shmoe You want: $$\lim_{n\rightarrow\infty}\frac1n\sum_{x\in X|x\le n}1=0$$ or equivalently here pi(n)/n->0 as n->infinity. In otherwords, the asymptotic density of the primes is zero.
Oops, you're absolutely right. That's what I meant.
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Linear & Abstract Algebra 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": 0, "mathjax_display_tex": 12, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9452594518661499, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/40224/probability-of-a-point-taken-from-a-certain-normal-distribution-will-be-greater | Probability of a point taken from a certain normal distribution will be greater than a point taken from another?
Let's say I have one point that will be taken randomly from a normal distribution with mean $\mu_1$ and standard deviation $\sigma_1$. Let's say I have another point that is taken much in the same way from another normal distribution with mean $\mu_2$ and standard deviation $\sigma_2$;.
How can I compute the probability, given $\mu_1$, $\mu_2$, $\sigma_1$, and $\sigma_2$, that my first point will be larger than the second?
I am sort of interested in the reasoning behind an "analytic" answer (or as analytic as you can possibly get with the normal distribution, which isn't that much), but I am more importantly looking for an algorithm of computing this probability, as it will be used in a simulation/model.
Does anyone know where I could get started on reasoning through this?
Note: For actual computation, having a table of values of the % of the curve within a given multiple of the standard deviation is feasible in my situation.
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– ogerard May 20 '11 at 9:07
@ogerard -- yes, they are independent in the manner you describe. – Justin L. May 20 '11 at 9:44
So my answer applies. – Shai Covo May 20 '11 at 10:17
1 Answer
Suppose that $X_1 \sim {\rm N}(\mu_1,\sigma_1^2)$ and $X_2 \sim {\rm N}(\mu_2,\sigma_2^2)$ are independent. Then, $${\rm P}(X_1 > X_2 ) = {\rm P}(X_1 - X_2 > 0) = 1 - {\rm P}(X_1 - X_2 \le 0).$$ Now, by independence, $X_1 - X_2$ is normally distributed with mean $$\mu := {\rm E}(X_1 - X_2) = \mu_1 - \mu_2$$ and variance $$\sigma^2 := {\rm Var}(X_1 - X_2) = \sigma_1^2 + \sigma_2^2.$$ Hence, $$\frac{{X_1 - X_2 - \mu}}{{\sigma}} \sim {\rm N}(0,1),$$ and so $${\rm P}(X_1 - X_2 \le 0) = {\rm P}\bigg(\frac{{X_1 - X_2 - \mu }}{\sigma } \le \frac{{0 - \mu }}{\sigma }\bigg) = \Phi \Big( \frac{-\mu }{\sigma }\Big),$$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Thus, $${\rm P}(X_1 > X_2 ) = 1 - {\rm P}(X_1 - X_2 \le 0) = 1 - \Phi \Big( \frac{-\mu }{\sigma }\Big).$$
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So, the probelm reduces to calculating the probability that a standard normal variable is less than a given $a$. – Shai Covo May 20 '11 at 9:46
I really like this answer; however, I can't seem to get the results to align with the empirical results I am generating. Perhaps it's an issue with my random number generator? For $\sigma=1$ for both and $\mu_1 = 0$ and $\mu_2 = 1$, I am consistently getting around 23.9% of all simulated selections resulting in the former being higher; however, your method seems to give an expected result of 15.9%; did I use yours wrong? – Justin L. May 20 '11 at 16:51
Nevermind, it turns out I was using the wrong value for $\sigma$ -- $1$ instead of $\sqrt{2}$. Everything works like a charm now :) – Justin L. May 20 '11 at 21:12 | {"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": 20, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9554454684257507, "perplexity_flag": "head"} |
http://math.stackexchange.com/questions/5166/algorithm-to-calculate-the-control-points-of-a-cubic-bezier-curve?answertab=votes | # algorithm to calculate the control points of a cubic Bezier curve
I have all points where my curve pass through, but I need to get the coordinates of the control points to be able to draw the curve. How can I do to calculate this points?
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## 2 Answers
When what you already have is a set of points where your curve must pass through, Bézier is not what you want; you should be using a parametric (cubic) spline.
Assuming that your points $\mathbf{P}_i=(x_i,y_i)$, $i=1\dots n$ represent a general curve (as opposed to a function $f(x)$, for which simply applying the usual cubic spline algorithm on your points suffices), Eugene Lee proposed centripetal parametrization to generate suitable parameter values $t_i$ associated with your $\mathbf{P}_i$. The prescription for generating the $t_i$ (in its most general form) is
where $\left\| \cdot \right\|$ is the (Euclidean) length, $e$ is an adjustable exponent in the interval $[0,1]$, and $t_1=0$. (A usual value for $e$ is 0.5, but $e=1$ is sometimes used as well.)
From this, one applies the usual cubic spline algorithm to the sets $(t_i,x_i)$ and $(t_i,y_i)$, from which you now have your parametric spline. (The periodic spline is recommended for closed curves, and the "not-a-knot" spline for all other cases.)
The MATLAB Spline Toolbox has support for this parametrization scheme, though it shouldn't be too hard to write your own implementation.
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It depends how many points you have.
If you have only 4 points, then you will need only one Bezier cubic segment. Suppose your known points (through which you want the curve to pass) are $Q_0, Q_1, Q_2, Q_3$. First you have to choose 4 parameter values $t_0,t_1,t_2,t_3$ to assign to these points. The centripedal approach described in the other answer is good. In simple situations, where your points are fairly equidistant, a much simpler choice is just $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$. Then you have to solve a system of 4 linear equations, as follows. Suppose $P_0,P_1, P_2, P_3$ are the (unknown) control points of the Bezier curve, and denote the curve by $C(t)$. Then we want $C(t_i) = Q_i$ for $i=0,1,2,3$. This means $$\sum_{j=0}^3{b_j(t_i)P_j} = Q_i$$ for $i=0,1,2,3$, where $b_0, b_1,b_2,b_3$ are the Bernstein polynomials of degree 3. Solve these equations to get $P_0,P_1, P_2, P_3$. If you always use $(t_0,t_1,t_2,t_3) = (0, 1/3, 2/3, 1)$, then the coefficient matrix of the linear system is fixed, and you can just invert it once (exactly), and save the answer. This gives you a simple formula relating the $P$'s and $Q$'s.
If you have more than 4 points, then you can use a Bezier curve with degree higher than 4. The calculations are analogous to those shown above. But a better approach is to use a "spline" that consists of several Bezier cubic segments joined end-to-end. There are many ways to compute splines. One of the easiest (and the most popular amongst graphics folks) is the Catmull-Rom spline. It gives you simple formulae for the control points in terms of the given points $Q_i$.
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http://dsp.stackexchange.com/questions/2654/what-is-the-difference-between-convolution-and-cross-correlation | # What is the difference between convolution and cross-correlation?
I've found on multiple sites that convolution and cross-correlation are similar (including the tag wiki for convolution), but I didn't find anywhere how they differ.
What is the difference between the two? Can you say that autocorrelation is also a kind of a convolution?
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## 2 Answers
The only difference between cross-correlation and convolution is a time reversal on one of the inputs. Discrete convolution and cross-correlation are defined as follows (for real signals; I neglected the conjugates needed when the signals are complex):
$$x[n] * h[n] = \sum_{k=0}^{\infty}h[k] x[n-k]$$
$$corr(x[n],h[n]) = \sum_{k=0}^{\infty}h[k] x[n+k]$$
This implies that you can use fast convolution algorithms like overlap-save to implement cross-correlation efficiently; just time reverse one of the input signals first. Autocorrelation is identical to the above, except $h[n] = x[n]$, so you can view it as related to convolution in the same way.
Edit: Since someone else just asked a duplicate question, I've been inspired to add one more piece of information: if you implement correlation in the frequency domain using a fast convolution algorithm like overlap-save, you can avoid the hassle of time-reversing one of the signals first by instead conjugating one of the signals in the frequency domain. It can be shown that conjugation in the frequency domain is equivalent to reversal in the time domain.
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This answer is fine for real signals, but Jason brought up complex-valued signals, in which case it is important to note that it is not quite the case that the "only difference is .... time reversal ..." Indeed, complex conjugates are needed on one of the two signals in the correlation formula (which one is conjugated is a matter of convention - some say to may to and some say to mah to - but both call a fruit a vegetable). On the other hand, neither signal is conjugated in the convolution formula. – Dilip Sarwate Jun 20 '12 at 2:44
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but what does it mean that they so similar? Using some deep intuitive words! – Diego Dec 14 '12 at 17:20
For continuous convolution $$[Hf](x) \equiv f(x) * h(x) \equiv \int\mathrm{d}x' h(x-x')f(x')$$ and continuous cross-correlation $$[Gf](x) \equiv f(x) \star h(x) \equiv \int \mathrm{d}x'h^*(x'-x)f(x')$$ It's easy to show that the cross-correlation operator $G$ is the adjoint operator of the the convolution operator $H$.
Also, the convolution operation is commutative$$f(x) * h(x) = h(x) * f(x),$$ while cross-correlation does not have such a property.
$\\\\\\\\$
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http://mathhelpforum.com/calculus/200624-finding-area-between-parabolas.html | 3Thanks
• 1 Post By Reckoner
• 1 Post By skeeter
• 1 Post By Plato
# Thread:
1. ## Finding Area between Parabolas
The question is:
A shape has been formed by three intersecting parabolas, the equations of these parabolas are:
1. $y=-x^2-6x-9$
2. $y=x^2-4.5$
3. $y=-x^2+6x-9$
Using calculus, find the area of the shape formed by the intersecting parabolas.
My attempt:
Well, firstly the graph of them:
I'm guessing that the question means to find that area in the middle underneath the red curve. Well, that's the assumption I've made in answering the question. Anyways, firstly is some info that isn't present on the graph that isn't hard to work out. The x-intercepts of the lower parabolas are -3 and 3. The point of intersection that the lower parabolas have with the middle one is at x=-1.5 and x=1.5. The x-intercepts of the red parabola are $\pm\sqrt{4.5}$.
I've worked out the area of that shape to be 3.022 sq. units. Firstly I found the area above the lower parabolas for $x=-3$ to $x=3$. I did this by finding the area from one of the x-intercepts to zero and then just doubling it. This yielded in 18.
Then I found the area bounded by the x-axis for the red parabola (the middle one). The area was 12.728 sq units.
Then I found the area from one of the x-intercepts to the point that the two parabolas touch ( $x=\pm1.5$). This gave a value of 1.125 sq units. So then I worked out the area of the shape as:
$Area=18-12.728-2(1.125)$
$Area=3.022$
Is this approach of finding the area correct? Thanks in advanced.
Attached Thumbnails
2. ## Re: Finding Area between Parabolas
Originally Posted by NewUser01
Is this approach of finding the area correct? Thanks in advanced.
It doesn't have to be that complicated. You can find the area using two integrals. We know the lower parabolas intersect the upper parabola at $\textstyle x=\pm\frac32,$ and the lower parabolas intersect each other at $x=0.$ Thus, the area of the middle region is simply
$A = \int_{-3/2}^0\left[\left(x^2 - \frac92\right) - \left(-x^2-6x-9\right)\right]\,dx + \int_0^{3/2}\left[\left(x^2 - \frac92\right) - \left(-x^2+6x-9\right)\right]\,dx$
And due to symmetry, we have
$A = 2\int_0^{3/2}\left[\left(x^2 - \frac92\right) - \left(-x^2+6x-9\right)\right]\,dx$
$= 2\int_0^{3/2}\left(2x^2 - 6x + \frac92\right)\,dx$
$= 2\left[\frac23x^3 - 3x^2 + \frac92x\right]_0^{3/2}$
$= 2\left(\frac94 - \frac{27}4 + \frac{27}4 - 0\right)$
$= \frac92.$
Your approach should work, but I'm not sure how you determined the area "from one of the x-intercepts to the point that the two parabolas touch" to be 1.125.
3. ## Re: Finding Area between Parabolas
using symmetry ...
$A = 2\int_0^{1.5} (x^2-4.5) - (-x^2+6x-9) \, dx = 4.5$
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4. ## Re: Finding Area between Parabolas
This $2\int_0^{1.5} {[{x^2} - 4.5 + {{\left( {x - 3} \right)}^2]}dx}$ will also give the answer.
5. ## Re: Finding Area between Parabolas
Wow, that approach is much easier. What was I thinking? Thanks for the help everyone.
Question solved. | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9478815793991089, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/201787/recursive-curve-fitting?answertab=votes | # Recursive curve fitting
I have a few points for which I have to obtain a best fit. (I tried to use the least squares curve fitting method as Robert says, however, since it seems to be fitting rather poorly, any other suitable method would also work for me.)
However, the function is recursive and looks like this -
$$f(x+1)= f(x) + k_1(f(x)-k_2)^3$$
where the starting point $f(1)$ is known and fitting needs to be done only for positive integral values of $x$, from $1$ upto some natural no. $n$
I don't understand how would I apply the matrix method of minimizing the norm $\min(\lVert Ax-B\rVert^2)$ since I would have to consider all possible combinations of powers of $k_1$ and $k_2$ which makes my $x$ matrix terribly complex with every iteration.
This suggests that least square curve fitting is probably not the best way of doing this.
I tried to find other questions related to recursive curve fitting here, but I couldn't find any similar situation. Please help me solve this.
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do you observe $f(x)$ for $x=1,2,...,n$? – mpiktas Sep 25 '12 at 10:33
## 1 Answer
The parameters enter nonlinearly into your equation. However, fortunately you can rewrite your equation in such a way that you have linear parameters:
$$(f(x+1) - f(x))^{1/3} = a f(x) - b$$ where $k_1 = a^3$ and $k_2 = b/a$.
So you can apply least squares to these linear equations in $a$ and $b$ for $x$ from $1$ to $n$.
EDIT: I tried some data that seem close to what you had in your picture:
$$y = [237,130,120, 113, 111, 110]$$
Linear least squares for the residuals $(y_{i+1} - y_i)^{1/3} - a y_i + b$, $i = 1 \ldots,5$, in Maple produces $$a = -.0273525754168209,\ b = -1.67458695987193$$ which corresponds to $$k_1 = - 0.0000204641953284227606,\ k_2= 61.2222774036158200$$ I then tried nonlinear least squares for the residuals $y_{i+1} - y_i - k_1 (y_i - k_2)^3$, $i=1 \ldots 5$, using the above as initial values (this is often necessary because the nonlinear least squares algorithms often provide only a local minimum rather than the global minimum). The result was $$k_1 =- 0.0000166852730695581124, k_2 = 51.1767746591434971$$
The recursion using initial value $237$ and parameters from the linear least squares produces the values $$237, 125.855961638565, 120.330463855443, 116.104385196179, 112.721501867265, 109.926405798693$$ while the recursion using parameters from the nonlinear least squares produces $$237, 129.938505804843, 121.786225982822, 115.912389815574, 111.385883230420, 107.744051206183$$
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Thanks Robert for your answer. However, could there be a method for solving this with higher accuracy? When I calculated $k_1$ and $k_2$ using this least square method, they were not as accurate as the solution I already have (and consequently they fit the actual data points rather poorly). I can give you the data points and the solution if you would like. – Guanidene Sep 25 '12 at 9:12
– Guanidene Sep 25 '12 at 12:23
You could try nonlinear least squares rather than linear least squares. In Maple this can be done using LSSolve in the Optimization package. – Robert Israel Sep 25 '12 at 18:43
Thanks Robert for the effort. I might take some time to verify this. I'll get back to you soon. This indeed looks promising. – Guanidene Sep 26 '12 at 9:05 | {"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": 24, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9553151726722717, "perplexity_flag": "head"} |
http://mathhelpforum.com/statistics/30976-can-any-one-help-normal-distribution-dont-have-much-time-left.html | # Thread:
1. ## Can any one help with Normal Distribution. dont have much time left!!
i have a last Q left to do for my maths coursework and it has completely stumped me. can anyone help. here is the whole Q:
A scientist was measuring oak trees in a woodland and noticed that their height was normally distributed. Unfortunately a sudden gust of wind caught her by surprise and most of her results were blown away. All she had left was the knowledge that she measured the height of 200 oak trees and found that 12 trees were less than 8 metres tall and 36 trees were less than 10 metres tall.
Fortunately the scientist knew the properties of normal distribution to work out the mean and standard deviation of the height of the oak trees. Do you?
By sketching a graph of the normal distribution and indicating appropraite areas, determine the mean height of the oak trees and their standard deviation.
If anyone can help quickly that would be great.
Thanks
2. Originally Posted by wicca16
i have a last Q left to do for my maths coursework and it has completely stumped me. can anyone help. here is the whole Q:
A scientist was measuring oak trees in a woodland and noticed that their height was normally distributed. Unfortunately a sudden gust of wind caught her by surprise and most of her results were blown away. All she had left was the knowledge that she measured the height of 200 oak trees and found that 12 trees were less than 8 metres tall and 36 trees were less than 10 metres tall.
Fortunately the scientist knew the properties of normal distribution to work out the mean and standard deviation of the height of the oak trees. Do you?
By sketching a graph of the normal distribution and indicating appropraite areas, determine the mean height of the oak trees and their standard deviation.
If anyone can help quickly that would be great.
Thanks
Pr(X < 8) = 12/200 = 0.06.
Pr(X < 10) = 36/100 = 0.18.
$z_{0.06} = -1.5548$
$z_{0.18} = -0.9154$
(correct to four decimal places).
$Z = \frac{X-\mu}{\sigma}$. Therefore:
$-1.5548 = \frac{8-\mu}{\sigma}$ .... (1)
$-0.9154 = \frac{10-\mu}{\sigma}$ .... (2)
Solve equations (1) and (2) simultaneously.
3. ## help a bit more please!
-0.5 = 8 - mue / s.d.
8 - mue = -0.5
B = 0.14 z= 0.0557
0.0557 = 10 - mue / s.d.
10 - mue = 0.0557
0.0557 x 100 = 5.57
dont know if this is right or not and we are quite stuck on how else to do it. could you help any more?
4. Originally Posted by wicca16
-0.5 = 8 - mue / s.d.
8 - mue = -0.5
B = 0.14 z= 0.0557
0.0557 = 10 - mue / s.d.
10 - mue = 0.0557
0.0557 x 100 = 5.57
dont know if this is right or not and we are quite stuck on how else to do it. could you help any more?
I have no idea what you've done here.
And I have no inclination to explain why it's wrong - all I can see is that you've completely ignored the approach I suggested.
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http://mathoverflow.net/questions/70493/dg-lie-structure-on-hh-and-koszul-duality | ## dg-lie structure on $HH^*$ and Koszul duality
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This is shamelessly close to my other question: http://mathoverflow.net/questions/70151/a-question-on-koszul-duality-and-b-infty-structures-on-hh. Maybe this one will get a better response. Rather than rewrite that one, I am going to ask about a specific aspect of it in more detail. As in that question, it is known that for a space simply connected space M:
`$HH^*(C_*(\Omega(M),\mathbb{Q}),C_*(\Omega(M),\mathbb{Q})) \cong HH^*(C^*(M,\mathbb{Q}), C^*(M,\mathbb{Q}))$`
as Gerstenhaber algebras.
In particular, this implies that `$HH^*(C_*(\Omega(M),\mathbb{Q}),C_*(\Omega(M),\mathbb{Q}))[1] \cong HH^*(C^*(M,\mathbb{Q}), C^*(M,\mathbb{Q}))[1]$` as Lie algebras.
Question:When are the dg-Lie algebra structures on Hochschild cochains: `$HCH^*(C_*(\Omega(M),\mathbb{Q}),C_*(\Omega(M),\mathbb{Q}))[1] \cong HCH^*(C^*(M,\mathbb{Q}), C^*(M,\mathbb{Q}))[1]$` quasi-isomorphic.
As I mentioned in that question: this follows from more general results of Keller in the case M is formal and coformal (i.e. the d.g. algebra `$C^*(M)$` is equivalent to a graded Koszul algebra).
Now suppose that g is a graded finite dimensional Lie algebra and work over $\mathbb{C}$(or $\mathbb{R}$), which corresponds to M be a $\mathbb{C}$ coformal space, with finite dimensional $\mathbb{C}$ homotopy groups. Let `$C^*(g)$` be the Chevalley complex which is a model for `$C^*(M)$`. Here is an approach for proving the result:
Step 1. We know from this MO question http://mathoverflow.net/questions/56145/extension-of-the-formality-theorem that `$HCH^*(C^*(g), C^*(g)) \cong (T_{poly},[v,])$` as $L(\infty)$ algebras. Here v is a vector field which corresponds to the d on `$C^*(g)$`(see that question for a detailed explanation of notation). These notes http://math.univ-lyon1.fr/~calaque/LectureNotes/LectETH.pdf by Damien Calaque are also extremely useful.
Step 2. now `$(T_{poly},[v,])$` is canonically isomorphic as a complex `$C^*(g,Sym(g))$`, that is the Chevalley complex in the Lie algebra module `$Sym(g)$`.
Step 3. By PBW $Sym(g) \cong Ug$ as g modules.
Step 4. Just as in Step 1, we have an isomorphism between `$C^*(g,Ug) \cong HH^*(U(g),U(g))$`. To obtain this we think of Ug as the deformation quantization of $Sym(g)$ given by the Kirillov Poisson structure on $g^*$. Ordinarily, this exists as a formal deformation but just like in Step 1, there is no problem setting the formal parameter t=1. Just as in that question, there is an induced $L(\infty)$ map on tangent cohomology groups that is an iso.
Question: Can these steps be generalized to dg-Lie algebras with finite dimensional homology? Note it follows from the cited question that step one generalizes. A generalization of Step 3 is given here in this paper of Baranovsky http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.1396v1.pdf, but it seems tricky to make this work out with the other steps above.
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## 1 Answer
Hi,
Your Question:When are the dg-Lie algebra structures on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1] quasi-isomorphic ?
this is always true.
Step 1: From my paper with Felix and Thomas, looking at the proof, you can see that dg-Lie algebra structures on Hochschild cochains: `$HCH∗(\Omega C_*(M),\Omega C_*(M))[1]≅HCH∗(C∗(M,Q),C∗(M,Q))[1]$` are quasi-isomorphic Here `$\Omega C_*(M)$` is the Adams Cobar construction on the coalgebra C_*(M).
Step 2: There is an quasi-isomorphism of chains algebras called Adams cobar equivalence `$\Theta:\Omega C_*(M)\rightarrow C∗(Ω(M)$`. In our paper, we prove (very short proof) that this quasi-isomorphism $\Theta$ induces an isomorphism of Gerstenhaber algebras between `$HH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))$` and `$HH∗(\Omega C_*(M),\Omega C_*(M))$`. In particular, we have an isomorphism of graded Lie algebras. You want a dg-Lie algebra isomorphism on Hochschild cochains: HCH∗(C∗(Ω(M),Q),C∗(Ω(M),Q))[1] and HCH∗(\Omega C_(M),\Omega C_(M)). This is true. One of my coauthor had a proof. But it is not in our paper, since I thought it was not interesting and too complicated. But if I remember well, Hamilton and Lazarev proved it in a paper following our paper. I think that Keller proved also in the paper you quote "Derived Invariance of Higher Structures of the Hochschild complex".
ps: There is two versions of my paper with Felix and Thomas, the published squezeed version valid only over a field, and the arxiv longer version with more details.
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Thanks for your answer! I was confused at Step 1 because the map you call $D_C: HCH^∗(C∗(Ω(M)),C∗(Ω(M)) \to HCH^∗(C∗(M),C∗(M))$ is not a map of dg Lie algebras. But looking at your Arxiv version I see that this is not a problem since you have the section you call $\Gamma$ which gives the conclusion. – Daniel Pomerleano Jul 18 2011 at 8:12 | {"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": 13, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8992086052894592, "perplexity_flag": "middle"} |
http://crypto.stackexchange.com/questions/2153/unit-testing-a-library-rng-and-its-seeding-for-practical-security/2203 | # Unit testing a library RNG and its seeding for practical security
Suppose I have a cryptography library that provides me with a black-box random number generator that is supposed to be suitable for long-lived key generation. (Under the hood it may be assumed to use some combination of `/dev/random` or equivalent, and stretching with a CSPRNG.)
I would like to write a unit test that can detect implementation errors that would result in practical failures of security. For instance, it should reliably detect the infamous CVE-2008-0166 (Debian OpenSSL weak key generation) if applied to a version of OpenSSL with that bug. Note that statistical tests as discussed e.g. in this earlier question will not do the job by themselves, because of the CSPRNG in between me and the guts of the black box (the CSPRNG itself has already been checked by use of test vectors).
How might this be accomplished?
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1
I'm not a Unix specialist, but isn't `/dev/random` itself a black-box random number generator that is supposed to be suitable for long-lived key generation, without any extra library? Is the library supposed to compensate for some weakness of some implementations of `/dev/random`, which one, and using what principle? Or is the objective of the library simply to limit the entropy gathered from `/dev/random`, for performance reasons? – fgrieu Mar 21 '12 at 8:08
My impression is that the main goal of the wrapper is to hide portability differences among various popular OS's secure-RNG APIs. Limiting the amount of entropy consumed by any given process may also have been a goal, I'm not sure. – Zack Mar 21 '12 at 14:49
Might be slightly off-topic in this question but I read somewhere that the output of `/dev/random` isn't cryptographically random (i.e. not indistinguishable from a true RNG). – recluze Mar 27 '12 at 3:22
1
@recluze Whoever told you that was wrong. All the OSes that have that run the output through a CSPRNG before revealing it, so it's no worse than a CSPRNG even when the entropy pool is totally exhausted. (If there's an OS that happens to use a broken CSPRNG, I'm not aware of it.) – Zack Mar 27 '12 at 3:41
@Zack Yes, after a bit of more study, I agree with you. Thanks for the clarification. – recluze Mar 28 '12 at 4:37
## 2 Answers
You're not going to like this answer, but strictly speaking, you can't.
In a number of specific cases you can do what you want. For example, in the specific case of the Debian PRNG debacle, you could detect that (and as a matter of fact, bad keys generated by this PRNG failure were tracked down). But that's because the RNG was not seeded much at all and ended up with some small number of possible states. (I'm saying RNG not CSPRNG because I'm not considering any cryptographically insecure pseudo-random number generators.)
But if the output stage of your PRNG comes from a suitable pseudo-random function, it will pass all your tests. AES in counter mode, an iterated hash function, and so on will all be just fine -- these outputs are themselves pseudo-random functions and therefore pass all your pseudo-random tests. It will be cryptographically secure from a mathematical reference point. If it is insecure, it will be insecure because of an operational failure, like the Debian problem, which was operational, not mathematical.
You are only going to find practical failures of a PRNG by white-box testing. You will have to have a reasonably smart crypto person to code-review it. You will have to look inside it to see if it is constructed correctly.
Consider this as a badly-built PRNG: imagine a chip with a PRNG that is seeded with a completely random 64-bit number to which is added the chip's serial number, and then a high-quality output stage (Yarrow, SP 800-90 DRBG, etc.) generates random bits from that seed from then on.
In other words, for the entire family of chips, there's a single 64-bit chunk of entropy E, and each chip has put into it E+1, E+2, ... E+N for the N chips we manufacture. Call this number E_n.
We'll presume that all re-seeding is broken and has no effect. You very likely won't be able to tell this from testing the output only.
You might in some cases -- for example, let's suppose that the output is a naïve iterated hash. You might be able to detect that the substring R[i..i+31] is equal to SHA256(R[i-32..i-1]). But be lucky to detect that, because you'd have to explicitly check for it.
Similarly, testing its output can tell you if the output function isn't a PRF, but not a lot more than that, and these days the errors aren't going to be because of the output function. Even dummies are going to get that right; look at the Debian debacle again. The problem was seeding, not pseudo-randomness.
You wouldn't find it if the bogus chip hashed in E_n along with each iterated hash, or if there was an HMAC. You wouldn't find it if there was a simple counter hashed in, either. You almost certainly wouldn't detect anything amiss if the PRNG used a cipher in counter mode with E_n as the key, either. (There are plenty of edge cases where you would -- for example, if you used DES in counter mode and happened to pick a weak key, you might detect that. Bear with me for the general case here.)
This completely bogus random number chip has only 64 bits of good randomness in the entire chip production, and yet no black box testing is going to detect that. You're only going to find the problem by auditing its design and implementation, not by testing its output.
I told you you weren't going to like my answer, but there it is. Black box testing will only detect a small class of stupid errors, and none of the evil ones.
Jon
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The CSPRNG output should be indistinguishable from a random distribution over $\{0,1\}^n$, where $n$ is the output length. The CSPRNG is responsible for ensuring that this occurs given its input domain of $\{0,1\}^m$, where $m$ is the length of the CSPRNG input. (Note that $n$ and $m$ need not be fixed for a given CSPRNG.)
The individual strings output by a CSPRNG should not be "reversible", aka, leak information about the input. The summary there is that, regardless of the input domain, the output domain should still "look random". This means that statistical tests on the individual output strings or on groups of output strings are useless, as you noted. The CSPRNG should always pass all of them, so long as it is executing properly.
If the CSPRNG output is detectably weak (assuming that obtaining the original RNG input itself is excluded) then it should violate one of the above assumptions (some of them were subtle). Here are the key ones:
• The CSPRNG is implemented correctly. - Here, KAT unit tests could ensure the implementation is valid. Seemingly innocent changes can sometimes cause other code to fail, so this is worth testing. (This can also help verify the sub-assumption that the CSPRNG is in fact deterministic.)
• The CSPRNG input domain is well-supplied. - Failure here limits the output domain. Unfortunately, the best way to test this, without access to the input, is to sample the outputs you generate and check for statistically unlikely collisions. With an output length of 32 bits, you should expect to see the first collision in about $2^{16}$ outputs. With an output length of 128 bits, you should be able to generate $2^{32}$ outputs with negligible collision. Significantly more collisions in a sample than expected suggests a limited input/output domain. How many samples you collect depends on how many time/space resources you can let the unit test use. You can toy around with what output lengths and sample sizes you want to use, so long as you can calculate what sort of collisions to expect.
• The CSPRNG output is properly used. - Not necessarily beyond the scope of your unit tests. Make sure that whatever APIs call into the CSPRNG know how to do it right. Ie, don't just write and test the CSPRNG, write and test the wrappers that actually generate the keys.
• The CSPRNG implementation does not leak internal information. - Obviously, we assume that the CSPRNG does not give away the information we feed into it. While seemingly mundane, it is a valid concern. I'm not sure how to properly encapsulate this in a unit test, but you might want to check for memory leaks or memory that is not overwritten after use. Simply leaving critical key-related data in memory is a valid vulnerability.
The second point is the most likely one, IMO, to go wrong in practice. While you can take a statistical sampling approach for the CSPRNG, this check is best suited to be put before the CSPRNG, rather than after. If possible, I would put the unit tests for that on the input to the CSPRNG, rather than the output. That is not only easier, but gives you the option of testing a lot more than just collisions.
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http://math.stackexchange.com/questions/146615/how-many-matrices-generate-2d-lie-algebra-i-e-c-1-c-2-k-1-c-1-k-2-c-2?answertab=active | # “How many” matrices generate 2d Lie algebra (i.e. $[c_1,c_2]=k_1 c_1 +k_2 c_2$)?
Consider a pair of matrices $(c_1, c_2)$. The words "it generates the 2-dimensional Lie algebra", means that there exists a pair of scalars $k_1$, $k_2$, such that $$[c_1, c_2] = k_1 c_1 + k_2 c_2,$$ where $[a,b]$ is the "commutator" $ab-ba$.
Not any pair generates 2-d Lie algebra.
Question: what is the dimension of the subset of matrices $(c_1,c_2)$ which generate 2-d Lie algebra? at least for $2\times 2$ matrices ?
It is clearly greater than $n^2+n$ since I can take - $c_1$ - arbitrary (so $n^2$) and $c_2$ - commuting with $c_1$, which gives ($+n$ for generic matrices). But in this way I get only $[c_1, c_2] =0$.
The motivation for the question comes from this question on MathOverflow.
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## 1 Answer
It seems the dimension is n^2+n. The subtlety is that this manifold is highly reducible, there are about n^2+1 components, and n^2 of them of small dimensions, the only one component of dimension n^2+n is [A,B]=0.
Details.
Consider [A,B]=aA+bB. Take A - diagonal matrix with different values on the diagonal. Consider a, b non-zeros.
Remark 1. [A, * ] - has always a trivivial diagonal.
Corollary diag(B) = - aA/b.
Remark 2. [A, *] on the space of matrices with trivial diagonal has eigen matrices E_{ij} with eigenvalues (a_i - a_j).
Corollary off-diag of B is equal to kE_{ij} for some i,j and b=(a_i-a_j).
Hence if we fix A - diagonal, and a,b (non-zeros), then matrix B is determined uniquely as B= -aA/b + E_{ij}/(a_i-a_j)
Hence the dimension of the of the variety for fixed a,b which are non-zero is n^2.
While for a,b=0 we have n^2+n .
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http://physics.stackexchange.com/questions/tagged/energy-conservation?sort=faq&pagesize=15 | Tagged Questions
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I'm working through a problem involving energy conservation. Unfortunately, I cannot calculate the work done by a spring. ... | {"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": 9, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9292864799499512, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/44774/difference-between-maximal-element-and-greatest-element?answertab=votes | difference between maximal element and greatest element
I know that it's very elementary question but I still don't fully understand difference between maximal element and greatest element. If it's possible, please explain to me this difference with some examples etc.
I tried to explain this difference to myself using only definition, but maximal element and greatest element still seems almost the same for me.
Thank you.
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1
You have asked several questions, all of them received several answers. If there is an answer which satisfies you please accept it by using the half-transparent check mark nearby the voting, if there is no satisfactory answer please edit the question to reflect what is missing from the current answers. (Note, you can accept only one answer per question) – Asaf Karagila Jun 19 '11 at 14:37
@Asaf Karagil: To be honest, I didn't know, that it's necessary to mark one answer. I mean - I had several answers which helped me and I didn't want to mark only one answer and I decided to be neutral, not marking any post. – exTyn Jun 20 '11 at 18:52
7 Answers
You are maximal when there is nobody above you.
You are greatest when you are above everyone else.
• If nobody has eaten you, it doesn't follow that you have eaten everyone else.
• If nobody is standing on your head, it doesn't follow that you are standing on everyone else's head.
• If you live on the top floor of your apartment building, it doesn't follow that you live above everyone else in the city.
So maximal elements need not be greatest.
(Incidentally, although a greatest element is always a unique maximal element, it is a fun exercise to come up with a partial orde having a unique maximal element, which is not greatest.)
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I gave an example in my answer for such partial order. – Asaf Karagila Jun 11 '11 at 15:01
Ah, yes indeed you did. – JDH Jun 11 '11 at 15:22
2
Your apartment illustration is very nice! – wildildildlife Jun 11 '11 at 16:33
Suppose $\langle A,R\rangle$ is a partially ordered set (i.e. $A$ is non-empty, $R$ is a partial order relation on $A$).
An element $a\in A$ is called maximal if $\forall b\in A(aRb\rightarrow b=a)$. That is, there is no one "above" $a$ (except perhaps $a$ itself).
An element $a\in A$ is called maximum or greatest if $\forall b\in A(bRa\lor b=a)$, that $a$ stands "above" everyone in $A$ in the relation $R$.
Note that both these definitions hold whether or not you require $R$ to be reflexive.
From this follows that a greatest element is by definition maximal, but not vice versa.
Consider the following case $A=\{a,b\}$ and $R$ is defined to be the identity relation, then both $a$ and $b$ are maximal, but neither is the greatest.
Another strong example with a single maximal element, but no greatest element is this: consider the set $\{a\}\cup\mathbb Z$, and the relation $R$ defined to be $<$ for integers and $aRa$ otherwise (i.e. $a$ does not stand with the integers in this relation). In this case $a$ is trivially a maximal element, no one is "above" it, however there is no maximum element, since no one stands over both $a$ and all the integers.
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2
I think your last "maximal" should be "greatest", if I'm understanding correctly. – Jason DeVito Jun 11 '11 at 15:56
@Jason: Oh, right. Sorry :-) – Asaf Karagila Jun 11 '11 at 16:49
The crucial issue is the difference between a partial order and a linear order.
A partial order is a set where sometimes you can say "this thing is bigger than that", but some things are just incomparable. In a linear order, you can always say "this thing is bigger than that."
For example, if you're working with the natural numbers (except $0$), $\{1,2,3,...\}$, and you say "I'm going to declare that $n \leq m$ iff $n|m$. Then, reflexivity is clear ($n|n$), antisymmetry is clear (if $n|m$ and $m|n$, then $n = m$. And transitivity is also easy to check ($n|m$ and $m|p$ implies $n|p$). So, this really is a partial order.
Now, in this notion of order, is $2\leq 3$? No, since $2$ doesn't divide evenly into 3. Well, is $3\leq 2$? Again, no, since $3$ doesn't divide evenly into $2$. The conclusion is that we simply have no way of comparing the sizes of $2$ and $3$ in this order.
Now, to answer your actual question, as Stefan has noted, in a linear order (where any two elements are comparable) the two notions coincide.
In a partial order, we can see the difference. An element is maximal if it's bigger than everything it can be compared to, but we're not claiming it can be compared to everything.
An element is greatest if it's bigger than everything it can be compared to, and we can compare it with everything.
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Good answer. Along with Stefan's, the only clear ones on the page, as I see it! Once one realises that the poset concept isn't nearly so intuitive as a totally-ordered set, one can separate these concepts much more readily. – Noldorin Sep 9 '12 at 0:25
In a totally ordered set those concepts are the same. So we need to consider partially ordered sets that are not totally ordered in order to find an example. The most common sets of that kind are sets of sets ordered by inclusion. Consider the set $\{\{0\},\{1\}\}$. It contains two elements, none of which is a subset of the other. Both elements are maximal, but the set has no greatest element.
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Maybe I would not take the degenerate case as an example :-) – Listing Jun 11 '11 at 16:40
Take an ordered set $(P,\leq)$. An element $g\in P$ is called greatest if for every $p\in P$ we have $p\leq a$. An element $m\in P$ is called maximal if there exists no $q\in P$ such that $m<q$. If $P$ is totally ordered then the two notions coincide. But in partially ordered sets the two notions are different. Here's an example:
Take all humans that have ever lived on earth (including those still alive) and order them as following $A<B$ if $B$ is a descendant of $A$ (this in fact is an ordering). Observe now that every person $P$ that doesn't have children yet is a maximal element of the aforementioned order, since there doesn't exist any human that is descendant of $P$. On the other hand this order doesn't have a greatest element. Such a person would need to be a descendant of every human that has ever lived, which is impossible.
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Consider the set ${2,3,4,6}$ under divisibility. This set has 2 maximal elements since $2|4$, $2|6$ and $3|6$ where we write $a|b$ to mean means $b$ is divisible by $a$. Since we do not have $3|4$ or $4|6$, both $4$ and $6$ act as Maximal elements, but it has no greatest element.
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Suppose you have a set of boxes. A box that does not fit into any other box is maximal. A box into which any other box fits is greatest.
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http://mathhelpforum.com/math-challenge-problems/127365-integral-over-head-light-contour-via-residue-theorem.html | # Thread:
1. ## Integral over head-light contour via Residue Theorem
This is a challenge question that I already know how to solve: Let $f(z)=\sqrt{1-z^2}$ and consider a smooth (differentiable) map $f(H(t))$ where $H$ is the general shape shown in the first plot below (loops around the branch-points,then looping around to close the contour). That is, the contour traverses an analytic path over multiple branches of the integrand. Show via the Residue Theorem:
$\mathop\oint\limits_{H(t)} f(z)dz=0$
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http://math.stackexchange.com/questions/162790/levy-measure-and-compensation | # Lévy measure and compensation
In the Lévy-Itô decomposition it's necessary to compensate small jumps. That's clear. The small jumps are perhaps non-summable. But why are the jumps quared summalbe?
In the "ordinary" proofs of the Lévy Khintchine Formula or the Lévy-Itô Decomposition i can't get the point where we exactly use THIS and where the proof fails if we would not use this. I think there is a connection in both proofs, because the integrand in the L-K formula is near the origin something to the power of 2.
Are there basic results where it's proven that the x^2 integrated with respect to a Lévy measure is finite without using the Lévy-Khintchine Formula or the L-I decomposition?
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I'm not sure what last part of the question means, but there are martingale ways of proving that the jumps must be square summable. If you only have small jumps, you can localize to get an $\mathbb L ^2$ martingale, and large jumps can be put in the bounded variation term, in which case you start with an $\mathbb L^2$ semi martingale with bounded jumps. – mike Jun 25 '12 at 12:07
But using this martingale ways it turns out that, substracting the big jumps, the rest is a $\mathbb{L}^p$ martingale for every $p \geq 1$. So also for $p=1$ and hence it should be also simple summable, but this don't work... – kuemmelsche Jun 25 '12 at 12:52
I should add that it's not alone needed to substract the big jumps $W_t=X_t-\sum_{s\leq t}\Delta X_t \mathbb{I}_{\left\{ \left|\Delta X_t \right| \geq 1 \right\}}$ but also compensate by $W_t - \mathbb{E}W_t$ to get a $\mathbb{L}^p$ martingale. But why this implys that $X$ has summable squared jumps? – kuemmelsche Jun 25 '12 at 13:04
I had in mind looking at the quadratic variation, which should have a component which is the sum of the squares of the jumps, but maybe it is not as simple as I thought. – mike Jun 26 '12 at 19:03
Oh okay, while looking at the quadratic variation there also appers the sum of small jumps. But how i can proof that the quadratic variation process of a Lévy process is always finite? – kuemmelsche Jun 28 '12 at 9:15
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http://math.stackexchange.com/questions/293766/beginners-in-calculus/293772 | beginners in calculus
i am using the Stewart calculus early transcendentals text and in chapter $2.4$there is a question:
use the given graph of f to find the number delta such that
if $0<|x-5|< \delta$ then $|f(x) -3|< 0.6$
Is the answer: right delta = 0.7 and left delta = 1.0? If so pick the smallest ,correct.....
Any help would be appreciated.
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You just use the provided graph: Draw a vertical strip, vertically centered at $y=3$, of height $1.2$. Now draw a horizontal strip centered about $x=5$. What width would this strip have to have in order for the graph of $f$, excluding the point $(5,f(5))$ possibly, to be contained in the rectangle formed by the vertical and horizontal strips? Find the maximum such value using the provided graph. $\delta$ will be half of the width found. – David Mitra Feb 3 at 17:25
If they give you $f(x)$ by a graph, presumably you should estimate the points where the condition just starts to fail by eye... – vonbrand Feb 3 at 17:25
2 Answers
How far can you get away from $x=5$ before the function's value exceeds 3.6 or is below 2.4?
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so right delta = 0.7 and left delta = 1.0? – codenamejupiterx Feb 3 at 21:50
I can't say without seeing the graph, but, you are correct (as you stated in your edit to the question) that you should pick the least of the two. – Jonathan Feb 4 at 0:36
Thank You for your help! – codenamejupiterx Feb 4 at 6:58
Maybe this will help ?
You want something in the nature of :
5 - delta < x < 5 + delta
and 3-0.6 = 2.4 < f(x) < 3 + 0.6 = 3.6
If they gave you the graph for f(x), I think that implies you can estimate the bounds (thus lead to the finding of delta) by some eye-balling (Since you already know the lower & upper bound for f(x), using the 2nd condition that I listed above). Using the strategy as in David's comment also works.
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so right delta = 0.7 and left delta = 1.0? – codenamejupiterx Feb 3 at 21:58 | {"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": 11, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8981708884239197, "perplexity_flag": "middle"} |
http://mathforum.org/mathimages/index.php?title=Law_of_cosines&diff=17460&oldid=17456 | # Law of cosines
### From Math Images
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| | Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure <math>B</math>, | | Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure <math>B</math>, |
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| | <math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> | | <math> b^{2} = a^{2} + c^{2} - 2ab \cos B </math> |
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| | | + | Substituting in the appropriate side lengths gives us |
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| | | + | <math> (6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B </math> |
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| | | + | Simplify for |
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| | | + | <math> 36 (2) = 36 + 36 - 72 \cos B </math> |
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| | | + | <math>72 = 72 - 72 \cos B </math> |
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| | | + | Subtracting <math>72</math> from both sides gives us |
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| | | + | <math>0 = - 72 \cos B </math> |
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| | | + | Dividing both sides by <math>-72</math> gives us |
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| | | + | <math>0 = \cos B </math> |
## Revision as of 11:21, 30 May 2011
The law of cosines is a formula that helps in triangulation when two or three side lengths of a triangle are known. The formula relates all three side lengths of a triangle to the cosine of a particular angle.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
When to use it: SAS, SSS.
## Proof
Let $\vartriangle ABC$ be oriented so that $C$ is at the origin, and $B$ is at the point$(a,0)$.
### Distance Formula
$distance = \sqrt {(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$
$c$ is the distance from $A$ to $B$.
Substituting the appropriate points into the distance formula gives us
$c = \sqrt {(a-b \cos C)^{2} + (0-b \sin C)^{2}}$
Squaring the inner terms, we have
$c = \sqrt {(a^{2}-2ab \cos C+b^{2} \cos^{2} C) + (b^{2} \sin^{2} C)}$
Since $\cos^{2} C + \sin^{2} C = 1$,
$c = \sqrt {(a^{2}+b^{2}-2ab \cos C+b^{2}}$
Square both sides for
$c^{2} = (a^{2}+b^{2}-2ab \cos C+b^{2}$
## Example Triangulation
Complete the triangle using the law of cosines.
$c^{2} = a^{2} + b^{2} - 2ab \cos C$
### Solution
To find the side length $c$,
$c^{2} = 6^{2} + (6 \sqrt{2})^{2} -2 (6) (6 \sqrt{2}) \cos 45^\circ$
Simplify for
$c^{2} =36 + 36 (2) - 72 \sqrt{2}) \cos 45^\circ$
Since $\cos 45^\circ = \frac{1}{\sqrt{2}}$, substitution gives us
$c^{2} =36 + 36 (2) - 72 \sqrt{2} (\frac{1}{\sqrt{2}})$
Simplify for
$c^{2} =36 + 72 - 72$
$c^{2} =36$
Taking the square root of both sides gives us
$c =6$
Now we can orient the triangle differently to get get a new version of the law of cosines so we can find angle measure $B$,
$b^{2} = a^{2} + c^{2} - 2ab \cos B$
Substituting in the appropriate side lengths gives us
$(6 \sqrt{2})^{2} = 6^{2} + 6^{2} - 2(6)(6) \cos B$
Simplify for
$36 (2) = 36 + 36 - 72 \cos B$
$72 = 72 - 72 \cos B$
Subtracting $72$ from both sides gives us
$0 = - 72 \cos B$
Dividing both sides by $-72$ gives us
$0 = \cos B$ | {"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": 32, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7701507806777954, "perplexity_flag": "middle"} |
http://math.stackexchange.com/questions/302444/groups-of-symmetries-and-presentation-of-semi-direct-products | # Groups of Symmetries and presentation of Semi-direct products
Let $L$ the tessellation of the euclidean plane by equilateral triangles. Prove that the group $G = Sym(L)$ of isometries of $\mathbb{R}^2$ leading $L$ in $L$ is is a semi-direct product $\mathbb{Z} \times \mathbb{Z} \rtimes D_6$, and obtain a presentation of this group.
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1
What have you tried? – Tobias Kildetoft Feb 13 at 20:54
$D_6$ denote the dihedral group with six elements – Agenor Andrade Feb 13 at 20:55
Do you know how to characterize the group of all isometries of the Euclidean plane as a semi-direct product? – Thomas Andrews Feb 13 at 21:13
I'm not sure you are correct about which $D_6$ you want - it seems like it ought to be the group of symmetries of a regular hexagon, which means $D_6$ has $12$ elements. Maybe I'm missing something. – Thomas Andrews Feb 13 at 21:19
Thomas Andrews I do not know how characterize the group of all isometries of the Euclidean plane as semi-direct product. And the notation for $D_6$ is the dihedral group with six elements ($D_{2.3} = D_6$). Thanks. – Agenor Andrade Feb 13 at 21:29
show 1 more comment
## 1 Answer
Hint/outline: Consider two subsets of the group of $\mathrm{Sym}(L)$, $T$ being the isometries that are pure translations, and $R$ the isometries that fix some $0\in L$.
Then you need to prove that $T$ and $R$ are subgroups of $\mathrm{Sym}(L)$, with $T\cong \mathbb Z\times \mathbb Z$ and $R\cong D_{12}$. (Here, I'm using $D_{12}$ as group of isometries on a hexagon, so it has $12$ elements.)
Then you need to prove every element of $\mathrm{Sym}(L)$ can be written as an isometry fixing $0$ follwed by a translation. (That is, $\mathrm{Sym}(L)=TR$ or $RT$, depending on whether you write the group action on the left or right.)
Then you need to show that $T$ is a normal subgroup. The previous step means that you only need to show that $rtr^{-1}\in T$ when $r\in R$ and $t\in T$. That's the hard trick, and it is easier to do for two generators of $R$.
All together, this would prove that $\mathrm{Sym}(L)\cong (\mathbb Z\times\mathbb Z)\rtimes D_{12}$.
Not sure if this can be used to prove that it isn't $\mathbb Z\times \mathbb Z\rtimes D_6$ - it might be possible to factor the group differently.
The "obvious" way to include $D_6$ in $\mathrm{Sym}(L)$ is to consider some atomic triangle, $a,b,c\in L$, and consider the isometries that send $\{a,b,c\}$ to itself.
However, then the other subgroup has to be something more complex than translations, since we need to be able to send $\{a,b,c\}$ to any triangle in $\mathrm{L}$, which can't happen since translation can send a triangle to only triangles oriented similarly.
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I don't see how it could be isomorphic to $({\mathbb Z} \times {\mathbb Z}) \rtimes D_6$, because no group with that structure could contain an element of order 6. – Derek Holt Feb 14 at 8:26
Yeah, that was my feeling, too, @DerekHolt, but when I tried to prove that, I hit stumbling blocks. – Thomas Andrews Feb 14 at 13:08
Where is the stumbling block? – Derek Holt Feb 14 at 15:01
Ah, I was just being brain-dead. Yeah, it is fairly easy to show. – Thomas Andrews Feb 14 at 15:20
Forgive my ignorance gentlemen, @ThomasAndrews but I can not prove that $Sym(L)$ can be written to an isometry fixing 0 follwed by a translation. Nor can show that $T$ is $\mathbb{Z} \times \mathbb{Z}$ and $R$ is $D_{12}$ .. Would greatly help if you could show me more this way. thank you – Agenor Andrade Feb 14 at 19:58
show 3 more comments | {"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": 46, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9472872614860535, "perplexity_flag": "head"} |
http://mathoverflow.net/questions/88233?sort=votes | ## how many consecutive integers $x$ can make $ax^2+bx+c$ square ?
### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
The following problem was raised in a Mathlinks thread:
If $a,b,c\in\mathbb Z$ such that $a\ne0$ and $b^2-4ac\ne 0$, for how many consecutive integers $x$ can $ax^2+bx+c$ ba a perfect square ?
The polynomial $-15x^2+64$ is obviously a square for the five numbers $x=-2,...,2$, but the method used for finding this in the above thread cannot be extended further. Should $5$ really be the best possible answer?
Has this problem been treated somewhere else?
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For starters $15w^2-14$ is a square for $w=\pm1,\pm3,\pm5$ so you get six by taking $w=2x+1$. But eventually you get to surfaces of general type. See my answer to mathoverflow.net/questions/73346 . – Noam D. Elkies Feb 11 2012 at 23:01
4
...and if I did it right, you can get to $w = \pm 7$ using points on an elliptic curve of (conductor 360 and) rank 1. That might be the maximum. FWIW $0,\pm1,\pm2,\pm3$ seems to give a curve of conductor 30 with 12 torsion points but rank zero. – Noam D. Elkies Feb 11 2012 at 23:12
## 1 Answer
To resonate with Noam Elkies' comments, it is conjectured that $8$ squares is the maximum for arbitrary $a$, and $4$ squares is the maximum for $a=1$. For $5$ symmetric squares the smallest known leading coefficients are $a=15$ and $a=-20$, while for $5$ increasing squares they are $a=60$ and $a=-56$. It is known that there are infinitely many examples with $5$ or $8$ symmetric squares, or with $6$ increasing squares. It is also known that there is no symmetric sequence of $7$ squares, and only finitely many of $10$ squares up to obvious equivalences (this one follows from Falting's theorem applied to a specific hyperelliptic curve).
Good starters are:
Browkin-Brzeziński: On sequences of squares with constant second differences, Canad. Math. Bull. 49 (2006), 481–491.
Bremner: On square values of quadratics, Acta Arith. 108 (2003), 95–111.
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4
A recent paper of Gonzales-Jimenez and Xarles [Acta Arith. 149 (2011)] (apologies for the lack of accents) gets a (sharp) upper bound of $8$ for those quadratics with an axis of symmetry half-way between two integers. – Mike Bennett Feb 12 2012 at 4:42
@Mike: That's great, thank you! – GH Feb 12 2012 at 7:19 | {"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": 31, "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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9268357753753662, "perplexity_flag": "middle"} |
http://www.physicsforums.com/showthread.php?p=3588462 | Physics Forums
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## The Lorentz Transformations and a Few Concerns
The derivation of the Lorentz transformations is based on the homogeneity[of space and time] and the isotropy of space.
Could one derive the same transformations wrt space which is not homogeneous or[not] isotropic?
You may consider a few chunks of dielectric strewn here and there. I am assuming for the sake of simplicity that they are at rest in some inertial frame. Such a distribution is not possible without introducing gravitational effects. Running of clocks is affected by gravity.Does the anisotropy of space itself have any effect on them[running of clocks]?
[Incidentally isotropy of space is connected with clocks in the derivation of the Lorentz transformation.Clocks placed symmetrically wrt the x-axis[and lying on the y-z plane as example] should record the same time. Otherwise isotropy of spece gets violated.This idea is commonly used in the derivations.You may consider the one given in "Introduction to Relativity" by Robert Resnick ]
Again the Lorentz transformations are embedded in [present in] Maxwell's equations. But they are the vacuum equations---homogeneity of space[and time] and isotropy are in due consideration.
The Lorentz Transformations are of course correct--only in the context of the homogeneity[of space and time] and isotropy of space. They are extremely useful, like frctionless planes we studied in our childhood days.Frictionless planes helped us in understanding mechanics--but it is extremely difficult to realize them in practice.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Blog Entries: 1 Recognitions: Science Advisor Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.
Quote by Bill_K Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.
Could you provide me with a sample derivation of the Lorentz Transformations in the "LOCAL CONTEXT"?
Hope you don't have to go back to Resnick!
## The Lorentz Transformations and a Few Concerns
A Basic Issue to be addressed:
Can the anisotropy of space itself affect clock rates[considered apart from gravity]?
[Even if you consider an infinitesimally small region of space[surrounded by matter] you can have milloins of directions emanating from a point in it,providing enough scope for anisotropy]
Quote by Bill_K Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime.
These are the priceless words of the century--from Bill_K
Mentor One way to find the Lorentz transformations is to consider a family of global coordinate systems $\{x_i:\mathbb R^4\rightarrow\mathbb R^4|i\in I\}$ such that each "coordinate change" function $x_i\circ x_j^{-1}$ takes straight lines to straight lines, and the set $\big\{x_i\circ x_j^{-1}|i,j\in I\big\}$ is a group, with a subgroup that's isomorphic to the translation group, and another that's isomorphic to the rotation group. To drop the requirement of isotropy or homogeneity would be to drop the requirement that the group of coordinate change functions must have the appropriate subgroups. I don't know what we get if you do. I suppose that we don't want to completely drop them, but rather weaken them somewhat, so that the group almost has the translation and rotation groups as subgroups. I don't even know how to make that statement precise.
Mentor
Quote by Bill_K Regardless of what Resnick may or may not have said, the Lorentz transformations in no way depend on the isotropy and homogeneity of spacetime. Nor do they depend on the presence or absence of material objects or EM or gravitational fields. Lorentz invariance is a local property of vacuum. All the equations of physics are local equations and rely on Lorentz invariance at every point for their consistency.
Lorentz transformations are permutations of $\mathbb R^4$ that satisfy a few additional conditions. The presence of matter in the actual universe obviously has no effect on $\mathbb R^4$, and therefore no effect on the Lorentz transformations. But you certainly do have to make assumptions of homogeneity and isotropy to be able to "derive" them from Einstein's postulates. (You can't actually derive them from Einstein's postulates. You derive them from mathematical statements that can be thought of as expressing aspects of Einstein's postulates mathematically. Are homogeneity and isotropy such aspects, or are they separate assumptions? I think that's actually a matter of taste. Einstein's postulates aren't very precise, so you can interpret them in more than one way).
Blog Entries: 47 Recognitions: Gold Member Homework Help Science Advisor In Special Relativity, the Lorentz Transformations appear in two places... in spacetime (transforming events or coordinates of events) and in the tangent-spaces associated with each event (transforming tensors or components of tensors).
Blog Entries: 4 Recognitions: Gold Member Some questions (1) What is your concept of the universe? (a) Space-time curvature embedded in an overall flat space-time, or (b) tiny little differential sections of flat space-time that are approximately flat, embedded in an overall curved universe?(2) Where does Lorentz Transformations apply? (a) Throughout any flat region of spacetime? (b) Only along a differential path element of space-time.(3) Which is the Lorentz Transformation more similar to?(a) A transformation from spherical to rectangular coordinates: $$\begin{align*} x &= r \cos(\phi) \sin(\theta)\\ y &= r \sin(\phi)\sin(\theta)\\ z &= r \cos(\theta) \end{align*}$$ (b) A rotation $$\begin{align*} x' &= x \cos(\theta)+y \sin(\theta) \\ y' &= -x \sin(\theta)+y \cos(\theta) \\ z' &= z \end{align*}$$ (4) What is the radial limitation on a rotation transformation?(a) Universal: If I turn my reference frame to the right, then even galaxies billions of light years away will move counterclockwise around me. (b) Local: Objects in my room will move counterclockwise around me, but galaxies billions of light years away will maintain their positions.
Blog Entries: 4 Recognitions: Gold Member I'd like to simplify question #3 to two dimensions, and delve into it more deeply. Let's look at the polar to rectangular transformation: $$\begin{align*} x &= r \cos(\theta)\\ y &= r \sin(\theta) \end{align*}$$ This is a global, but nonlinear expression. There is no way to express these equations in the form: $$\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} \square & \square \\ \square & \square \end{pmatrix}\begin{pmatrix} r\\ \theta \end{pmatrix}$$ However, if we take the derivatives, $$\begin{align*} dx &= \cos(\theta)dr - r \sin(\theta) d\theta \\ dy &= \sin(\theta)dr + r \cos(\theta) d\theta \end{align*}$$ this becomes an expression which can be expressed in a form which at least resembles a linear equation: $$\begin{pmatrix} dx\\ dy \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -r \sin(\theta)\\ \sin(\theta) & r \cos(\theta) \end{pmatrix}\begin{pmatrix} dr\\ d\theta \end{pmatrix}$$ However, since r and θ are functions of x and y, there's not any global linearity. The function is only "almost" linear in a "sufficiently small" region of space. One could say that the r,θ coordinates are almost linear in a small enough region. This should be contrasted with the rotation transformation, which has no such restrictions about small regions. $$\begin{align*} x'&=x \cos(\theta)-y \sin(\theta)\\ y'&=x \sin(\theta) + y\cos(\theta) \end{align*}$$ Unlike the polar-to-rectangular transformation which is a global but NONlinear transformation, this is a global LINEAR transformation. So when we try to change the form, as follows: $$\begin{pmatrix} x'\\ y' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}$$ There is no difficulty. And if we take the derivatives: $$\begin{pmatrix} dx'\\ dy' \end{pmatrix} =\begin{pmatrix} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta) \end{pmatrix}\begin{pmatrix} dx\\ dy \end{pmatrix}$$ There is also no difficulty. The key difference here is that in the polar-to-rectangular case, the value of r and θ actually are functions of position. In the rotation case, however, the value of θ is a global parameter that does not vary. So what I'm trying to convince you of here, is that the rotation transformation is NOT a local transformation but it applies to every particle in the universe. If I turn my head to the left, everything in the universe moves clockwise around my head. If I turn my head to the right, everything in the universe moves counterclockwise around my head. There is no situation where the stuff in my room moves counterclockwise, but the stuff outside my room remains stationary, unless I'm not really rotating but my room is. There's no way to say that the rotation only applies locally. It is a linear global transformation that affects all particles in the universe. I know some people will laugh at me, saying that my turning my head has no effect on distant particles in the universe, but that is not my argument. I'm not saying that turning my head exerts a force on galaxies billions of light years away causing them to go around in a circle. When you do a rotation transformation, you are changing the observers' reference frame. And thus everything in the universe moves according to the transformation with respect to the observer. Finally, the Lorentz Transformation equations are NOT ANALOGOUS to the polar-to-cartesian coordinate transformations. They are analogous to the rotation transformation equations. They are global linear equations that can be expressed as $$\begin{align*} ct'&=ct \cosh(\varphi)-x \sinh(\varphi)\\ x'&=-ct \sinh(\varphi) + x\cosh(\varphi) \end{align*}$$ and $$\begin{pmatrix} c t'\\ x' \end{pmatrix}= \begin{pmatrix} \cosh(\varphi) & -\sinh(\varphi)\\ -\sinh(\varphi) & \cosh(\varphi) \end{pmatrix}\begin{pmatrix} c t\\ x \end{pmatrix}$$ The parameter φ is not a function of t or x; it is an independent global parameter, just like θ in the rotation transformation. I don't know for sure, but it seems to me, these two types of transformations; global but nonlinear, and GLOBALLY LINEAR transformations are confused. The transformation that makes working with spherical coordinates convenient is nonlinear, and on the scale where the coordinates appear linear, it is local. The Lorentz transformations, on the other hand, have nothing to do with converting from spherical to polar. They have to do with changing velocities; changing rapidity. The Lorentz Transformation equations are a GLOBAL LINEAR TRANSFORMATION, just like rotation.
Quote by JDoolin So what I'm trying to convince you of here, is that the rotation transformation is NOT a local transformation but it applies to every particle in the universe.
Yes, I have to agree with that. If you analyze what the LT does on the basis of its original intent by Voigt and H. A. Lorentz (Voigt's use of the transformation preceded that of Lorentz by several years) then what you say (and more) becomes even clearer. As applied to produce both a change in form of the wave equation and the proper change of initial conditions to accompany the wave equation, it results in initial conditions which can never be reconciled between adjacent points of time or space unless a stretch or shift of all other points is accomplished. But each time that is done the situation remains recursively the same as if you're chasing something that can never be caught. There is no way to make the operation of the LT local unless you freeze the transformation and step back into a Galilean frame of reference.
If we look from the reverse direction[instead of the deductive side] the issue is apparently becoming much simpler. The manifold[described by the metric coefficients] may take care of all anisotropies and inhomogeneities.A metric can include charges[eg:Reissner-Nordstrom metric]. If a metric can include charges it may include strong charges ,spin and other properties of matter with their distributions. One could think in the direction of a generalized unified metric incorporating all known/conceivable properties of matter. These should have their independent effects on clock rates and change of other physical dimensions. In fact the temporal part of the Reissner-Nordstrom metric contains charge as well as mass. [The same is true for the spatial part] One could just imagine reducing the mass and keeping the charge same[in this metric]. The manner in which the clock rate would be affected by charge alone could be examined theoretically. One could try out similar exercises with other properties like spin etc in other types of metrics. The distribution[redistribution] of mass will change the value of the metric coefficients. Distribution[redistribution] of chargees etc will also alter the manifold properties taking care of inhomogeneties and anisotropies.At the same time,one must note, that the clock rates and the spatial dimensions will depend on the distribution/redistribution of these properties in respect of space and time So by writing a general type of a metric one should be able to explain refraction of light. But these metrics will fail to explain properties like the slowing down of light in a medium.Perhaps inclusion of refractive index in the metric could improve the situation.But this[ie,RI] is not a fundamental property like charge or mass. I am very much confused on this issue. Even if RI is included the metric will not interpret the change of the speed of light [in a medium]if it has a form: ds^2=g(00)dt^2-g(11)dx1^2- g(22)dx^2-g(33)dx3^2 Again if two frames are in relative motion at a uniform rate in curved space-time full of anisotropies and inhomogenities, should one apply the usual lorentz Transformations as we know them?
The Local Context: [On Local Inertial Frames] We may consider a local region in strong curved spacetime. It may be surrounded by inhomgenities and anisotropies. How do we get the Lorentz Transformations here? Actually a local transformation to flat spacetime is not a physical transformation.We are doing it on paper--in our imagination. So to that little bit of flat spacetime[we got by transformation] add the rest of it---in your imagination. It is something remote from reality--a workspace in our imagination.A freely falling lift in the earths field can have sufficient inhomogenities and anisotropies around it and does not seem to match with the workspace. That describes our "frictionless plane" We can derive the Lorentz Transformations with sufficient conditions of homogenity of space [and time] and isotropy of space[in the "ideal workspace"] from the Postulates of Special Relativity. [In this process[local transformation from curved space to flat spacetime] we are considering the transformation of a small region of spacetime in total alienation from its surroundings[in so far as anisotropy is concerned], hoping to get correct interpretations from the reverse transformations, when the surroundings are present] In the transformed flat spacetime[local transformation] the un-transformed anisotropies[and heterogeneities] of the surroundings [of the original curved space] are neglected/ignored. Rather we ASSUME "a Global Transformation" for the rest of the flat space when we consider the lorentz Transformations [The gravitational field itself, in general will have an anisotropic and a heterogeneous distribution of the metric coefficients wrt some arbitrary point---one may consider this point in the small transformed space in this context] The Lorentz transformations are of course correct-----only in the proper context of their application
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Quote by Anamitra The Local Context: [On Local Inertial Frames] We may consider a local region in strong curved spacetime. It may be surrounded by inhomgenities and anisotropies. How do we get the Lorentz Transformations here?
I would say the Lorentz Transformations should be applied in exact analogy to the Rotation Transformations. That is, if we have some inhomogeneity or anisotropy in spacetime, and we turn and look the other direction, then the whole thing moves, out of our sight.
What coordinates are used when I do this rotation?
Let's look for a moment at a simple example; the Schwarzschild metric:
$$c^2 d\tau^2 = \left( 1 - \frac{2 G M}{c^2 r} \right)c^2 dt^2 - \left( 1-\frac{2 G M}{c^2 r}\right)^{-1} dr^2 - r^2 (d\theta^2 + sin^2(\theta)d\varphi^2)$$
(Thanks for the help finding the parsing error, Fredrik)
It describes a transformation from dt to dτ. And if written in its negative form, it makes a transformation from ds to a function of dr, dθ, and dφ.
My suggestion would be, instead of treating t, r, θ, and φ as "meaningless" variables, we should treat them as spherical coordinates which can easily map to Minkowski coordinates, and are in fact, the domain and range of the Lorentz Transformation equation.
Again, people will laugh at me, and say that there is no place where matter is not present, so, we have schwarzchild metric embedded in another schwarzschild metric, embedded in another schwarzschild metric. In effect, they will say there is no place in the universe where I can get rid of the curvature altogether. And therefore we might as well not try?
It's a fallacious argument. We should always be aware that there may be masses and motion that we have not accounted for, and so lines we think are straight may be curved after all. But the question is not whether we can know for certain that lines are straight. The question is whether the straight lines exist at all.
As long as we have a concept of straightness. Can we not compare one geodesic path to another geodesic path, and unambiguously declare which of those two paths is straighter? Yes, of course. The straight line is the direction that an object would travel if there were no forces acting on it at all. The geodesic is the direction that an object travels when a force is acting on it.
In real life, we know the difference. But once you start reading a lot of General Relativity books, you start calling geodesics straight lines. And for some reason, you ignore t, r, θ, and φ as "meaningless" variables, and you only look at τ, and say. The object in a geodesic only has τ changing, and is otherwise motionless (with respect to itself).
There are some great things you can do with that dτ; in particular, setting it equal to zero, to see what the path of a photon through your curved space is. But that path is through the t, r, θ, and φ coordinates. Those coordinates are not meaningless. They are the overlying GLOBAL LORENTZ FRAME COORDINATES IN SPHERICAL FORM, COMOVING WITH THE CENTRAL MASS.
If you rotate the coordinate system, r, θ, and φ are affected. If you Lorentz Transform the coordinate system, t, r, θ, and φ are affected.
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Quote by JDoolin (sorry, I can't find the parsing error in the LaTeX.)
There's a ^ followed by a space.
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Quote by JDoolin Some questions (1) What is your concept of the universe? (a) Space-time curvature embedded in an overall flat space-time, or (b) tiny little differential sections of flat space-time that are approximately flat, embedded in an overall curved universe?
If I'm not mistaken, the General Relativity "textbook" answer to this question is (b). But I want to make the argument that the actual answer is (a).
For the Schwarzschild metric, in particular, let's discuss the meanings of these partial derivatives:
$$\left (\frac{\partial \tau}{\partial t} \right )_{\theta,\phi,r \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial r} \right )_{\theta,\phi,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \phi} \right )_{\theta,r ,t \, \mathrm{const}}$$
$$\left (\frac{\partial s}{\partial \theta} \right )_{\phi,r ,t \, \mathrm{const}}$$
What are the textbook definitions of of t, r, θ, and φ? And what are the labels for
Typically, I find textbooks tend to slide over the subject with a great deal of ambiguity.
Quote by CarrollLecturesPage8 "But inspiration aside, it is important to think of these vectors as being located at a single point, rather than stretching from one point to another. (Although this won't stop us from drawing them as arrows on spacetime diagrams.)"
I don't know exactly what Carroll's point is here, but to contrast with Carroll, I will say it is important to think of these vectors as stretching between one event and another event, rather than located at a single point. If you are doing derivatives, with a dt, a dr, a dθ, and a dφ, you can't PUT that at a single point. You can't have a variation in positional variables t, r, θ, and φ "at a single point."
There is an s-component parallel to the r-direction (space-like interval between events), an s-component parallel to the θ direction (space-like interval between events), and an s-component parallel to the φ direction (space-like interval between events), and another component in the t-direction (time-like interval between events), called τ (tau).
There are two different coordinate systems here, and they overlap. You have the three unnamed s-components in the same direction as r, θ, and φ, and the named tau component, in the same direction as t.
The three unnamed s-components and tau make up the curved coordinate system. But there is a continuous bijection from these curved coordinates to the t, r, θ, and φ coordinates, which are flat. The curved coordinates are embedded in the flat coordinates. You can do things easily with the flat coordinates that perhaps you can't easily do with the curved coordinates. For instance, you can do the rotation transformation.
Carroll goes on to say:
A vector is a perfectly well-dened geometric object, as is a vector field, defined as a set of vectors with exactly one at each point in spacetime.
I say there is not one, but two sets of vectors defined at each point in spacetime.
(1) There are Δt, Δθ, Δφ, and Δr which are the flat coordinates, quite easily mapped into Cartesian coordinates, which can then be readily rotated or Lorentz Transformed to whatever angle and rapidity you like,
(2) then there are the "curved" components
$$\begin{matrix} \Delta \tau_{\left (r,\theta,\phi\;\mathrm{const} \right )}\\ \Delta s_\left ({r,\theta, t\;\mathrm{const}} \right )\\ \Delta s_\left ({r,\phi, t\;\mathrm{const}} \right )\\ \Delta s_\left ({\theta,\phi, t\;\mathrm{const}} \right ) \end{matrix}$$
It is these curved components which control the local acceleration and behavior of matter. It is within these curved components where you can set dτ=0 and find the geodesic of a photon, for instance, and all of the neat cool stuff you can do with General Relativity.
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Quote by PhilDSP Yes, I have to agree with that. If you analyze what the LT does on the basis of its original intent by Voigt and H. A. Lorentz (Voigt's use of the transformation preceded that of Lorentz by several years) then what you say (and more) becomes even clearer. As applied to produce both a change in form of the wave equation and the proper change of initial conditions to accompany the wave equation, it results in initial conditions which can never be reconciled between adjacent points of time or space unless a stretch or shift of all other points is accomplished. But each time that is done the situation remains recursively the same as if you're chasing something that can never be caught. There is no way to make the operation of the LT local unless you freeze the transformation and step back into a Galilean frame of reference.
A Galilean Transformation "warps" space-time like this:
where the events move along lines of constant t.
whereas a Lorentz Transformation "warps" space-time like this:
and the events move along hyperbolic arcs of constant $c t^2 - x^2$
I suppose the Galilean Transformation could be said to be "local" if by "local" you mean that it does not affect events which are occurring "now." But it has a big effect on events which occur in the far future or the far past.
And the Lorentz Transformation could be said to be "local" if by "local" you mean that it does not effect THE event which occurs "here" and "now".
(Either way calling a transformation local based on the only events that it doesn't affect seems backward.)
Accelerating toward an event in the future causes it to lean toward you. But accelerating toward an event in the past causes it to lean away.
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| | Special & General Relativity | 8 | | {"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": 6, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9238621592521667, "perplexity_flag": "middle"} |
http://mathhelpforum.com/geometry/41460-pythag-question.html | # Thread:
1. ## Pythag Question
Hi,
This question is an extract from a book which i took a photo of
as I couldn't be bothered replicating it on the computer
http://i116.photobucket.com/albums/o...chet/Maths.jpg
Currently baffled as to what the answer is.
A Nudge in the right direction will be appreciated.
2. Originally Posted by immunofort
Hi,
This question is an extract from a book which i took a photo of
as I couldn't be bothered replicating it on the computer
http://i116.photobucket.com/albums/o...chet/Maths.jpg
Currently baffled as to what the answer is.
A Nudge in the right direction will be appreciated.
First find the hypotenuse of $\triangle ABC$. Then use that hypotenuse value to help you determine the leg $\overline{CD}$ in $\triangle ADC$.
Does that give you enough info to continue on with the problem?
3. ## errrr?
No Sorry, didn't help me much.
I know how to create the long version of the equation which has 2 squareroots but i cant to be able to figure the simplified version.
More Help would be appreciated
4. Hello, immunofort!
Chris is absolutely correct . . . and he did give you a nudge, right?
Code:
``` A *
** * 2
* * *
x * * * D
* * *
* * *
B * * * *
1 C```
Use Pythagorus in $\Delta ABC$
$AC^2 \:=\:AB^2 + BC^2\quad\Rightarrow\quad AC^2 \:=\:x^2+1\quad\Rightarrow\quad AC \:=\:\sqrt{x^2+1}$
Use Pythagorus in $\Delta ADC$
$CD^2 + AD^2 \:=\:AC^2\quad\Rightarrow\quad CD^2 + 2^2 \:=\:(\sqrt{x^2+1})^2 \quad\Rightarrow\quad CD^2 + 4 \:=\:x^2+1$
Therefore: . $CD^2 \:=\:x^2-3 \quad\Rightarrow\quad CD \:=\:\sqrt{x^2-3}$ | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9384283423423767, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/83808?sort=newest | ## What is the relation between a ‘'homotopy fiber bundle’' and a Serre fibration?
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Since I got no responses to this question at Stack Exchange, please let me try my luck here.
Call a continuous map $\pi:E\to B$ between CW complexes a homotopy fiber bundle if for any $x$ in the image of $\pi$, there is an open neighbourhood $U\subset B$ of $x$ and homotopy equivalence $\pi^{-1}(U)→U\times F$ over $U$.
I don't know if this has a different name in the literature or even if it is reasonable. Replacing ''homotopy equivalence'' by ''homeomorphism'' should be the definition of an ordinary fiber bundle.
How relates a ''homotopy fiber bundle'' to the notion of a Serre fibration?
At least both properties imply that the fibers over connected components are all weakly homotopy equivalent.
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Do you want $F$ to be $\pi^{-1}(x)$? Or is $F$ some fixed space independent of $x$? – unknown (google) Dec 18 2011 at 19:49
## 2 Answers
There are many "homotopy fiber bundles" which are not fibrations. For example take any homotopy equivalence $E \to B$ that is not a fibration, then it is a "homotopy fiber bundle" with a one-point fiber.
On the other hand the other implication is (almost) true. The following works for Hurewicz fibrations. I don't whether it is true that a fiber of a Serre fibration between CW-complexes has a homotopy type of a CW-complex. If this is the case, then the proof works also for Serre fibrations.
Let $\pi : E \to B$ be a Hurewicz fibration between CW-complexes and $x \in B$. CW-complexes are locally contractible, so there is a contractible neighborhood $U$ of $x$. Let $\pi_U : E_U \to U$ be the restriction of $\pi$ to $U$. If $E_x$ is the fiber of $\pi$ at $x$, then the inclusion $E_x \to E_U$ is a pullback of the inclusion `$\{x\} \to U$` along a Hurewicz fibration, so it is a homotopy equivalence and admits a homotopy inverse $f : E_U \to E_x$. Thus $(\pi_U, f) : E_U \to U \times E_x$ is a homotopy equivalence over $U$.
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There is a nice theorem in geometric topology that a Serre fibration between actual CW complexes (not just homotopy types thereof) is in fact a Hurewicz fibration. It's in a paper by Steinberger and West (Covering homotopy properties of maps between CW complexes or ANRs, Proc AMS92(1984), 573-577), with a correction in a paper by R. Cauty (Sur les ouverts des CW-complexes et les fibr\'es de Serre, Colloq. Math. 63(1992), 1-7).
Then the rest of your answer applies. But of course the proof of the cited theorem proceeds by checking homotopical local triviality and then quoting the local to global characterization of Hurewicz fibrations.
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http://mathoverflow.net/revisions/120548/list | ## Return to Answer
3 deleted 11 characters in body
The empty structure is ''disconnected'' and the definition of a tree is ''connected acyclic graph''.
For example to get from the generating series $G(t)$ of a connected species to the generating series of the set of its structures $G^*(t) F(t)$ the formula is $$G^*(t) F(t) = \exp(G^*(t)) exp(G(t)) = 1 + \dots .$$ .. $. This '$1$'$1\$ at degree zero is the empty structure.
2 fix markup
The empty structure is ''disconnected'' and the definition of a tree is ''connected acyclic graph''.
For example to get from the generating series $G(t)$ of a connected species to the generating series of the set of its structures $G^(t) G^*(t)$ the formula is $$G^(t) G^*(t) = \exp(G^*(t)) = 1 + \dots ...$$. This'$1$ This '$1$' at degree zero is the empty structure.
1
The empty structure is ''disconnected'' and the definition of a tree is ''connected acyclic graph''.
For example to get from the generating series $G(t)$ of a connected species to the generating series of the set of its structures $G^(t)$ the formula is $$G^(t) = \exp(G^*(t)) = 1 + ...$$. This'$1$' at degree zero is the empty structure. | {"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": 12, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "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}, "english_confidence": 0.9303515553474426, "perplexity_flag": "middle"} |
http://mathoverflow.net/questions/118339?sort=oldest | ## Is there a general process for conditioning a stochastic process above a boundary?
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$(X_t, Y_t)$ is a two-dimensional Markov stochastic process that runs on time interval $[0, t_f]$. Given its transition function $a(x, y | x', y')$, I would like to condition the process on $\inf_{s \in [0, t_f]} X_s \ge k$ and find the new transition function.
Can the problem be solved at this level of generality? Or must we dig into the specifics of $a$ to find a solution on a case-by-case basis?
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Have you tried applying Doob's h-transforms? – Bati Jan 8 at 18:10
## 2 Answers
More generally, let $A$ denote some subset of the state space of a homogenous Markov process $(Z_t)_{0\leqslant t\leqslant t_0}$ with transition kernel $(a_{t-s})_{0\leqslant s\leqslant t\leqslant t_0}$, $T=\inf\{0\leqslant t\leqslant t_0\mid Z_t\in A\}$, $\mathbb Q=\mathbb P(\ \mid T=+\infty)$, and $h_t(z)=\mathbb P(T\gt t\mid Z_0=z)$. By an elementary conditioning, with respect to $\mathbb Q$, the process $(Z_t)_{0\leqslant t\leqslant t_0}$ has inhomogenous transition kernel $$\mathbb Q(Z_t=z\mid Z_s=z')=a_{t-s}(z\mid z')h_{t_0-t}(z)h_{t_0-s}(z')^{-1},$$ for every $0\leqslant s\leqslant t\leqslant t_0$ and every $z$ and $z'$ not in $A$.
Apply this to $Z=(X,Y)$ and $A=(-\infty,k)\times\mathfrak Y$, where $\mathfrak Y$ is the state space of $(Y_t)_t$.
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www.math.harvard.edu/~alexb/rm/Doob.pdf
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http://math.stackexchange.com/questions/299081/do-numbers-of-this-form-have-a-name | # Do numbers of this form have a name?
$$48625 = 4^{5} + 8^{2} + 6^{6} + 2^{8} + 5^{4}$$
Notice that the digits of the number are in digit order, while the exponents are the digits of the number in reverse order.
Do numbers of this form have a name?
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## 1 Answer
They are "tabulated" at OEIS. I use the quotes because there are only two of them, the other one being $397612$ (I don't count $1$). No name is given at that site, but there are some links there that you might follow up. If you find anything, be sure to let us know.
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Thanks. I thought there might possibly be only one, so at least there is another example and some things to investigate. Regards – Amzoti Feb 10 at 2:54 | {"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}, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9562785029411316, "perplexity_flag": "head"} |
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