url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://coertvonk.com/category/math/pre-calc
|
Proofs Euler's formula using the MacLaurin series for sine and cosine. Introduces Euler's identify and Cartesian and Polar coordinates.
Describes Oliver Heaviside's method to decompose rational functions of polynomials as they occur when using the Laplace or Z-transforms.
Introduces complex numbers in an intuitive way, drawing a parallel to negative numbers.
Introduce the functions that operate on complex arguments. Shows how arithmetic functions extend from the 1D number line into the 2D complex-plane.
A quadratic equation is any equation that can be rearranged in standard form as ax2+bx+c=0
|
2022-12-03 08:08:09
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.90966796875, "perplexity": 534.5146187276441}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710926.23/warc/CC-MAIN-20221203075717-20221203105717-00496.warc.gz"}
|
http://www.pytables.org/release-notes/RELEASE_NOTES_v1.0.html
|
# What’s new in PyTables 1.0¶
Author: Francesc Altet faltet@carabos.com Ivan Vilata i Balaguer ivilata@carabos.com
This document details the modifications to PyTables since version 0.9.1. Its main purpose is help you ensure that your programs will be runnable when you switch from PyTables 0.9.1 to PyTables 1.0.
• The new Table.col() method can be used to get a column from a table as a NumArray or CharArray object. This is preferred over the syntax table['colname'].
• The new Table.readCoordinates() method reads a set of rows given their indexes into an in-memory object.
• The new Table.readAppend() method Append rows fullfilling the condition to a destination table.
## Backward-incompatible changes¶
• Trying to open a nonexistent file or a file of unknown type raises IOError instead of RuntimeError. Using an invalid mode raises ValueError instead of RuntimeError.
• Getting a child node from a closed group raises ValueError instead of RuntimeError.
• Running an action on the wrong type of node now (i.e. using file.listNodes() on a leaf) raises a TypeError instead of a NodeError.
• Removing a non-existing child now raises a NoSuchNodeError, instead of doing nothing.
• Removing a non-empty child group using del group.child fails with a NodeError instead of recursively doing the removal. This is because of the potential damage it may cause when used inadvertently. If a recursive behavior is needed, use the _f_remove() method of the child node.
• The recursive flag of Group._f_walkNodes() is True by default now. Before it was False.
• Now, deleting and getting a non-existing attribute raises an AttributeError instead of a RuntimeError.
• Swapped last two arguments of File.copyAttrs() to match the other methods. Please use File.copyNodeAttrs() anyway.
• Failing to infer the size of a string column raises ValueError instead of RuntimeError.
• Excessive table column name length and number of columns now raise ValueError instead of IndexError and NameError.
• Excessive table row length now raises ValueError instead of RuntimeError.
• table[integer] returns a numarray.records.Record object instead of a tuple. This was the original behavior before PyTables 0.9 and proved to be more consistent than the last one (tables do not have an explicit ordering of columns).
• Specifying a nonexistent column in Table.read() raises a ValueError instead of a LookupError.
• When start >= stop an empty iterator is returned by Table.iterrows() instead of an empty RecArray. Thanks to Ashley Walsh for noting this.
• The interface of isHDF5File() and isPyTablesFile() file has been unified so that they both return true or false values on success and raise HDF5ExtError or errors. The true value in isPyTablesFile() is the format version string of the file.
• Table.whereIndexed() and Table.whereInRange() are now private methods, since the Table.where() method is able to choose the most adequate option.
• The global variables ExtVersion and HDF5Version have been renamed to extVersion and hdf5Version, respectively.
• whichLibVersion() returns None on querying unavailable libraries, and raises ValueError on unknown ones.
The following modifications, though being (strictly speaking) modifications of the API, will most probably not cause compatibility problems (but your mileage may vary):
• The default values for name and classname arguments in File.getNode() are now None, although the empty string is still allowed for backwards compatibility. File hierarchy manipulation and attribute handling operations using those arguments have changed to reflect this.
• Copy operations (Group._f_copyChildren(), File.copyChildren(), File.copyNode()…) do no longer return a tuple with the new node and statistics. Instead, they only return the new node, and statistics are collected via an optional keyword argument.
• The copyFile() function in File.py has changed its signature from:
copyFile(srcfilename=None, dstfilename=None, title=None, filters=None,
copyuserattrs=True, overwrite=False, stats=None)
to:
copyFile(srcfilename, dstfilename, overwrite=False, **kwargs)
Thus, the function allows the same options as File.copyFile().
• The File.copyFile() method has changed its signature from:
copyFile(self, dstfilename=None, title=None, filters=None,
copyuserattrs=1, overwrite=0, stats=None):
to:
copyFile(self, dstfilename, overwrite=False, **kwargs)
This enables this method to pass on arbitrary flags and options supported by copying methods of inner nodes in the hierarchy.
• The File.copyChildren() method has changed its signature from:
copyChildren(self, wheresrc, wheredst, recursive=False, filters=None,
copyuserattrs=True, start=0, stop=None, step=1,
overwrite=False, stats=None)
to:
copyChildren(self, srcgroup, dstgroup, overwrite=False, recursive=False,
**kwargs):
Thus, the function allows the same options as Group._f_copyChildren().
• The Group._f_copyChildren() method has changed its signature from:
_f_copyChildren(self, where, recursive=False, filters=None,
copyuserattrs=True, start=0, stop=None, step=1,
overwrite=False, stats=None)
to:
_f_copyChildren(self, dstgroup, overwrite=False, recursive=False,
**kwargs)
This enables this method to pass on arbitrary flags and options supported by copying methods of inner nodes in the group.
• Renamed srcFilename and dstFilename arguments in copyFile() and File.copyFile() to srcfilename and dstfilename, respectively. Renamed whereSrc and whereDst arguments in File.copyChildren() to wheresrc and wheredst, respectively. Renamed dstNode argument in File.copyAttrs() to dstnode. Tose arguments should be easier to type in interactive sessions (although 99% of the time it is not necessary to specify them).
• Renamed object argument in EArray.append() to sequence.
• The rows argument in Table.append() is now compulsory.
• The start argument in Table.removeRows() is now compulsory.
## API refinements¶
• The isHDF5() function has been deprecated in favor of isHDF5File().
• Node attribute-handling methods in File have been renamed for a better coherence and understanding of their purpose:
• getAttrNode() is now called getNodeAttr()
• setAttrNode() is now called setNodeAttr()
• delAttrNode() is now called delNodeAttr()
• copyAttrs() is now called copyNodeAttrs()
They keep their respective signatures, and the old versions still exist for backwards compatibility, though they issue a DeprecationWarning.
• Using VLArray.append() with multiple arguments is now deprecated for its ambiguity. You should put the arguments in a single sequence object (list, tuple, array…) and pass it as the only argument.
• Using table['colname'] is deprecated. Using table.col('colname') (with the new col() method) is preferred.
## Bug fixes (affecting API)¶
• Table.iterrows() returns an empty iterator when no rows are selected, instead of returning None.
Enjoy data!
—The PyTables Team
|
2019-02-21 00:24:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22397078573703766, "perplexity": 9167.452448273552}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247496855.63/warc/CC-MAIN-20190220230820-20190221012820-00303.warc.gz"}
|
http://clay6.com/qa/52114/if-the-density-of-some-lake-water-is-1-25g-ml-and-contains-92-g-of-na-ions-
|
Browse Questions
# If the density of some lake water is $1.25g mL^{–1}$ and contains 92 g of $Na^+$ ions per kg of water, calculate the molality of $Na^+$ ions in the lake.
$\begin{array}{1 1}4m\\5m\\6m\\7m\end{array}$
Number of molality = $\large\frac{\text{Mole of }Na^+\text{ ions}}{\text{Mass of water(kg)}}$
92g of $Na^+$ ions =$\large\frac{929}{23gmol^{-1}}$
$\Rightarrow 4$ mole
Therefore molality = 4m
|
2017-01-20 16:06:32
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7555987238883972, "perplexity": 3524.0736468385353}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280835.60/warc/CC-MAIN-20170116095120-00105-ip-10-171-10-70.ec2.internal.warc.gz"}
|
https://dsp.stackexchange.com/questions/74434/the-plot-of-instantaneous-power-of-the-dirac-function
|
# The plot of instantaneous power of the Dirac function
I am very confused. I have tried researching this question for the last two weeks and I cannot get a conclusive answer.
I was wondering how would I go about plotting the instantaneous power in the time domain of the Dirac input signal. The problem becomes as I try to calculate the power in time domain, $$P(t) = \left\lvert\delta(t)\right\rvert^2$$. I know how to calculate power by using Rayleigh's theorem to convert the the signal to frequency domain where delta(f) = 1. to get infinity over an infinite period which is undefined value for power, but I just want to plot power over time for the Dirac function.
delta isn't square-able since as x goes to 0 the delta function goes to infinity and you cannot square infinity, also something about the delta function being a distribution and not able to be square-able.
I was wondering if someone could point me in the right direction to be able to plot this function.
Thank you
• Is there a reason you want to do this? A Dirac impulse is not a function. What could be the possible interpretation or utility of calculating its "power"?
– MBaz
Apr 12 at 18:38
• How are you trying to use the delta functional that you need to calculate its power? For engineering purposes, the delta functional is there to make your life easier -- if it's making your life impossible, you're misusing it. Apr 12 at 20:07
• the other thing that you should be aware of is that, while the notion of White Noise exists, the reality of White Noise does not. White Noise has equal power per Hz of bandwidth and an unlimited bandwidth. so White Noise (which has a power spectrum that is flat all the way to infinity and a dirac delta as an autocorrelation) has more power output than the sun. Apr 12 at 21:17
• @MBaz It's part of an assignment I was asked to plot the instantaneous power of the signal. Apr 14 at 7:01
• @lowFrequencyLearning Then I suggest asking your instructor to clarify the meaning of the assignment, given what you have learned here.
– MBaz
Apr 14 at 12:45
The problem becomes as I try to calculate the power in time domain, $$P(t) = \left\lvert\delta(t)\right\rvert^2$$
As you figured out yourself, that's impossible: the power of the Dirac impulse isn't defined, it diverges (goes to infinity).
Hence, you especially can't plot the power "correctly"; it's zero anywhere but at $$t=0$$, where it diverges.
But quite honestly, this should come as no surprise: you can't correctly plot $$\delta(t)$$, either (the value at $$t=0$$ isn't defined just as much as its square isn't).
So, whatever question can be answered by that plot: I'm afraid you'll have to find a "better" question.
• Would you say the function |δ(t)|^2 could be similar to the plot of the δ(t) function? Apr 14 at 7:04
• again, none of these have a plot. They are not functions. So, "is it similar" can be answered with "what are you even talking about?" Apr 14 at 9:54
• thanks for replying Apr 15 at 4:31
|
2021-09-21 19:52:51
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 5, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.750892162322998, "perplexity": 380.86619753041674}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057227.73/warc/CC-MAIN-20210921191451-20210921221451-00332.warc.gz"}
|
http://www.bio.net/bionet/mm/neur-sci/2004-November/059526.html
|
# Futures of Man: Total DNA Control Path to Immortality
Fabrizio J. Bonsignore djbonsignore at beethoven.com
Mon Nov 29 08:35:30 EST 2004
Sent to some UN delegations:
Myelination, the process of cnnecting nerves and axons to expedite the
transmissio neuroelectric signal. Hallucinogens peform this process,
either permanently by remyelinizing the sheaths or temporarily. They
connext different areas of the brain that are not generally connected,
producing cinestesia, the mix of sensorial input and interpretation,
like flavoring colors or smelling sounds. A bad trip, the permanent
trip, the freak, and also the backflashes can be explained by a
permanent reconnectio of the brain tissue due to the direct or
indirect action of hallucinogens. A totally connected brain (in its
functional areas) would require an impregnation, probably gradual, to
achieve the state of illuminatio of the old mystic disciplines (which
can be explained as the myelinization or similar processes by the
semiconscious effect of meditation). The process is similar to thatof
contring respiration, which is automatic but can be subjected to
conscious control (incidentallly an imporatnt part of all meditatio
and mystical techniques). With conscious control of the totality of
our brain we can achieve the same kind of control of other automatic
processes, down to the cellular and intracellular level, as the brain
has enough computing power due to the high order of interconnections
of the billions of neurones we have. With such awareness it is
possible to control our DNA and consciously fix it to prevent cancers
(entropy) and extend our life span, though it would require knowledge
and intelligence to avoid monstruous mistakes. It would also permit
the conscious control of the interaction of the magnetic neurofields
with the surrounding environment, much in the way MRI scannings work.
Note: it is to be noted that our brains lack this total
interconnection quality despite the fact that it has a lot of
plasticity (recoves after brain damage), as if evolution has been
stopped just before this conditio was acquired. But, if we take
seriously the old myths, it is more like we *have* lost this
abilities, we we able to do this knd of total awareness feats but have
lost the ability, as if we were turning back to animality... The
original sin of knowledge being gradually erased from the genetic
human pool; instead of acquiring intelligence we are losing it and our
current state of development is due more to the systematic application
of the scientific method than by a slow but steady gain in
intelligence. Human evolution is stopped compared to conscience,
Reason, but our genetic pool may be experimenting more rapid changes
tha we have been able to notice given the short span we have been
using the scientific method...
http:\\Fab\ri\zio\j\bon\sig\nore
http:\\Da\nilo\J\Bonsig\nore
http:\\not\an\url\if\something\it\is\censure
http:\\in\new\york
> Bob ono nsn sis igi gng non oro rer ree
> Dad ana nin ili lol lo
>
> 1111 4444 7777 E 4444 3333 zc 11111 00000 00000 11111 77777
>
> nyn yny unu n
|
2014-11-23 08:52:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5291702151298523, "perplexity": 8896.247860333408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-49/segments/1416400379404.25/warc/CC-MAIN-20141119123259-00160-ip-10-235-23-156.ec2.internal.warc.gz"}
|
http://golem.ph.utexas.edu/category/2008/08/bergman_on_infinityvector_bund.html
|
August 17, 2008
Bergman on Infinity-Vector Bundles Coupled to Topological Strings
Posted by Urs Schreiber
As Aaron Bergman kindly pointed out here, he has an interesting article
Aaron Bergman
New Boundaries for the B-Model
arXiv:0808.0168
which gives a direct interpretation of the derived categories of coherent sheaves appearing in topological strings in terms of connections on $\infty$-(Chan-Paton)-vector bundles, using a crucial insight by Jonathan Block.
Formerly, the way to see that the branes (= boundary conditions) of the topological B-model string are objects in a derived category of coherent sheaves was somewhat involved and a bit mysterious. Once upon a time I had tried to summarize the main steps involved as reviewed by Aspinwall here
Now, Jonathan Block showed that these derived categories of coherent sheaves are actually equivalent to homotpy categories of representations of holomorphic tangent Lie algebroids on chain complexes – but these are special cases of flat linear (as opposed to principal) $\infty$-connections.
Aaron Bergman takes this theorem at face value and concludes that therefore it should be true that there is a direct way to see that the boundary conditions of the topological B-string come from such flat $\infty$-vector bundles with connection, pretty much analogous to how an ordinary conformal string (the “physical string”) couples on D-branes to ordinary (maybe $\mathbb{Z}_2$-graded) vector bundles with connection.
As an ansatz, he considers boundary insertions in the path integral that should correspond to the generalization of the familiar holonomy along the String’s boundary of the pulled back connection to the $\infty$ setup. He derives that this satisfies the B-string’s topological invariance precisely if the flat $\infty$-connection satisfies the appropriate axioms it should satisfy. He then derives that, similarly, imposing the required topological invariance on the boundary field insertions interpolating between two such flat $\infty$-connections yields the right notion of morphism in the category of flat $\infty$-connections that Jonathan Block considered.
Posted at August 17, 2008 12:46 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1766
Re: Bergman on Infinity-Vector Bundles Coupled to Topological Strings
I should say that it turns out that the coupling had already been derived in a different way in Herbst, Hori and Page.
Posted by: Aaron Bergman on August 17, 2008 3:07 PM | Permalink | Reply to this
Re: Bergman on Infinity-Vector Bundles Coupled to Topological Strings
these derived categories of coherent sheaves are actually equivalent to homotopy categories of representations of holomorphic tangent Lie algebroids on chain complexes
Wow that sounds very elegant.
Posted by: Bruce Bartlett on August 17, 2008 5:33 PM | Permalink | Reply to this
Re: Bergman on Infinity-Vector Bundles Coupled to Topological Strings
these derived categories of coherent sheaves are actually equivalent to homotopy categories of representations of holomorphic tangent Lie algebroids on chain complexes
Wow that sounds very elegant.
I guess in concrete applications it may depend which of the two points of view is more useful, but the interpretation in terms of reps of $L_\infty$-algebroids seems to clarify a bit more what is really going on here.
By the way, do you rememebr when we were all sitting in the chinese restaurant in Toronto during a break at the Fields institute workshop on higher categories and their applications with Aaron Lauda and a couple of other guys? We were talking about algebras assigned to boundaries in 2d TFT and how these can be interpreted as placeholders for 2-vector spaces. I think I said that we should try to understand how $\infty$-vector spaces come into this picture, because these are the ones that appear in the phyiscally interesting models. I still remember how you nodded. So here is Aaron Bergman making a bit of progress towards this goal.
Posted by: Urs Schreiber on August 17, 2008 8:13 PM | Permalink | Reply to this
Re: Bergman on Infinity-Vector Bundles Coupled to Topological Strings
Indeed. I can see there is some serious progress here; it’s just that I have to pump some serious iron in the mathematical gym before I can get my technical level up to the point where I can understand it.
Posted by: Bruce Bartlett on August 17, 2008 9:50 PM | Permalink | Reply to this
Post a New Comment
|
2014-09-02 18:52:49
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 10, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7572511434555054, "perplexity": 802.0231791356072}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1409535922763.3/warc/CC-MAIN-20140909050933-00244-ip-10-180-136-8.ec2.internal.warc.gz"}
|
https://cracku.in/13-for-a-sequence-of-real-numbers-x_1x_2x_n-if-x_1-x_-x-cat-2021-slot-2-quant
|
Question 13
# For a sequence of real numbers $$x_{1},x_{2},...x_{n}$$, If $$x_{1}-x_{2}+x_{3}-....+(-1)^{n+1}x_{n}=n^{2}+2n$$ for all natural numbers n, then the sum $$x_{49}+x_{50}$$ equals
Solution
Now as per the given series :
we get $$x_1=1+2\ =3$$
Now $$x_1-x_2=\ 8$$
so$$x_2=-5$$
Now $$x_1-x_2+x_3\ =\ 15$$
so $$x_3\ =7$$
so we get $$x_n\ =\left(-1\right)^{n+1}\left(2n+1\right)$$
so $$x_{49}\ =\ 99$$ and $$x_{50}\ =-101$$
Therefore $$x_{49\ }+x_{50}\ =-2$$
|
2022-08-13 10:08:57
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9602159857749939, "perplexity": 212.73981478101288}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882571911.5/warc/CC-MAIN-20220813081639-20220813111639-00015.warc.gz"}
|
https://eco-news.space/2021/ecology/testing-of-how-and-why-the-terpios-hoshinota-sponge-kills-stony-corals/
|
in
# Testing of how and why the Terpios hoshinota sponge kills stony corals
### Experiment 1: Sponge fragments
Evidence of bleaching first occurred 3 days after the treatment and was only evident in the group with fragments of T. hoshinota. No bleaching was detected in the other 2 groups with the black cloth (to block light) and white cloth (control) (Table 1). Chi-square tests confirmed that the occurrence of bleaching depended on the treatments (p < 0.001 in both tests: sponge fragment vs. black cloth, and sponge fragment vs. white cloth).
### Experiment 2: Sponge mixture
Significant difference in CICI was detected among the 3 treatments (p < 0.001; Friedman Test, n = 19; Fig. 1). The treatment with sponge and a black cap had the greatest effect, whereas the treatment with the transparent cap demonstrated the smallest response. All 3 pairwise comparisons were significant [CICI (sponge and black caps) > CICI (black caps) > CICI (transparent caps), n = 19, p < 0.01 in all 3 pair-wise comparisons, Wilcoxon signed-rank tests].
### Experiment 3: Sponge supernatant
Two days after the treatment, a significant difference was detected between the sponge + black cap, and fish-meat + black cap groups (Fig. 2) with the former showing significantly greater effects.
Significant difference in CICI was observed among the 3 treatments, i.e., only the sponge supernatant, only black caps, and both sponge supernatant and black caps (n = 17; p < 0.01, Friedman Test, Fig. 3), 4 days after the treatment. Further pair-wise analyses indicated that both factors combined had stronger effects than when single factors were applied [p = 0.055 (against the black cap alone), p < 0.01 (against sponge alone), Wilcoxon signed-rank test, n = 17]. No significant difference was observed between the 2 single factors (p = 0.12; Wilcoxon signed-rank test, n = 17).
### Stable isotope experiment
Six pieces of transplanted corals, 3 from the isotope labeling group, and 3 from the control group, which had been covered by T. hoshinota were successfully retrieved. The other samples either lost their labels or were not covered by the sponge in the field.
The control corals had δ13C: − 14.4 ± 1.2‰, δ15N: 4.8 ± 0.6‰ (n = 3). The error terms were 95% confidence intervals of the means throughout this study. The sponges grown on the control corals had lower values, i.e., δ13C: − 21.4 ± 0.8‰, δ15N: 3.6 ± 0.1‰ (n = 3). The control corals had a significantly higher heavy stable isotopes content than the sponges grown on them (δ13C, p < 0.01; δ15N; p < 0.05; t-tests, n = 3, Table 2).
The enriched corals had δ13C: 31.8 ± 12.1‰, δ15N: 71.0 ± 29.5‰, and the T. hoshinota grown on them had lower values: δ13C: − 16.0 ± 1.0‰, δ15N: 14.7 ± 5.3‰ (n = 3). In fact, the stable isotope composition of Terpios on the enriched corals was closer to the isotope compositions of the sponges on the control corals, than to the coral tissues underneath them (Fig. 4).
Analysis of the sponges indicated that a significant difference in δ13C along the growth axis was evident only in the samples with the enriched treatment (ANOVA, F2,6 = 8.33; p < 0.05), but not in the sponges grown on the control corals (ANOVA, F2,6 = 0.01; p = 0.99; Fig. 5). A Fisher’s PLSD test indicated that the sponge δ13C was the highest directly on top of the enriched corals, intermediate at the junction, and the lowest at the > 5-cm positions (p < 0.01; n = 3), and the > 5-cm and the < 5-cm positions were not significantly different (p = 0.13; n = 3). In comparing the sponge δ13C between the enriched and control corals of the equivalent positions, a significant difference was only observed in the positions right above the transplanted corals (Fig. 5). This was a clear indication that the incorporated heavy stable C of the sponge was not translocated to the more proximal part of the sponges.
Concerning δ15N in the control group, significant difference was observed among the different positions of the sponge (ANOVA, F2,6 = 6.93; p < 0.05), although the difference was small. Lighter nitrogen was discovered near the expanding fronts (Fisher’s PLSD test, p = 0.01; n = 3; Fig. 6). In the enriched group, significant difference was observed among different positions of the sponge (ANOVA, F2,6 = 6.91; p < 0.01), but the trend was the opposite because the expanding fronts that covered the coral had higher levels of nitrogen. The junction was heavier than the more proximal part. By comparing the δ15N that grew on the 2 groups of corals, the sponges on the enriched corals had higher δ15N than those on the control corals in the newly grown tissues. However, the 2 groups of sponge tissues did not vary significantly in the more proximal parts that were far from the new growth (Figs. 5, 6). This result was consistent with the result of δ13C and did not support the suggestion that the frontal sponge tissues translocate acquired new materials to more proximate parts of the sponge.
### Estimation of underlying coral tissue contribution to sponge
Three possible sources contributed to the stable isotope compositions of the T. hoshinota sponge, namely the coral tissues underneath, microbes and food particles filtered from the water column, and the translocation of tissues or materials from the proximal part of the sponge. The stable isotope of the sponge could be expressed as the following equation:
$$updelta _{{{text{sponge}}}} =updelta _{{{text{coral}}}} times f_{{{text{coral}}}} +updelta _{{{text{water}}}} times f_{{{text{water}}}} +updelta _{{text{back sponge}}} times f_{{text{back sponge}}}$$
(1)
δsponge, Isotope composition of new T. hoshinota tissues; δcoral, Isotope composition of coral tissues underneath T. hoshinota; δwater, Isotope composition of sponge food in the water column; δback sponge, Isotope composition of the proximal part of T. hoshinota; fcoral, Fraction contributed by underlying coral to new T. hoshinota tissues; fwater, Fraction contributed by food particles in the water column to new T. hoshinota tissues; fback sponge, Fraction contributed by proximal part of the tissue to new T. hoshinota tissues.
$${text{and}},f_{{{text{coral}}}} + f_{{{text{water}}}} + f_{{text{back sponge}}} = , 1$$
Assuming that δwater × fwater + δback sponge × fback sponge remained unchanged between treatments, the formula could be transformed to
$$updelta _{{{text{sponge}}}} = f_{{{text{coral}}}} timesupdelta _{{{text{coral}}}} + {text{Constant}}$$
(2)
Therefore, the fraction of new sponge tissues contributed by the underlying coral tissues (fcoral) could be estimated when the δsponge values under different δcoral were available. Because the constant in Eq. (2) was not known, at least 2 pairs of data were required to estimate fcoral. Using 3 paired samples of enriched coral tissues and the sponge tissues that covered them, the fcoral estimated from δ13C and δ15N were 9.5% and 16.9%, respectively, using regression. In Eq. (2), the enrichment of the heavy isotope composition along the food chain was not considered.
Source: Ecology - nature.com
|
2021-04-17 04:30:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.48419293761253357, "perplexity": 3648.895072127408}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038101485.44/warc/CC-MAIN-20210417041730-20210417071730-00025.warc.gz"}
|
https://www.aimsciences.org/article/doi/10.3934/naco.2022013
|
# American Institute of Mathematical Sciences
doi: 10.3934/naco.2022013
Online First
Online First articles are published articles within a journal that have not yet been assigned to a formal issue. This means they do not yet have a volume number, issue number, or page numbers assigned to them, however, they can still be found and cited using their DOI (Digital Object Identifier). Online First publication benefits the research community by making new scientific discoveries known as quickly as possible.
Readers can access Online First articles via the “Online First” tab for the selected journal.
## Optimization method in counter terrorism: Min-Max zero-sum differential game approach
1 Basic Science Department, Faculty of Computers and informatics, Suez Canal university, Ismailia, Egypt 2 Department of Mathematics, Faculty of Science, Tanta University, Tanta, Egypt
*Corresponding author: heba.tallah@ci.suez.edu.eg
Received May 2021 Revised April 2022 Early access May 2022
One of the most critical problems facing governments at present is terrorism. Most recent studies are striving to find an optimal solution to this problem that threatens the security and stability of people. To combat terrorism, government uses various means, such as improving education quality, providing labor opportunities, seeking social justice, creating religious awareness, and building security arrangements. This study aims to evaluate the optimum strategy for both government and terrorist organization using a min-max zero-sum differential game approach. In addition, it analyses the dynamics of the government's activities separately and explain the impact of the government's counter-terror measures on the activity and strength of international terror organization.
Citation: Abd El-Monem A. Megahed, Ebrahim A. Youness, Hebatallah K. Arafat. Optimization method in counter terrorism: Min-Max zero-sum differential game approach. Numerical Algebra, Control and Optimization, doi: 10.3934/naco.2022013
##### References:
[1] E. Ahmed, A. Elgazzar and A. Hegazi, On complex adaptive systems and terrorism, Physics Letters A, 337 (2005), 127-129. [2] J. Caulkins, G. Feichtinger, D. Grass and G. Tragler, Optimal control of terrorism and global reputation: A case study with novel threshold behavior, Operations Research Letters, 37 (2009), 387-391. doi: 10.1016/j.orl.2009.07.003. [3] J. Caulkins, D. Grass, G. Feichtinger and G. Tragler, Optimizing counter-terror operations: Should one fight fire with "fire" or "water"?, Computers & Operations Research, 35 (2008), 1874-1885. [4] H.-J. Chu, J.-G. Hsieh, K.-H. Hsia and L.-W. Chen, Fuzzy differential game of guarding a movable territory, Information Sciences, 91 (1996), 113-131. doi: 10.1016/0020-0255(95)00299-5. [5] S. Hegazy, A. Megahed, E. Youness and A. Elbann, Min-max zero-sum two persons fuzzy continuous differential games, International Journal of Applied Mathematics, 21 (2008), 1-16. [6] K.-H. Hsia and J.-G. Hsieh, A first approach to fuzzy differential game problem: guarding a territory, Fuzzy Sets and Systems, 55 (1993), 157-167. doi: 10.1016/0165-0114(93)90128-5. [7] A. Megahed, A differential game related to terrorism: min-max zero-sum two persons differential game, Neural Computing and Applications, 30 (2016), 865-870. doi: 10.1016/j.joems.2017.03.007. [8] A. Megahed, The development of a differential game related to terrorism: Min-max differential game, Journal of the Egyptian Mathematical Society, 25 (2017), 308-312. doi: 10.1016/j.joems.2017.03.007. [9] A. Megahed, The stackelberg differential game for counter-terrorism, Quality & Quantity, 53 (2018), 207-220. [10] A. Megahed, A differential game related to terrorism: Stackelberg differential game of e-differentiable and e-convex function, European Journal of Pure and Applied Mathematics, 12 (2019), 654-667. doi: 10.29020/nybg.ejpam.v12i2.3375. [11] A. Megahed and S. Hegazy, Min-max zero sum two persons continuous differential game with fuzzy controls, Asian Journal of Current Engineering and Maths, 2 (2013), 86-98. [12] A. Novak, G. Feichtinger and G. Leitmann, A differential game related to terrorism: Nash and stackelberg strategies, Journal of Optimization Theory and Applications, 144 (2010), 533-555. doi: 10.1007/s10957-009-9643-z. [13] A. Roy and J. A. Paul, Terrorism deterrence in a two country framework: strategic interactions between r & d, defense and pre-emption, Annals of Operations Research, 211 (2013), 399-432. doi: 10.1007/s10479-013-1431-3. [14] J. Wang and P. Wang, Counterterror measures and economic growth: A differential game, Operations Research Letters, 41 (2013), 285-289. doi: 10.1016/j.orl.2013.02.008. [15] E. Youness, J. Hughes and N. El-Kholy, Parametric nash coalitive differential games, Mathematical and Computer Modelling, 26 (1997), 97-105. doi: 10.1016/S0895-7177(97)00125-8. [16] E. Youness and A. Megahed, A study on fuzzy differential game, Le Matematiche, 56 (2001), 97-107. [17] E. Youness and A. Megahed, A study on large scale continuous differential games, Bull. Culcutta. Math. Soc., 94 (2002), 359-368.
show all references
##### References:
[1] E. Ahmed, A. Elgazzar and A. Hegazi, On complex adaptive systems and terrorism, Physics Letters A, 337 (2005), 127-129. [2] J. Caulkins, G. Feichtinger, D. Grass and G. Tragler, Optimal control of terrorism and global reputation: A case study with novel threshold behavior, Operations Research Letters, 37 (2009), 387-391. doi: 10.1016/j.orl.2009.07.003. [3] J. Caulkins, D. Grass, G. Feichtinger and G. Tragler, Optimizing counter-terror operations: Should one fight fire with "fire" or "water"?, Computers & Operations Research, 35 (2008), 1874-1885. [4] H.-J. Chu, J.-G. Hsieh, K.-H. Hsia and L.-W. Chen, Fuzzy differential game of guarding a movable territory, Information Sciences, 91 (1996), 113-131. doi: 10.1016/0020-0255(95)00299-5. [5] S. Hegazy, A. Megahed, E. Youness and A. Elbann, Min-max zero-sum two persons fuzzy continuous differential games, International Journal of Applied Mathematics, 21 (2008), 1-16. [6] K.-H. Hsia and J.-G. Hsieh, A first approach to fuzzy differential game problem: guarding a territory, Fuzzy Sets and Systems, 55 (1993), 157-167. doi: 10.1016/0165-0114(93)90128-5. [7] A. Megahed, A differential game related to terrorism: min-max zero-sum two persons differential game, Neural Computing and Applications, 30 (2016), 865-870. doi: 10.1016/j.joems.2017.03.007. [8] A. Megahed, The development of a differential game related to terrorism: Min-max differential game, Journal of the Egyptian Mathematical Society, 25 (2017), 308-312. doi: 10.1016/j.joems.2017.03.007. [9] A. Megahed, The stackelberg differential game for counter-terrorism, Quality & Quantity, 53 (2018), 207-220. [10] A. Megahed, A differential game related to terrorism: Stackelberg differential game of e-differentiable and e-convex function, European Journal of Pure and Applied Mathematics, 12 (2019), 654-667. doi: 10.29020/nybg.ejpam.v12i2.3375. [11] A. Megahed and S. Hegazy, Min-max zero sum two persons continuous differential game with fuzzy controls, Asian Journal of Current Engineering and Maths, 2 (2013), 86-98. [12] A. Novak, G. Feichtinger and G. Leitmann, A differential game related to terrorism: Nash and stackelberg strategies, Journal of Optimization Theory and Applications, 144 (2010), 533-555. doi: 10.1007/s10957-009-9643-z. [13] A. Roy and J. A. Paul, Terrorism deterrence in a two country framework: strategic interactions between r & d, defense and pre-emption, Annals of Operations Research, 211 (2013), 399-432. doi: 10.1007/s10479-013-1431-3. [14] J. Wang and P. Wang, Counterterror measures and economic growth: A differential game, Operations Research Letters, 41 (2013), 285-289. doi: 10.1016/j.orl.2013.02.008. [15] E. Youness, J. Hughes and N. El-Kholy, Parametric nash coalitive differential games, Mathematical and Computer Modelling, 26 (1997), 97-105. doi: 10.1016/S0895-7177(97)00125-8. [16] E. Youness and A. Megahed, A study on fuzzy differential game, Le Matematiche, 56 (2001), 97-107. [17] E. Youness and A. Megahed, A study on large scale continuous differential games, Bull. Culcutta. Math. Soc., 94 (2002), 359-368.
The relation between $u$ and $\nu$
The relation between $Y$ and $\nu$
[1] Hermes H. Ferreira, Artur O. Lopes, Silvia R. C. Lopes. Decision Theory and large deviations for dynamical hypotheses tests: The Neyman-Pearson Lemma, Min-Max and Bayesian tests. Journal of Dynamics and Games, 2022, 9 (2) : 123-150. doi: 10.3934/jdg.2021031 [2] Valery Y. Glizer, Oleg Kelis. Singular infinite horizon zero-sum linear-quadratic differential game: Saddle-point equilibrium sequence. Numerical Algebra, Control and Optimization, 2017, 7 (1) : 1-20. doi: 10.3934/naco.2017001 [3] José M. Amigó, Ángel Giménez. Formulas for the topological entropy of multimodal maps based on min-max symbols. Discrete and Continuous Dynamical Systems - B, 2015, 20 (10) : 3415-3434. doi: 10.3934/dcdsb.2015.20.3415 [4] Jinchuan Zhou, Changyu Wang, Naihua Xiu, Soonyi Wu. First-order optimality conditions for convex semi-infinite min-max programming with noncompact sets. Journal of Industrial and Management Optimization, 2009, 5 (4) : 851-866. doi: 10.3934/jimo.2009.5.851 [5] Roberto Livrea, Salvatore A. Marano. A min-max principle for non-differentiable functions with a weak compactness condition. Communications on Pure and Applied Analysis, 2009, 8 (3) : 1019-1029. doi: 10.3934/cpaa.2009.8.1019 [6] Meixia Li, Changyu Wang, Biao Qu. Non-convex semi-infinite min-max optimization with noncompact sets. Journal of Industrial and Management Optimization, 2017, 13 (4) : 1859-1881. doi: 10.3934/jimo.2017022 [7] Abbas Ja'afaru Badakaya, Aminu Sulaiman Halliru, Jamilu Adamu. Game value for a pursuit-evasion differential game problem in a Hilbert space. Journal of Dynamics and Games, 2022, 9 (1) : 1-12. doi: 10.3934/jdg.2021019 [8] Eduardo Espinosa-Avila, Pablo Padilla Longoria, Francisco Hernández-Quiroz. Game theory and dynamic programming in alternate games. Journal of Dynamics and Games, 2017, 4 (3) : 205-216. doi: 10.3934/jdg.2017013 [9] Astridh Boccabella, Roberto Natalini, Lorenzo Pareschi. On a continuous mixed strategies model for evolutionary game theory. Kinetic and Related Models, 2011, 4 (1) : 187-213. doi: 10.3934/krm.2011.4.187 [10] Anna Lisa Amadori, Astridh Boccabella, Roberto Natalini. A hyperbolic model of spatial evolutionary game theory. Communications on Pure and Applied Analysis, 2012, 11 (3) : 981-1002. doi: 10.3934/cpaa.2012.11.981 [11] Pierre Cardaliaguet, Chloé Jimenez, Marc Quincampoix. Pure and Random strategies in differential game with incomplete informations. Journal of Dynamics and Games, 2014, 1 (3) : 363-375. doi: 10.3934/jdg.2014.1.363 [12] Nidhal Gammoudi, Hasnaa Zidani. A differential game control problem with state constraints. Mathematical Control and Related Fields, 2022 doi: 10.3934/mcrf.2022008 [13] Marianne Akian, Stéphane Gaubert, Antoine Hochart. A game theory approach to the existence and uniqueness of nonlinear Perron-Frobenius eigenvectors. Discrete and Continuous Dynamical Systems, 2020, 40 (1) : 207-231. doi: 10.3934/dcds.2020009 [14] Serap Ergün, Bariş Bülent Kırlar, Sırma Zeynep Alparslan Gök, Gerhard-Wilhelm Weber. An application of crypto cloud computing in social networks by cooperative game theory. Journal of Industrial and Management Optimization, 2020, 16 (4) : 1927-1941. doi: 10.3934/jimo.2019036 [15] Tinggui Chen, Yanhui Jiang. Research on operating mechanism for creative products supply chain based on game theory. Discrete and Continuous Dynamical Systems - S, 2015, 8 (6) : 1103-1112. doi: 10.3934/dcdss.2015.8.1103 [16] Amina-Aicha Khennaoui, A. Othman Almatroud, Adel Ouannas, M. Mossa Al-sawalha, Giuseppe Grassi, Viet-Thanh Pham. The effect of caputo fractional difference operator on a novel game theory model. Discrete and Continuous Dynamical Systems - B, 2021, 26 (8) : 4549-4565. doi: 10.3934/dcdsb.2020302 [17] Juan Pablo Pinasco, Mauro Rodriguez Cartabia, Nicolas Saintier. Evolutionary game theory in mixed strategies: From microscopic interactions to kinetic equations. Kinetic and Related Models, 2021, 14 (1) : 115-148. doi: 10.3934/krm.2020051 [18] Yadong Shu, Ying Dai, Zujun Ma. Evolutionary game theory analysis of supply chain with fairness concerns of retailers. Journal of Industrial and Management Optimization, 2022 doi: 10.3934/jimo.2022098 [19] Ovide Arino, Eva Sánchez. A saddle point theorem for functional state-dependent delay differential equations. Discrete and Continuous Dynamical Systems, 2005, 12 (4) : 687-722. doi: 10.3934/dcds.2005.12.687 [20] Vianney Perchet, Marc Quincampoix. A differential game on Wasserstein space. Application to weak approachability with partial monitoring. Journal of Dynamics and Games, 2019, 6 (1) : 65-85. doi: 10.3934/jdg.2019005
Impact Factor:
|
2022-08-16 01:32:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3774752914905548, "perplexity": 7654.13810459369}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00724.warc.gz"}
|
https://planetmath.org/vertlangleTvvranglevertleqmuVertVVert2ForAllVImpliesVertTVertleqmu
|
$\delimiter"026A30C\delimiter"426830ATv,v\delimiter"526930B\delimiter"026A30C% \leq{\mu}\delimiter"026B30Dv\delimiter"026B30D^{2}$ for all $v$ implies $\delimiter"026B30DT\delimiter"026B30D\leq{\mu}$
Theorem.
Let $H$ be a unitary space, $T$ be a self-adjoint linear operator and $\mu\geq 0$. If $|\left\langle Tv,v\right\rangle|\leq\mu\|v\|^{2}$ for all $v\in H$ then $T$ is a bounded operator and $\|T\|\leq\mu$.
Proof.
We will show that $\left\|Tv\right\|\leq\mu\left\|v\right\|$ for all $v\in H$. This is trivial if $\left\|Tv\right\|$ or $\left\|v\right\|$ is zero, so assume they are not. Let $\lambda$ be any positive number.
$\displaystyle\left\|Tv\right\|^{2}$ $\displaystyle=\left\langle Tv,Tv\right\rangle$ $\displaystyle=\frac{1}{4}\left[\left\langle T\left(\lambda v+\frac{1}{\lambda}% Tv\right),\left(\lambda v+\frac{1}{\lambda}Tv\right)\right\rangle-\left\langle T% \left(\lambda v-\frac{1}{\lambda}Tv\right),\left(\lambda v-\frac{1}{\lambda}Tv% \right)\right\rangle\right]$ $\displaystyle\leq\frac{\mu}{4}\left[\left\|\lambda v+\frac{1}{\lambda}Tv\right% \|^{2}+\left\|\lambda v-\frac{1}{\lambda}Tv\right\|^{2}\right]$ $\displaystyle\leq\frac{\mu}{2}\left[\lambda^{2}\left\|v\right\|^{2}+\frac{1}{% \lambda^{2}}\left\|Tv\right\|^{2}\right]$
Now if we put $\displaystyle\lambda^{2}=\frac{\left\|Tv\right\|}{\left\|v\right\|}$ we get $\left\|Tv\right\|^{2}\leq\mu\left\|Tv\right\|\left\|v\right\|$ hence $\left\|Tv\right\|\leq\mu\left\|v\right\|$. ∎
Reference:
F. Riesz and B. Sz-Nagy, Functional Analysis, F. Ungar Publishing, 1955, chap VI.
Title $\delimiter"026A30C\delimiter"426830ATv,v\delimiter"526930B\delimiter"026A30C% \leq{\mu}\delimiter"026B30Dv\delimiter"026B30D^{2}$ for all $v$ implies $\delimiter"026B30DT\delimiter"026B30D\leq{\mu}$ vertlangleTvvranglevertleqmuVertVVert2ForAllVImpliesVertTVertleqmu 2013-03-22 15:25:33 2013-03-22 15:25:33 Gorkem (3644) Gorkem (3644) 16 Gorkem (3644) Theorem msc 46C05
|
2023-01-29 15:33:05
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 26, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9868408441543579, "perplexity": 320.7292705860731}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499744.74/warc/CC-MAIN-20230129144110-20230129174110-00140.warc.gz"}
|
http://clay6.com/qa/37717/in-a-class-of-50-students-15-are-girls-5-of-the-girls-and-large-frac-of-the
|
Browse Questions
Home >> AIMS
# In a class of 50 students,15 are girls,5 of the girls and $\large\frac{2}{7}$ of the boys were chosen to represent their class in a game.The total number of students chosen is
$\begin{array}{1 1}(A)\;12&(B)\;15\\(C)\;19&(D)\;25\end{array}$
The total number of students chosen is 15
Hence (B) is the correct answer.
|
2017-03-27 04:39:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9850996732711792, "perplexity": 531.2162438991722}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189403.13/warc/CC-MAIN-20170322212949-00146-ip-10-233-31-227.ec2.internal.warc.gz"}
|
https://www.semanticscholar.org/paper/Noise-sensitivity-of-second-top-eigenvectors-of-and-Lee/74d61cdd6a269eb29526bdf53d7dbf2ff37c97d9
|
Corpus ID: 210156949
# Noise sensitivity of second-top eigenvectors of Erd\H{o}s-R\'{e}nyi graphs and sparse matrices
@article{Lee2020NoiseSO,
title={Noise sensitivity of second-top eigenvectors of Erd\H\{o\}s-R\'\{e\}nyi graphs and sparse matrices},
author={Jaehun Lee},
journal={arXiv: Probability},
year={2020}
}
• Jaehun Lee
• Published 2020
• Mathematics
• arXiv: Probability
We consider eigenvectors of adjacency matrices of Erd\H{o}s-R\'{e}nyi graphs and study the variation of their directions by resampling the entries randomly. Let $\mathbf{v}$ be the eigenvector associated with the second-largest eigenvalue of the Erd\H{o}s-R\'{e}nyi graphs. After choosing $k$ entries of the given matrix randomly and resampling them, we obtain another eigenvector $\mathbf{w}$ corresponding to the second-largest eigenvalue of the matrix obtained from the resampling procedure. We… Expand
#### References
SHOWING 1-10 OF 14 REFERENCES
Sparse random matrices have simple spectrum
• Mathematics
• 2018
Let $M_n$ be a class of symmetric sparse random matrices, with independent entries $M_{ij} = \delta_{ij} \xi_{ij}$ for $i \leq j$. $\delta_{ij}$ are i.i.d. Bernoulli random variables taking the valueExpand
Noise sensitivity of the top eigenvector of a Wigner matrix
• Mathematics
• 2019
We investigate the noise sensitivity of the top eigenvector of a Wigner matrix in the following sense. Let v be the top eigenvector of an $$N\times N$$ N × N Wigner matrix. Suppose that k randomlyExpand
Largest eigenvalues of sparse inhomogeneous Erdős–Rényi graphs
• Mathematics
• The Annals of Probability
• 2019
We consider inhomogeneous Erd\H{o}s-R\'enyi graphs. We suppose that the maximal mean degree $d$ satisfies $d \ll \log n$. We characterize the asymptotic behavior of the $n^{1 - o(1)}$ largestExpand
Local law and complete eigenvector delocalization for supercritical Erdős–Rényi graphs
• Mathematics, Physics
• The Annals of Probability
• 2019
We prove a local law for the adjacency matrix of the Erdős-Renyi graph $G(N, p)$ in the supercritical regime $pN \geq C\log N$ where $G(N,p)$ has with high probability no isolated vertices. In theExpand
The Largest Eigenvalue of Sparse Random Graphs
• Mathematics, Computer Science
• Combinatorics, Probability and Computing
• 2003
We prove that, for all values of the edge probability $p(n)$, the largest eigenvalue of the random graph $G(n, p)$ satisfies almost surely $\lambda_1(G)=(1+o(1))\max\{\sqrt{\Delta}, np\}$, where Δ isExpand
Spectral Statistics of Erdős-Rényi Graphs II: Eigenvalue Spacing and the Extreme Eigenvalues
• Mathematics, Physics
• 2012
We consider the ensemble of adjacency matrices of Erdős-Rényi random graphs, i.e. graphs on N vertices where every edge is chosen independently and with probability p ≡ p(N). We rescale the matrix soExpand
Spectral radii of sparse random matrices
• Mathematics
• 2017
We establish bounds on the spectral radii for a large class of sparse random matrices, which includes the adjacency matrices of inhomogeneous Erdős-Renyi graphs. Our error bounds are sharp for aExpand
Eigenvector Statistics of Sparse Random Matrices
• Mathematics
• 2016
We prove that the bulk eigenvectors of sparse random matrices, i.e. the adjacency matrices of Erdős-Renyi graphs or random regular graphs, are asymptotically jointly normal, provided the averagedExpand
Spectral statistics of Erdős–Rényi graphs I: Local semicircle law
• Mathematics, Physics
• 2013
We consider the ensemble of adjacency matrices of Erdős–Renyi random graphs, that is, graphs on N vertices where every edge is chosen independently and with probability p≡p(N). We rescale the matrixExpand
The Isotropic Semicircle Law and Deformation of Wigner Matrices
• Mathematics, Physics
• 2011
We analyze the spectrum of additive finite-rank deformations of N × N Wigner matrices H. The spectrum of the deformed matrix undergoes a transition, associated with the creation or annihilation of anExpand
|
2021-10-21 12:08:23
|
{"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9525493383407593, "perplexity": 1113.6375812404904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585405.74/warc/CC-MAIN-20211021102435-20211021132435-00228.warc.gz"}
|
http://mymathforum.com/algebra/11641-mathematic-induction.html
|
My Math Forum mathematic induction
Algebra Pre-Algebra and Basic Algebra Math Forum
March 7th, 2010, 03:44 AM #1 Newbie Joined: Mar 2010 Posts: 2 Thanks: 0 mathematic induction Hello, I don't know how to continue: Prove by mathematical induction that http://img230.imageshack.us/img230/6089/46951789.png is divisible by 6. I thought you have to use n = n+1 but I don't know how to continue. Does somebody know? thanks
March 7th, 2010, 04:10 AM #2 Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0 Re: mathematic induction If you show me what you have so far, I'll do my best to help. $6 \mid (5^{2n-1} + 1)$
March 7th, 2010, 05:28 AM #3 Newbie Joined: Mar 2010 Posts: 2 Thanks: 0 Re: mathematic induction Firstly, I did n = 1 so 5^1 + 1 = 6 Then I rewrote it to 5^2(n+1)-1 > (25)5^2n + 1 and then I kinda got stuck lol..
March 7th, 2010, 09:07 AM #4 Senior Member Joined: Apr 2008 Posts: 435 Thanks: 0 Re: mathematic induction You're pretty much there. I refer to two things: First, if a divides the product bc, but the gcd(a,b) = 1, what do we know about a and c? Second, what does your inductive hypothesis state? And how do we use that? For any proof by induction, we must use our hypothesis in some form or fashion. Is that clear enough? If not, I would recommend writing out your inductive hypothesis explicitly, and see if you can come up with something where it becomes useful. (easier than it sounds)
March 7th, 2010, 01:29 PM #5 Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689 Re: mathematic induction 5^(2n+1) +1=25[5^(2n-1)] +1=25[5^(2n-1)+1] -24. [5^(2n-1)+1] is divisible by 6 (induction hypothesis). 24 is divisible by 6. Result is divisible by 6.
Tags induction, mathematic
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post danoc93 Applied Math 16 October 2nd, 2013 03:47 PM maple_leaf Elementary Math 9 May 25th, 2013 06:41 AM Monox D. I-Fly Academic Guidance 0 February 24th, 2011 10:49 PM lulzimfazlija Algebra 2 December 31st, 1969 04:00 PM medos Applied Math 3 December 31st, 1969 04:00 PM
Contact - Home - Forums - Cryptocurrency Forum - Top
|
2019-04-21 16:30:33
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 1, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6730465292930603, "perplexity": 2906.8864587215166}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578531994.14/warc/CC-MAIN-20190421160020-20190421182020-00295.warc.gz"}
|
https://www.physicsforums.com/threads/pauli-hamiltonian.278960/
|
# Homework Help: Pauli Hamiltonian
1. Dec 11, 2008
### diegzumillo
1. The problem statement, all variables and given/known data
The Hamiltonian of an electron with mass m, electric charge q and spin
of $$\frac{\hbar }{2}\vec{\sigma}$$ in a magnetic field described by the
potential vector $$\vec{A}\left( \vec{r},t\right)$$ and a scalar potential $$U\left( \vec{r},t\right)$$ is given by
$$$H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\right] ^{2}+qU-\frac{q\hbar }{2m}\vec{ \sigma}.\vec{B}$$$
where $$\vec{B}=\vec{\nabla}\times \vec{A}$$. Show that this Hamiltonian can
also be obtained from Pauli Hamiltonian:
$$$H=\frac{1}{2m}\left\{ \vec{\sigma}.\left[ \vec{P}-q\vec{A}\right] \right\}^{2}+qU$$$
2. Relevant equations
I belive this one is useful here:
$$$\left( \vec{\sigma}.\vec{A}\right) \left( \vec{\sigma}.\vec{B}\right) =\vec{A}.\vec{B}I+i\vec{\sigma}.\left( \vec{A}\times \vec{B}\right)$$$
Wich in our case, we can rewrite it as
$$$\left( \vec{\sigma}.\vec{A}\right) ^{2}=A^{2}I+i\vec{\sigma}.\left( \vec{A}\times \vec{A}\right)$$$
(it's not the same vector A of the problem statement, of course)
3. The attempt at a solution
Using the above identity, we end up with a term like this:
$$$\left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] \times \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] =\vec{P}\times \vec{P}-\vec{P}\times q\vec{A}\left( \vec{R},t\right) -q\vec{A}\left( \vec{R},t\right)\times \vec{P}+q^{2}\vec{A}\left( \vec{R},t\right) \times \vec{A}\left( \vec{R},t\right)$$$
Wich is... almost nice. If I knew what to do with all of these guys! I can see that if we consider only the second term we can solve the problem. What does this mean?..
Using
$$$\vec{P}\rightarrow i\hbar \vec{\nabla}$$-\vec{P}\times q\vec{A}\left( \vec{R},t\right) =-i\hbar q\vec{\nabla}\times \vec{A}\left( \vec{R},t\right) =-i\hbar q\vec{B}$$$
And using this result in the Hamiltonian..
$$$H=\frac{1}{2m}\left\{ \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] ^{2}+i\vec{\sigma}.\left[ -i\hbar q\vec{B}\right] \right\} +qU\left( \vec{R},t\right)$$$$ $$H=\frac{1}{2m}\left\{ \left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right]^{2}+\hbar q\vec{\sigma}.\vec{B}\right\} +qU\left( \vec{R},t\right)$$$
$$$H=\frac{1}{2m}\left[ \vec{P}-q\vec{A}\left( \vec{R},t\right) \right] ^{2}+ \frac{\hbar q}{2m}\vec{\sigma}.\vec{B}+qU\left( \vec{R},t\right)$$$
2. Dec 11, 2008
### dextercioby
Actually
$$\hat{\vec{P}} = - i\hbar \vec{\nabla} \hat{1}$$ in ondulatory mechanics.
3. Dec 11, 2008
### diegzumillo
Oops!!
Thanks :] that lead to the correct answer, but still no reason why to cancel all those other terms!
I mean, the last term of the first equation in the 'attempt at a solution' is easy.. any regular vector (not the case with the derivative operator) the cross product with itself is 0.
I'm not sure this is the right path.. but I think there must be an argumento to justify this equation:
$$\vec{P}\times \vec{P}-q\vec{A}\times \vec{P}=0$$
4. Dec 12, 2008
### turin
Yeah, it's what you said. You're making the vanishing of the first and last terms too complicated. Of course, you have probably been viciously trained by now to hesitate when applying an old math rule to a new situation, and, for instance, you are probably afraid of things like PxP because you know that P is an operator, and you also remember that JxJ is NOT zero. OK, so good for you; this should be your initial attitude. In fact, calling A a "regular vector" (i.e. c-number vector) is now a big no-no. However, now let's decide what these cross products are.
Hints:
What are the commutators of X, Y, and Z among themselves?
What are the commutators of Px, Py, and Pz among themselves?
Does A depend on any operators other than X, Y, and Z?
Does P depend on any operators other than Px, Py, and Pz?
Write the cross products in terms of components and relate this expression to commutators.
Is it possible for any of these commutators to be nonzero?
OK, that takes care of first and last term. However, there are TWO terms left over, not just the one. You seem to have no problem with this. Can you explain?
Last edited: Dec 12, 2008
5. Dec 12, 2008
### diegzumillo
Hi Turin!
Thanks for taking the time :]
Well, I was indeed a little afraid of P! Because there was a little contradiction in my head... While I considered P as an general operator that its components commute with each other, the cross product with itself should vanish. But When I considered it to be the gradient operator the commutation didnt seem obvious anymore.
But now that you rubbed this in my nose :D I can think clearer. Partial derivatives on different variables will commute if the function has continuous second partial derivatives. (there's probably a theorem for that) So, PXP will vanish in every way I approach this.
Now, for the term qAXP, I don't really know. Wich lead me to your third question: "Does A depend on any operators other than X, Y, and Z?". er... I dont know! The only thing I know about it is that it's curl is the magnetic field... Sometimes I feel the answer is right below my nose.
6. Dec 15, 2008
### turin
Hint 1: You are definitely on the right track when you consider that P is actually a derivative operator, and how it should operate on functions. This gets at a very important point in QM: the operators are to some extent arbitrary, but their matrix elements had better behave. In particular, try to consider not only the right-action, but also the left-action of the operators.
Hint 2: Is it really true that PxA is proportional to B? (Use hint 1 to approach this question.)
BTW, I feel your pain. This is on my top ten list for most difficult issues in QM. Fortunately, this one has an explanation that I (think I) understand.
7. Feb 12, 2009
### Jezuz
You must be very careful when working with operators. Remember that you should always assume that it is multiplied with a function. This means that
$$\mathbf p \times \mathbf A \neq -i\hbar (\nabla \times \mathbf A).$$
Rather we have
$$\mathbf p \times \mathbf A \psi = - i\hbar \nabla \times (\mathbf A \psi) = - i\hbar \mathb B \psi - i\hbar \mathbf A \times \nabla \psi .$$
|
2018-11-14 19:53:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000094175338745, "perplexity": 8480.989917272796}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742263.28/warc/CC-MAIN-20181114191308-20181114213308-00431.warc.gz"}
|
http://gmatclub.com/forum/kay-began-a-certain-game-with-x-chips-on-each-of-the-next-145373.html?sort_by_oldest=true
|
Kay began a certain game with x chips. On each of the next : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack
It is currently 18 Jan 2017, 17:05
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Kay began a certain game with x chips. On each of the next
Author Message
TAGS:
### Hide Tags
Director
Joined: 29 Nov 2012
Posts: 898
Followers: 14
Kudos [?]: 1044 [2] , given: 543
Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 04:27
2
KUDOS
6
This post was
BOOKMARKED
00:00
Difficulty:
55% (hard)
Question Stats:
60% (02:16) correct 40% (01:42) wrong based on 256 sessions
### HideShow timer Statistics
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
[Reveal] Spoiler: OA
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077
Kudos [?]: 93113 [1] , given: 10552
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 04:52
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
fozzzy wrote:
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
On the first play she lost $$\frac{x}{2}+1$$ chips and she was left with $$x-(\frac{x}{2}+1)=\frac{x-2}{2}$$ chips.
On the second play she lost $$\frac{x-2}{4}+1$$ chips.
So, we have that $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.
We can also solve this question backward:
At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5.
The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.
Hope it's clear.
_________________
Moderator
Joined: 01 Sep 2010
Posts: 3090
Followers: 783
Kudos [?]: 6520 [0], given: 1012
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 04:52
well here is difficult to attack the question upfront, but reading carefully the stem you can do.
she lost one more than half the number of chips she had at the beginning of that play and she ends with 5 chips.
Work in reverse engineering. If she finishes with 5 chips and after each game she lost the half of chip + 1. 5*2= 10 + 1 =11 then 11*2=22 +1 =23
At the beginning she has 23 chips. So answer must be D
Regards
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077
Kudos [?]: 93113 [1] , given: 10552
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 04:58
1
KUDOS
Expert's post
carcass wrote:
well here is difficult to attack the question upfront, but reading carefully the stem you can do.
she lost one more than half the number of chips she had at the beginning of that play and she ends with 5 chips.
Work in reverse engineering. If she finishes with 5 chips and after each game she lost the half of chip + 1. 5*2= 10 + 1 =11 then 11*2=22 +1 =23
At the beginning she has 23 chips. So answer must be D
Regards
At the beginning she had 26 chips not 23. Check here: kay-began-a-certain-game-with-x-chips-on-each-of-the-next-145373.html#p1165407
Notice that x cannot be odd, because she lost one more than half the number of chips, which means that the number of chips must be even.
_________________
Moderator
Joined: 01 Sep 2010
Posts: 3090
Followers: 783
Kudos [?]: 6520 [0], given: 1012
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 05:04
Bunuel wrote:
carcass wrote:
well here is difficult to attack the question upfront, but reading carefully the stem you can do.
she lost one more than half the number of chips she had at the beginning of that play and she ends with 5 chips.
Work in reverse engineering. If she finishes with 5 chips and after each game she lost the half of chip + 1. 5*2= 10 + 1 =11 then 11*2=22 +1 =23
At the beginning she has 23 chips. So answer must be D
Regards
At the beginning she had 26 chips not 23. Check here: kay-began-a-certain-game-with-x-chips-on-each-of-the-next-145373.html#p1165407
Notice that x cannot be odd, because she lost one more than half the number of chips, which means that the number of chips must be even.
Got it $$N+1$$.........
But here is a risk to be a bit careless or depends on the question posed ?? I mean the logic was correct, after all
Thanks for suggesting
_________________
Director
Joined: 29 Nov 2012
Posts: 898
Followers: 14
Kudos [?]: 1044 [0], given: 543
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 05:27
Bunuel wrote:
fozzzy wrote:
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
On the first play she lost $$\frac{x}{2}+1$$ chips and she was left with $$x-(\frac{x}{2}+1)=\frac{x-2}{2}$$ chips.
On the second play she lost $$\frac{x-2}{4}+1$$ chips.
So, we have that $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.
We can also solve this question backward:
At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5.
The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.
Hope it's clear.
I didn't understand this step did you divide by 2?
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
Math Expert
Joined: 02 Sep 2009
Posts: 36548
Followers: 7077
Kudos [?]: 93113 [0], given: 10552
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 05:34
fozzzy wrote:
Bunuel wrote:
fozzzy wrote:
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
On the first play she lost $$\frac{x}{2}+1$$ chips and she was left with $$x-(\frac{x}{2}+1)=\frac{x-2}{2}$$ chips.
On the second play she lost $$\frac{x-2}{4}+1$$ chips.
So, we have that $$x-(\frac{x}{2}+1)-(\frac{x-2}{4}+1)=5$$ --> $$x=26$$.
We can also solve this question backward:
At the end Kay had 5 chips, so before that she had (5+1)*2=12 chips: 12-(12/2+1)=5.
The same way, she had 12 chips, so before that she had (12+1)*2=26 chips: 26-(26/2+1)=12.
Hope it's clear.
I didn't understand this step did you divide by 2?
On the second play she also lost one more than half the number of chips she had at the beginning of that play. Since at the begging of the second play she had $$\frac{x-2}{2}$$ chips, then she lost $$\frac{(\frac{x-2}{2})}{2}+1=\frac{x-2}{4}+1$$ chips.
Hope it's clear.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7121
Location: Pune, India
Followers: 2133
Kudos [?]: 13640 [5] , given: 222
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
07 Jan 2013, 09:12
5
KUDOS
Expert's post
fozzzy wrote:
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
The most important thing to understand here is this: she loses one more than half the number she has. This implies that at the end of a play, she is left with one less than half the number she has at the beginning.
After two plays, she is left with 5 (which is 1 less than half of what she had at the beginning of the second play). So she had 6*2 = 12 at the end of the first play. Since 12 is one less than half of what she had at the beginning of the first play, she must have had 13*2 = 26 at the beginning of the first play.
Hence, x = 26
_________________
Karishma
Veritas Prep | GMAT Instructor
My Blog
Get started with Veritas Prep GMAT On Demand for \$199
Veritas Prep Reviews
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13439
Followers: 575
Kudos [?]: 163 [0], given: 0
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
05 Dec 2014, 10:19
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13439
Followers: 575
Kudos [?]: 163 [0], given: 0
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
16 Dec 2015, 07:33
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Intern
Joined: 03 Jul 2016
Posts: 1
Followers: 0
Kudos [?]: 0 [0], given: 24
Re: Kay began a certain game with x chips. On each of the next [#permalink]
### Show Tags
02 Nov 2016, 18:18
fozzzy wrote:
Kay began a certain game with x chips. On each of the next two plays, she lost one more than half the number of chips she had at the beginning of that play. If she had 5 chips remaining after her two plays, then x is in the interval:
A. 7<x<12
B. 13<x<18
C. 19<x<24
D. 25<x<30
E. 31<x<35
Here's how I solved it:
At first she had x chips
After the first play, she had x/2 -1 chips
After the second play, she had 1/2 (x/2 -1)-1 which equals 5.
Solving for x, we get:
1/2 (x/2 -1)-1 =5
1/2 (x/2 -1) = 6
x/2 -1 =12
x/2 =13
x =26
Re: Kay began a certain game with x chips. On each of the next [#permalink] 02 Nov 2016, 18:18
Similar topics Replies Last post
Similar
Topics:
In the game of Dubblefud, red chips, blue chips and green chips are 3 28 Apr 2016, 02:46
6 In a certain game, each player scores either 2 points or 5 points. If 22 03 Apr 2015, 04:05
20 For each players turn in a certain board game, a card is 6 02 Nov 2012, 13:43
For a certain game, three numbers are drawn, each from 1 to 2 07 May 2012, 03:50
12 For each player's turn in a certain board game, a card is 7 06 May 2012, 09:29
Display posts from previous: Sort by
|
2017-01-19 01:05:23
|
{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6302865743637085, "perplexity": 4014.623716476811}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560280410.21/warc/CC-MAIN-20170116095120-00502-ip-10-171-10-70.ec2.internal.warc.gz"}
|
https://tex.stackexchange.com/questions/419017/operator-font-with-newtx
|
# Operator font with newtx
I am using \usepackage{newtxmath,newtxtext}. But, \operatorname{} seems to use another font.
How to set up accordingly?
• \usepackage{newtxtext,newtxmath} – egreg Mar 7 '18 at 23:51
• @egreg, ohhh I never imagined that the order could be a solution. Simpler than this, impossible. – Sigur Mar 7 '18 at 23:53
The correct order is the reverse, first newtxtext and then newtxmath.
\documentclass{article}
\usepackage{newtxtext,newtxmath}
\usepackage{amsmath}
\begin{document}
notations: $Y=N-\operatorname{int}(U)$
\end{document}
In newtxmath.sty we find
\DeclareSymbolFont{operators}{\tx@enc}{\rmdefaultB}{m}{n}
and \rmdefaultB is \rmdefault, which still is Computer Modern if newtxtext has not yet been loaded,
• Do you mean "if newtxtext has not yet been loaded"? – user94293 Mar 8 '18 at 0:34
• @user94293, I think so. – Sigur Mar 8 '18 at 1:01
• @user94293 Yes, probably I inverted Sigur’s inversion and the inverted again… :-) – egreg Mar 8 '18 at 7:34
|
2020-01-24 16:34:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8848610520362854, "perplexity": 6827.652128823251}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250624328.55/warc/CC-MAIN-20200124161014-20200124190014-00269.warc.gz"}
|
https://math.stackexchange.com/questions/2452040/why-the-expansion-of-functions-of-2-variables-in-legendre-polynomials-does-not-t
|
# Why the expansion of functions of 2 variables in legendre polynomials does not take into account all products Pi (x) Pj (y)?
My question arises from this reading link (In this article, in equation 6 on page 3, also expands a function of 2 variables in this way Parametric estimate of intensity inhomogeneities applied to MRI), where on page 10 expand a function of 2 variables in a sum of two-dimensional legendre polynomials:
My question is why in this case they do not use all the orthogonal combinations Pi (x) Pj (y) in the summation as is usual in other cases such as for example the discrete expansion of fourier:
• In English, no need for an interrogation point in front of a sentence. – Jean Marie Sep 30 '17 at 21:11
• This approach results in a polynomial of maximal degree $n$, i.e. the powers of $x$ and $y$ add to at most $n$. If one uses all combinations $P_i(x)P_j(x)$ with $i, j \le n$, then there could be terms such as $x^4y^4$ if $n = 4$, but no terms $x^6y^2$. – Hans Engler Sep 30 '17 at 21:22
• yes, it is true what you mention, in fact in the expansion in series of taylos in 2 variables for example, the polynomials that appear in each of the summands have the same degree, for example in the component of second order appear the polynomials $x^{2}$,$xy$ and $y^{2}$. – Roger Figueroa Quintero Sep 30 '17 at 22:15
• I will have to study a little more about why this way of combining the polynomials results in a base generating the vector space of the real functions. – Roger Figueroa Quintero Sep 30 '17 at 22:23
|
2019-08-24 20:39:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8005021214485168, "perplexity": 318.79857897120706}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321696.96/warc/CC-MAIN-20190824194521-20190824220521-00114.warc.gz"}
|
https://mathoverflow.net/questions/289368/uniform-convergence-of-resolvents-of-markov-processes
|
Uniform convergence of resolvents of Markov processes
I have a question about convergence of resolvents of Markov processes.
Let $X$, $X^n$ be Markov processes on a locally compact separable metric space $E$. We denote $\{ R_{\alpha}\}_{\alpha>0}$ and $\{ R_{\alpha}^n\}_{\alpha>0}$ by the resolvents of $X$ and $X^n$, respectively.
We assume the following:
• for any bounded measurable function $f$ and $\alpha>0$, $R_{\alpha}f$ and $R_{\alpha}^nf$ are continuous functions on $E$.
• for any bounded measurable function $f$ and $\alpha>0$, $\lim_{n \to \infty}\sup_{x \in E}|R_{\alpha}f(x)-R_{\alpha}^nf(x)|=0$.
My question
Let $\{p_t\}_{t>0}$ and $\{p_t^n\}$ be the semigroups of $X$ and $X^n$, respectively. Then, can we have the following?
• for any bounded measurable function $f$ and $t>0$, $\lim_{n \to \infty}\sup_{x \in E}|p_tf(x)-p_t^nf(x)|=0$.
There exists a related result in Theorem 3.4.2 of Pagy's book. However we cannot apply this results to my question since the space of continuous functions on $E$ is not a Banach space. If you know any other related results, please let me know.
1 Answer
In general: no.
Let $X$ be the uniform motion to the right with speed $1$ (so $X_t = X_0 + t$), and let $X^n$ be the uniform motion to the right with speed $1$ plus an independent Brownian motion with variance $t/(2 n^2)$. Then $R_\alpha$ is the convolution operator with kernel $e^{\alpha x} \mathbf{1}_{(-\infty, 0)}(x)$, while $R_\alpha^n$ is the convolution operator with kernel $$\frac{n}{\sqrt{\alpha + n^2}} \left( e^{2 n ((\alpha + n^2)^{1/2} - n) x} \mathbf{1}_{(-\infty, 0)}(x) + e^{-2 n ((\alpha + n^2)^{1/2} + n) x} \mathbf{1}_{(0, \infty)}(x) \right) .$$ Your first condition is satisfied: convolution of a bounded function with an integrable kernel is always continuous. It is also easy to see that the kernels of $R_\alpha^n$ converge in $L^1$ to the kernel of $R_\alpha$, which implies your second condition.
On the other hand it is clear that the distribution of $X_t^n$ is absolutely continuous, so $p_t^n f$ is continuous as a convolution of a bounded measurable function with an integrable kernel. It follows that if $f$ is discontinuous, $p_t^n f(x)$ cannot converge uniformly to $p_t f(x) = f(x + t)$.
|
2019-10-19 00:53:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9774511456489563, "perplexity": 66.17947055173333}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986685915.43/warc/CC-MAIN-20191018231153-20191019014653-00426.warc.gz"}
|
https://socratic.org/questions/how-do-you-solve-y-x-1-and-y-2x-1
|
# How do you solve y = x+1 and y = 2x -1?
Aug 3, 2015
$\textcolor{red}{x = 2 , y = 3}$
#### Explanation:
One way is to use the method of substitution.
Step 1. Enter the equations.
[1] $y = x + 1$
[2] $y = 2 x - 1$
Step 2. Substitute the value of $y$ from Equation 1 into Equation 2.
$y = x + 1$
$2 x - 1 = x + 1$
$2 x - x = 1 + 1$
[3] $x = 2$
Step 3. Substitute Equation 3 into Equation 1.
$y = x + 1$
$y = 2 + 1$
$y = 3$
Solution: $x = 2 , y = 3$
Check: Substitute the values of $x$ and $y$ in Equations 1 and2.
(a) In Equation 1
$y = x + 1$
$3 = 2 + 1$
$3 = 3$
(b) In Equation 2
$y = 2 x - 1$
$3 = 2 \left(2\right) - 1$
$3 = 4 - 1$
$3 = 3$
It checks!
Our solution is correct.
|
2019-09-18 03:08:13
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 21, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8316300511360168, "perplexity": 1380.2314462781976}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573176.51/warc/CC-MAIN-20190918024332-20190918050332-00167.warc.gz"}
|
http://atekihcan.github.io/CLRS/E04.03-03/
|
We saw that the solution of $T(n) = 2T(\lfloor n/2 \rfloor) + n$ is $O(n \lg n)$. Show that the solution of this recurrence is also $\Omega(n \lg n)$. Conclude that the solution is $\Theta(n \lg n)$.
Let us assume $T(n) \ge c (n + b) \lg (n + b)$ for all $n \ge n_0$, where $b$, $c$, and $n_0$ are positive constants.
The last step holds as long as $(n - cn - dc) \ge 0 \implies n \ge cd / (1 - c)$. For example, $d = 0$, $n \ge 1$, and $% $.
|
2017-08-17 23:03:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 13, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9925447702407837, "perplexity": 23.653864289526553}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886104172.67/warc/CC-MAIN-20170817225858-20170818005858-00071.warc.gz"}
|
https://www.sparrho.com/item/isotropisation-of-flat-homogeneous-universes-with-scalar-fields/920adb/
|
# Isotropisation of flat homogeneous universes with scalar fields
Research paper by Stephane Fay, Jean-Pierre Luminet
Indexed on: 18 Feb '04Published on: 18 Feb '04Published in: General Relativity and Quantum Cosmology
#### Abstract
Starting from an anisotropic flat cosmological model(Bianchi type $I$), we show that conditions leading to isotropisation fall into 3 classes, respectively 1, 2, 3. We look for necessary conditions such that a Bianchi type $I$ model reaches a stable isotropic state due to the presence of several massive scalar fields minimally coupled to the metric with a perfect fluid for class 1 isotropisation. The conditions are written in terms of some functions $\ell$ of the scalar fields. Two types of theories are studied. The first one deals with scalar tensor theories resulting from extra-dimensions compactification, where the Brans-Dicke coupling functions only depend on their associated scalar fields. The second one is related to the presence of complex scalar fields. We give the metric and potential asymptotical behaviours originating from class 1 isotropisation. The results depend on the domination of the scalar field potential compared to the perfect fluid energy density. We give explicit examples showing that some hybrid inflation theories do not lead to isotropy contrary to some high-order theories, whereas the most common forms of complex scalar fields undergo a class 3 isotropisation, characterised by strong oscillations of the $\ell$ functions.
|
2021-05-16 02:12:50
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.46410611271858215, "perplexity": 800.1007247706375}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991659.54/warc/CC-MAIN-20210516013713-20210516043713-00104.warc.gz"}
|
http://mathhelpforum.com/advanced-algebra/82024-how-determine-coefficient-irreducible-polynomial-help.html
|
# Math Help - How to determine coefficient in irreducible polynomial? HELP!!
1. ## How to determine coefficient in irreducible polynomial? HELP!!
Here is my problem:
Say if I have a irreducible polynomial for $GF(2^2)$ that is $x^2 + Nx + 1$, (N = elements of $GF(2)$) how do I determine the possible values of N ? Given the irreducible polynomial for $GF(2)$ is $x^2 + x + 1$
thanks alot!
2. Originally Posted by classic_phohe
Here is my problem:
Say if I have a irreducible polynomial for $GF(2^2)$ that is $x^2 + Nx + 1$, (N = elements of $GF(2)$) how do I determine the possible values of N ? Given the irreducible polynomial for $GF(2)$ is $x^2 + x + 1$
thanks alot!
Here is a field of order 4, $F=\mathbb{Z}_2[t]/(t^2+t+1)$. Therefore, $F = \{ 0,1,\alpha,\alpha+1\}$ where $\alpha = t + (t^2+t+1)$ and $\alpha^2 = \alpha + 1$.
For $x^2 + Nx+1$ to be irreducible it means it cannot have any zeros in $F$. The possible polynomials are: $x^2+1,x^2+x+1,x^2+\alpha x+1, x^2+(\alpha+1)x+1$. Which of these polynomials has zeros in $F$? The ones that do not are the irreducible ones.
3. Originally Posted by ThePerfectHacker
Here is a field of order 4, $F=\mathbb{Z}_2[t]/(t^2+t+1)$. Therefore, $F = \{ 0,1,\alpha,\alpha+1\}$ where $\alpha = t + (t^2+t+1)$ and $\alpha^2 = \alpha + 1$.
For $x^2 + Nx+1$ to be irreducible it means it cannot have any zeros in $F$. The possible polynomials are: $x^2+1,x^2+x+1,x^2+\alpha x+1, x^2+(\alpha+1)x+1$. Which of these polynomials has zeros in $F$? The ones that do not are the irreducible ones.
can I conclude that N can only either be 1 or 3 which is $\alpha$ and $\alpha +1$?
but I realize if I would like to derive elements of field 16 using that irreducible polynomial $x^2+\alpha x+1, x^2+(\alpha+1)x+1$, it cant be done.
unless I use $x^2+\alpha x+\alpha, x^2+(\alpha+1)x+(\alpha+1)$ then only I can produce 16 elements from the polynomials.
can you guide/advise me with this?
4. Originally Posted by classic_phohe
can I conclude that N can only either be 1 or 3 which is $\alpha$ and $\alpha +1$?
but I realize if I would like to derive elements of field 16 using that irreducible polynomial $x^2+\alpha x+1, x^2+(\alpha+1)x+1$, it cant be done.
unless I use $x^2+\alpha x+\alpha, x^2+(\alpha+1)x+(\alpha+1)$ then only I can produce 16 elements from the polynomials.
can you guide/advise me with this?
The polynomial $x^2+1$, it has a zero $x=1$.
The polynomial $x^2+x+1$, it has a zero $x=\alpha$.
The polynomial $x^2 + \alpha x + 1$, it has no zero - so irreducible.
The polynomial $x^2+(\alpha+1)x+1$, it has no zero - so irreducible.
Therefore, $F[x]/(x^2+\alpha x + 1)\text{ and }F[x]/(x^2+(\alpha+1)x+1)$ give rise to finite fields of 16 elements.
5. Ok! I agree what you explained. Now let sat I choose $x^2 + \alpha x + 1$ to be the irreducible polynomial, hence I use this to produce the elements in field of 16:
$x^0 = 1$
$x^1 = x$
$x^2 = \alpha x+ 1$
$x^3 = \alpha x + \alpha$
$x^4 = x + \alpha$
$x^5 = 1$
From there Im stuck. Can you give some tips how should I proceed with this? or I have already done wrongly at the first place?!
6. Originally Posted by classic_phohe
Ok! I agree what you explained. Now let sat I choose $x^2 + \alpha x + 1$ to be the irreducible polynomial, hence I use this to produce the elements in field of 16:
$x^0 = 1$
$x^1 = x$
$x^2 = \alpha x+ 1$
$x^3 = \alpha x + \alpha$
$x^4 = x + \alpha$
$x^5 = 1$
From there Im stuck. Can you give some tips how should I proceed with this? or I have already done wrongly at the first place?!
Any element in $F[x]/(x^2+\alpha x + 1)$ is $A+Bx + (x^2+\alpha x + 1)$. Let $\beta = x + (x^2+\alpha x + 1)$. Then $\text{GF}(2^4)$ would be $A + B\beta$ where $A,B\in \{0,1,\alpha,\alpha+1\}$ and $\beta$ has the property that $\beta^2 = \alpha \beta + 1$ (since it solves $x^2+\alpha x + 1$).
7. Originally Posted by ThePerfectHacker
Any element in $F[x]/(x^2+\alpha x + 1)$ is $A+Bx + (x^2+\alpha x + 1)$. Let $\beta = x + (x^2+\alpha x + 1)$. Then $\text{GF}(2^4)$ would be $A + B\beta$ where $A,B\in \{0,1,\alpha,\alpha+1\}$ and $\beta$ has the property that $\beta^2 = \alpha \beta + 1$ (since it solves $x^2+\alpha x + 1$).
thanks for the reply, but honestly I dont quite get it. Can you please explain further?
$A+Bx + (x^2+\alpha x + 1)$, with $\beta = x + (x^2+\alpha x + 1)$ how to become $A + B\beta$? wont this make the element become $A + Bx + B(x^2+\alpha x + 1)$ ?
8. Originally Posted by classic_phohe
thanks for the reply, but honestly I dont quite get it. Can you please explain further?
$A+Bx + (x^2+\alpha x + 1)$, with $\beta = x + (x^2+\alpha x + 1)$ how to become $A + B\beta$? wont this make the element become $A + Bx + B(x^2+\alpha x + 1)$ ?
The field with 16 elements is the set $\{ [A+Bx]:A,B\in \mathbb{F}_4 \}$ where $[ A + Bx]$ is the equivalent class modulo $x^2+\alpha x+1$. Notice that $[A+Bx] = A[1] + B[x]$ where $[1]$ is the equivalence class of $1$ modulo $x^2+\alpha x +1$ and $[x]$ is the equivalence class of $x$ modulo $x^2+\alpha x + 1$. Thus, what I am saying is that let $\beta = [x]$ then every element has form $A + B\beta$ (here we thinking of $A$ as an entire equivalence class).
9. Originally Posted by ThePerfectHacker
The field with 16 elements is the set $\{ [A+Bx]:A,B\in \mathbb{F}_4 \}$ where $[ A + Bx]$ is the equivalent class modulo $x^2+\alpha x+1$. Notice that $[A+Bx] = A[1] + B[x]$ where $[1]$ is the equivalence class of $1$ modulo $x^2+\alpha x +1$ and $[x]$ is the equivalence class of $x$ modulo $x^2+\alpha x + 1$. Thus, what I am saying is that let $\beta = [x]$ then every element has form $A + B\beta$ (here we thinking of $A$ as an entire equivalence class).
thank you very much!!!
so field with 16 elements is the set $\{ [A+Bx]:A,B\in \mathbb{F}_4 \}$so I will have 16 combination of A and B and hence produce 16 elements of field 16 with modulo the irreducible polynomial ; $\beta^2 = \alpha \beta + 1$ as $\beta = [x]$.
let say I proceed to field 256, I have the irreducible polynomial as $x^2 + x + \nu$ and I just use the same method you mentioned earlier reply to determine the possible value of $\nu$ ?
then I shall have the elements define as $\{ [A+Bx]:A,B\in \mathbb{F}_{16} \}$ ?
10. I believe the one you explain to me are all in polynomial basis. What if I would like to use Normal basis? Can you brief me how can it be done??
And also, the 16 elements produced by $x^2 + \alpha x + 1$ are in polynomial forms, each of them represent number from 0 to 15 right? let say i would like to identify number 2 and 14, is there any direct way to find which element representations are they in?
|
2014-12-28 17:38:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 114, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9309629201889038, "perplexity": 223.80184008623192}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447558371.2/warc/CC-MAIN-20141224185918-00057-ip-10-231-17-201.ec2.internal.warc.gz"}
|
https://www.shaalaa.com/question-bank-solutions/with-a-neat-labelled-diagram-show-that-all-harmonics-are-present-in-an-air-column-contained-in-a-pipe-open-at-both-the-ends-define-end-correction-study-vibrations-air-columns_3106
|
# With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction. - Physics
With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction.
#### Solution
Stationary waves are produced due to the superposition of two identical simple harmonic progressive waves traveling through the same part of the medium in opposite direction. In the case of pipe open at both ends, the air molecules near the open end are free to vibrate, hence they vibrate with maximum amplitude. Therefore the open end becomes an antinode.
The simplest mode of vibration is the fundamental mode of vibration. (in above fig a ) In this case one node & two antinodes are formed. If λ1 be the corresponding wavelength &
"L be length of the pipe" "L" = lambda_1 / 2
lambda_1 = 2 "L"
If n1 be corresponding frequency and v be velocity of sound in air v = "n"_1 lambda_1
"n"_1 = "v"/lambda_1
n_1= "v"/(2"L")......(1) This is called 1st harmonic.
The next possible mode of vibration is called the 1st overtone. (in above fig) In this case, two nodes and three antinodes are formed. If λ2 and n2 be the corresponding wavelength and frequency,
L = λ2
λ2 = L
But
v n2λ2 i.e = λ2 = v/λ_2 n2 = "v"/"L"
This is called 2nd harmonic.
The next possible mode of vibration is called 2nd overtone.(in above fig c) In this case three nodes & four antinodes are formed. If λ3 and n3 be the corresponding wavelength & frequency ,
L=3lambda_3/2
lambda_3=(2L)/3
v=n_3 lambda_3
i.e
n_3=v/lambda_3
n3 = 3"v"/2L .
This is called 3rd harmonic.
n_1 : n_2 : n_3 : : 1 : 2 : 3
Thus in case of pipe open at both end, all harmonics are present.
In this case, the node is forming at the closed-end i.e. at the water surface & antinode is forming at the open end. But antinode is not formed exactly at the open end but slightly outside it since air molecules are free to vibrate. The correction has to be applied is called end correction.
Concept: Study of Vibrations of Air Columns
Is there an error in this question or solution?
#### APPEARS IN
Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 6 Superposition of Waves
Exercises | Q 9 | Page 156
|
2021-10-24 04:11:23
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8192406296730042, "perplexity": 1630.0657743387533}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585837.82/warc/CC-MAIN-20211024015104-20211024045104-00250.warc.gz"}
|
https://bigdave1583.wordpress.com/2015/08/
|
# Taxi from HELL!!
Ok so I’m running a little late for work so I’m like fuck it! I don’t want to take the subway because it will take to long so I thought I’ll just jump into a cab so I can get there faster. I WAS SOOOOOO WRONG!!!!!!!!!!! This dumb ass cab took the Westside highway at rush hour. FUCKING DUMB ASSSSSSSSSS!!!!!!!! I’m like what the FUCK!! Are you doing? he’s like this is the quickest way. Quickest way my ass this is why Urber is taking over and these cabs are going to be long forgotten.
|
2021-10-28 05:27:48
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8418068885803223, "perplexity": 3277.7850437024726}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323588257.34/warc/CC-MAIN-20211028034828-20211028064828-00080.warc.gz"}
|
https://zbmath.org/?q=an%3A0535.49010
|
zbMATH — the first resource for mathematics
Strongly nonlinear unilateral problems. (English) Zbl 0535.49010
The authors prove the existence of a solution for the obstacle problem of the type $u\in K,\quad g(.,u)\in L^ 1(\Omega),\quad g(.,u)u\in L^ 1(\Omega),$ $<A(u)-f,v-u>+\int_{\Omega}g(.,u)(v-u)\geq 0\quad for\quad all\quad v\in K\cap L^{\infty}(\Omega),$ where A is a nonlinear differential operator of order 2m ($$m\geq 1)$$ satisfying the classical Leray-Lions conditions and giving rise to a ”good” operator on $$H^{m,p}(\Omega) (1<p<\infty)$$, g is a strongly nonlinear term satisfying essentially the sign condition, V is either $$H^{m,p}(\Omega)$$ or $$H_ 0\!^{m,p}(\Omega)$$, $$f\in V^*$$ and $$K=\{v\in V : v\geq \psi$$ in $$\Omega\}$$ with $$\psi \in V\cap L^{\infty}(\Omega)$$ and $$\Omega$$ some domain in $${\mathbb{R}}^ n.$$
An essential tool in the proof is the approximation result by L. J. Hedberg [Acta Math. 147, 237-264 (1981; Zbl 0504.35018)]. Some further properties of the solution are also proved and the bilateral obstacle problem is discussed. It is not hard to see that the restrictive assumption $$\psi\in V$$ can be removed.
Reviewer: V.Mustonen
MSC:
49J40 Variational inequalities 35J60 Nonlinear elliptic equations 35J85 Unilateral problems; variational inequalities (elliptic type) (MSC2000) 47J05 Equations involving nonlinear operators (general) 46E35 Sobolev spaces and other spaces of “smooth” functions, embedding theorems, trace theorems
Full Text:
References:
[1] Br?zis H (1968) Equations et in?quations non lin?aires dans les espaces vectoriels en dualit?. Ann Inst Fourier (Grenoble) 18:115-175 · Zbl 0169.18602 [2] Br?zis H, Browder FE (1978) Strongly nonlinear elliptic boundary value problems. Ann Sc Norm Sup Pisa 5:587-603 · Zbl 0453.35029 [3] Br?zis H, Broder FE (to appear) Some properties of higher order Sobolev spaces. J Math Pures Appl [4] Browder FE (1977) Pseudomonotone operators and nonlinear elliptic boundary value problems on unbounded domains. Proc Nat Acad Sci 74:2659-2661 · Zbl 0358.35034 · doi:10.1073/pnas.74.7.2659 [5] Edmunds DE, Webb JRL (1973) Quasilinear elliptic problems in unbounded domains. Proc Roy Soc London, Ser A, 337:397-410 · Zbl 0263.35034 [6] Hedberg LI (1978) Two approximation problems in function spaces. Ark Mat 16:51-81 · Zbl 0399.46023 · doi:10.1007/BF02385982 [7] Hedberg LI (to appear) Spectral synthesis in Sobolev spaces. Acta Math [8] Hess P (1973) Variational inequalities for strongly nonlinear elliptic operators. J Math Pures Appl 52:285-298 · Zbl 0222.47020 [9] Hess P (1974) On a class of strongly nonlinear elliptic variational inequalities. Math Ann 211:289-297 · Zbl 0285.49003 · doi:10.1007/BF01418226 [10] Mustonen V (1978) A class of strongly nonlinear variational inequalities. J London Math Soc (2) 18:157-165 · Zbl 0392.35022 · doi:10.1112/jlms/s2-18.1.157 [11] Mustonen V (1979) A class of strongly nonlinear variational inequalities in unbounded domains. J. London Math Soc (2) 19:319-328 · Zbl 0404.35035 · doi:10.1112/jlms/s2-19.2.319 [12] Webb JRL (1980) Boundary value problems for strongly nonlinear elliptic equations. J London Math Soc (2) 21:123-132 · Zbl 0438.35029 · doi:10.1112/jlms/s2-21.1.123 [13] Stein EM (1970) Singular integrals and differentiability properties of functions. Princeton University Press, Princeton · Zbl 0207.13501
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
|
2021-06-20 03:18:45
|
{"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6326325535774231, "perplexity": 3539.682970132848}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487655418.58/warc/CC-MAIN-20210620024206-20210620054206-00023.warc.gz"}
|
https://doc.sagemath.org/html/en/reference/modfrm/sage/modular/modform/j_invariant.html
|
# q-expansion of j-invariant¶
sage.modular.modform.j_invariant.j_invariant_qexp(prec=10, K=Rational Field)
Return the $$q$$-expansion of the $$j$$-invariant to precision prec in the field $$K$$.
If you want to evaluate (numerically) the $$j$$-invariant at certain points, see the special function elliptic_j().
Warning
Stupid algorithm – we divide by Delta, which is slow.
EXAMPLES:
sage: j_invariant_qexp(4)
q^-1 + 744 + 196884*q + 21493760*q^2 + 864299970*q^3 + O(q^4)
sage: j_invariant_qexp(2)
q^-1 + 744 + 196884*q + O(q^2)
sage: j_invariant_qexp(100, GF(2))
q^-1 + q^7 + q^15 + q^31 + q^47 + q^55 + q^71 + q^87 + O(q^100)
|
2019-06-27 08:12:23
|
{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9954420328140259, "perplexity": 2921.534325187602}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560628001014.85/warc/CC-MAIN-20190627075525-20190627101525-00068.warc.gz"}
|
https://mathematica.stackexchange.com/questions/89749/styling-the-edges-of-a-graph-according-to-the-multiplicities-of-the-edges
|
Styling the edges of a graph according to the multiplicities of the edges
I used the following code to make a simple triangle, in which the nodes have different effects each other:
data = {{0, 1, 0}, {0, 0, 2}, {3, 0, 0}};
That gave me three edges between nodes 1 and 3, and two edges between nodes 2 and 3. How can I show these multiplicities with edges that range from thin to thick.
I want to transform my graph into one which gives me single edge between the nodes. See the following picture:
for each different edge value of a different style. For example, for +3 dark black, for -3 dark red, etc.
Can you help to make a graph from my adjacency matrix that has these (minus and plus) labels?
• Are you asking how to transform your graph into one that has a single edge between each connected vertex, but indicates the multiplicity of the corresponding edge in your graph by increased thickness? – m_goldberg Aug 2 '15 at 19:15
• There are some answers here which will also answer your question. – Szabolcs Aug 3 '15 at 16:54
Taking into account the new information from your edit, I propose the following:
Function to produce the edge labels according to multiplicity indicated in the adjacency matrix.
edgeLbl[multipliciy_] :=
Style[StringJoin @ ConstantArray["+", multipliciy], Background -> White]
Function to style the edges according to multiplicity.
edgeStyle[multipliciy_] := AbsoluteThickness[multipliciy]
My reason for defining these functions is to make it easy to modify the look of graph edges without having to modify the main graph making function, which is:
gr[adjMatrix_] :=
Module[{m, edges, style, lbls},
m = Map[Boole[# > 0] &, adjMatrix, {2}];
style = Rule[#, edgeStyle[adjMatrix[[#[[1]], #[[2]]]]]] & /@ edges;
lbls = Rule[#, edgeLbl[adjMatrix[[#[[1]], #[[2]]]]]] & /@ edges;
EdgeStyle -> style,
EdgeLabels -> lbls,
VertexLabels -> "Name"]]
gr[{{0, 1, 0}, {0, 0, 2}, {3, 0, 0}}]
Update
The above doesn't adjust the size of the arrow heads according to the multiplicity. If you think that should happen, then a another function is needed.
arrow[multipliciy_] :=
GraphElementData[{"FilledArrow", "ArrowSize" -> .018 + .01 multipliciy}]
Module[{m, edges, style, arrows, lbls},
m = Map[Boole[# > 0] &, adjMatrix, {2}];
style = Rule[#, edgeStyle[adjMatrix[[#[[1]], #[[2]]]]]] & /@ edges;
lbls = Rule[#, edgeLbl[adjMatrix[[#[[1]], #[[2]]]]]] & /@ edges;
arrows = Rule[#, arrow[adjMatrix[[#[[1]], #[[2]]]]]] & /@ edges;
EdgeStyle -> style,
EdgeLabels -> lbls,
EdgeShapeFunction -> arrows,
VertexLabels -> "Name"]]
gr[{{0, 1, 0}, {0, 0, 2}, {3, 0, 0}}]
If I understand your question correctly, you want the thickness of the edge to depend on the vertex the edge is directed towards?
If so, then you will need to use EdgeShapeFunction:
data = {{0, 1, 0}, {0, 0, 2}, {3, 0, 0}};
EdgeShapeFunction ->
({
Thickness[Last[#2]*0.01],
Arrow[#1]} &
)]
EdgeShapeFunction provides two parameters:
1. The coordinates of the vertices in the first argument #1, which is used to give the points between which the arrow should be drawn.
2. The name of the vertices between which the arrow is directed in the second argument #2. In this code I take the second value of #2 using Last[#2], that is the name of the vertex the arrow is going towards.
It is the second argument that allows us to define the thickness of the edge using Thickness. I multiply the vertex number by 0.01 to scale down the thickness to an appropriate size. I also have to define the size of the 'Arrowheads', and I find that taking the square of the values used for Thickness works well.
If you wanted to define the thickness according to the value of the vertex the edges are coming from you would use First[#2] rather than Last[#2].
Note also that the brackets () enclosing the EdgeShapeFunction list are important to making the code work. If I remember correctly this is because we are using DirectedEdges: see 'Possible Issues' in: http://reference.wolfram.com/language/ref/EdgeShapeFunction.html#
P
|
2020-01-18 03:06:47
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2568590044975281, "perplexity": 1118.6904565679959}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250591763.20/warc/CC-MAIN-20200118023429-20200118051429-00031.warc.gz"}
|
https://www.gradesaver.com/textbooks/math/calculus/calculus-10th-edition-anton/chapter-1-limits-and-continuity-1-2-computing-limits-exercises-set-1-2-page-70/35
|
## Calculus, 10th Edition (Anton)
Let $f(x) = x, \, g(x) = x^2$. $\lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} g(x) = \infty$. But, $$\lim\limits_{x \to 0} \frac{f(x)}{g(x)} = \lim\limits_{x \to 0} \frac {x}{x^2} = \lim\limits_{x \to 0} \frac 1x = \infty$$ That is, $\lim\limits_{x \to 0} \frac{f(x)}{g(x)} \ne 1$.
|
2019-11-15 15:46:48
|
{"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9975611567497253, "perplexity": 3151.479549850326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668682.16/warc/CC-MAIN-20191115144109-20191115172109-00527.warc.gz"}
|
https://www.quantopian.com/posts/trading-volume-data-inconsistency
|
There may be a bug in the most commonly used Quantopian data for at least some stocks, or there is a gap in my understanding of how I am to use these data.
I've been trying to understand why I get so many partial orders with some strategies.
This naturally caused me to take a look at my trade sizes vs the historical daily volumes.
I do not understand the apparent inconsistency between data presented by pipeline and that presented outside of pipeline.
Here are some observations:
1) There is likely an inconsistency among price and volume data used by pipeline features like AverageDailyVolume() and that used to make/manage orders
2) The daily volume reported through pipeline and the normal interface do not match. The degree of mismatch is large and not constant in offset or multiple
3) Neither of the internally reported volumes match data published on yahoo.com
Attached is a backtest for a single stock [MTEX] from 7/1/2003 through 7/31/2003
7/1: 2963 shares ordered. 2963 purchased
7/8: -2963 shares ordered. 2963 sold
7/15: 5184 shares ordered. 1537 purchased 3647 shares were not filled
7/22: -1537 shares ordered. 1537 sold
7/29: 5509 shares ordered. 2239 purchased 3270 shares were not filled
Here is where my confusion begins.
The logged messages for total trading activity on the above trading days are below
These show that 6500 shares were traded on 7/15 and 3800 on 7/29.
Why then were my trades limited to 1537 and 2239, respectively?
2003-07-02 yesterdays_news:110 INFO yesterday MTEX traded 4600 shares at $7.41 2003-07-09 yesterdays_news:110 INFO yesterday MTEX traded 16400 shares at$ 6.00
2003-07-16 yesterdays_news:110 INFO yesterday MTEX traded 6500 shares at $6.62 2003-07-23 yesterdays_news:110 INFO yesterday MTEX traded 12300 shares at$ 8.49
2003-07-30 yesterdays_news:110 INFO yesterday MTEX traded 3800 shares at 8.06 Investigating further I find that the daily volume data in pipeline and the standard, nonpipeline interface do not match. Note: that the data are for the trading date prior to the logging date, so they don't align with the data above 2003-07-01 periodic_rebalance:93 INFO Vpipe = 638965 Vquant = 29170 2003-07-08 periodic_rebalance:93 INFO Vpipe = 246630 Vquant = 46400 2003-07-15 periodic_rebalance:93 INFO Vpipe = 111950 Vquant = 3900 2003-07-22 periodic_rebalance:93 INFO Vpipe = 332663 Vquant = 10200 2003-07-29 periodic_rebalance:93 INFO Vpipe = 61250 Vquant = 6200 From yahoo.com I find that the actual trading volumes as shown below. pipeline and nonpipeline values are show for comparison 2003-06-30 V(yahoo) = 64,900 Vpipe = 638965 Vquant = 29170 2003-07-01 V(yahoo) = 46,300 Vpipe not logged Vquant = 4600 2003-07-07 V(yahoo) = 26,400 Vpipe = 246630 Vquant = 46400 2003-07-08 V(yahoo) = 40,000 Vpipe not logged Vquant = 16400 2003-07-14 V(yahoo) = 13,000 Vpipe = 111950 Vquant = 3900 2003-07-15 V(yahoo) = 7,700 Vpipe not logged Vquant = 6500 2003-07-21 V(yahoo) = 43,500 Vpipe = 332663 Vquant = 10200 2003-07-22 V(yahoo) = 62,100 Vpipe not logged Vquant = 12300 2003-07-28 V(yahoo) = 6,200 Vpipe = 61250 Vquant = 6200 (6200 matches yahoo) 2003-07-29 V(yahoo) = 9,600 Vpipe not logged Vquant = 3800 This problem persists into the present with MTEX as seen in this data from 7/1/2016 through 7/31/2016 2016-07-06 periodic_rebalance:93 INFO Vpipe = 666 Vquant = 0 2016-07-12 periodic_rebalance:93 INFO Vpipe = 1926 Vquant = 893 2016-07-19 periodic_rebalance:93 INFO Vpipe = 34112 Vquant = 600 2016-07-26 periodic_rebalance:93 INFO Vpipe = 4474 Vquant = 161 4 Loading... Backtest from to with initial capital Total Returns -- Alpha -- Beta -- Sharpe -- Sortino -- Max Drawdown -- Benchmark Returns -- Volatility -- Returns 1 Month 3 Month 6 Month 12 Month Alpha 1 Month 3 Month 6 Month 12 Month Beta 1 Month 3 Month 6 Month 12 Month Sharpe 1 Month 3 Month 6 Month 12 Month Sortino 1 Month 3 Month 6 Month 12 Month Volatility 1 Month 3 Month 6 Month 12 Month Max Drawdown 1 Month 3 Month 6 Month 12 Month # Backtest ID: 57b73e68c464d7100bf0dc22 There was a runtime error. 6 responses Hi Peter, First off, the discrepancy between Pipeline and data.history data is resulting from the way data.history works when used during the trading day. As is explained in the documentation here, if used during the trading day requesting '1d' bars, data.history's last returned bar is a partial bar for the current day. So when you call yesterdays_news right after market open and look at the last daily bar returned by data.history, you get a partial bar which only contains data for the one minute that has elapsed so far on the current day. So the volume figure you get there is just the volume for that one minute. To properly get data for the previous day, you have to use the second-to-last bar given by data.history. So here's how you get yesterday's volume: data.history(context.test_stock, "volume", 2, '1d')[-2] So if you fix that code on lines 90, 107, and 108 you'll see the Pipeline-data.history discrepancy go away. As for your trade quantities being limited, our default slippage model limits trades to 2.5% of volume, and you'll find the order amounts that were able to be filled are about 2.5% of the volume for that day. Finally, the discrepancy with Yahoo: Note that MTEX had a 1:10 reverse split on 2012-01-17. While our volume values reflect the actual share volume from 2003, Yahoo's volume values are split-adjusted, since they're intended to be used from the perspective of today. So Yahoo's volume values are about 1/10 of ours. I hope this explains everything; let me know if you have more questions. Disclaimer The material on this website is provided for informational purposes only and does not constitute an offer to sell, a solicitation to buy, or a recommendation or endorsement for any security or strategy, nor does it constitute an offer to provide investment advisory services by Quantopian. In addition, the material offers no opinion with respect to the suitability of any security or specific investment. No information contained herein should be regarded as a suggestion to engage in or refrain from any investment-related course of action as none of Quantopian nor any of its affiliates is undertaking to provide investment advice, act as an adviser to any plan or entity subject to the Employee Retirement Income Security Act of 1974, as amended, individual retirement account or individual retirement annuity, or give advice in a fiduciary capacity with respect to the materials presented herein. If you are an individual retirement or other investor, contact your financial advisor or other fiduciary unrelated to Quantopian about whether any given investment idea, strategy, product or service described herein may be appropriate for your circumstances. All investments involve risk, including loss of principal. Quantopian makes no guarantees as to the accuracy or completeness of the views expressed in the website. The views are subject to change, and may have become unreliable for various reasons, including changes in market conditions or economic circumstances. Nathan, Thanks for taking the time to give such a thorough response. Your fixes did resolve the difference between the pipeline and nonpipeline data. Two concerns remain: First, the 2.5% limit in the slippage model did not work exactly as I'd expect. I modified my messages to display 2.5% of the day's trade volume. The actual limits imposed are slightly less than 2.5% of the reported daily volume. 2003-07-15 WARN Your order for 5184 shares of MTEX has been partially filled. 1537 shares were successfully purchased. 3647 shares were not filled by the end of day and were canceled. 2003-07-16 yesterdays_news:126 INFO yesterday MTEX traded 66200 shares so max trade size is 1655 ==> 1537 vs 1655 is about 7% less than expected 2003-07-29 WARN Your order for 5509 shares of MTEX has been partially filled. 2239 shares were successfully purchased. 3270 shares were not filled by the end of day and were canceled. 2003-07-30 yesterdays_news:126 INFO yesterday MTEX traded 96623 shares so max trade size is 2416 ==> 2239 vs 12416 is about 7% less than expected Second, the 10:1 split explains the general behavior (Quantopian volumes are ~10x Yahoo volumes) The ratio is not consistent, however. Google Finance volumes are very similar to the Yahoo data and are slightly higher. Perhaps they include trades outside of the session. Regardless of this the Quantopian/Google ratios would be similar to those shown for Quantopian/Yahoo. I'm puzzled as to why the ratio varies so much 2003-06-30 V(yahoo) = 64,900 Vpipe = 638965 ==> ratio = 9.85, V(google) = 65,056 2003-07-07 V(yahoo) = 26,400 Vpipe = 246630 ==> ratio = 9.34, V(google) = 26,503 2003-07-14 V(yahoo) = 13,000 Vpipe = 111950 ==> ratio = 8.61, V(google) = 13,315 2003-07-21 V(yahoo) = 43,500 Vpipe = 332663 ==> ratio = 7.65, V(google) = 43,576 2003-07-28 V(yahoo) = 6,200 Vpipe = 61250 ==> ratio = 9.87, V(google) = 6,275 Nathan, If understand you right, While our volume values reflect the actual share volume from 2003 means that Quantopian historical volume and price data is not split adjusted. Yes. I am puzzled too. You need to adjust both. I am confused about Nathan's comment about split adjustment of data. I ran a quick test that indicates all is behaving in a split adjusted manner. Backtest: buys 1000 shares of AAPL on 6/2/2014 and holds them until 5% return is achieved Background: AAPL had a 7-to-1 split over the weekend between 6/6/2014 and 6/9/2014 I picked AAPL for this because if there were a problem with it, then others would have yelled loudly in June of 2014 Logged data results: The logged data show that this split is captured in the reported price and volume history data of Quantopian 2014-06-05 my_rebalance:36 INFO yesterday AAPL traded 8964095 shares at 644.82 2014-06-06 my_rebalance:36 INFO yesterday AAPL traded 7841964 shares at 647.35 2014-06-09 my_rebalance:36 INFO yesterday AAPL traded 67368599 shares at 92.23 2014-06-10 my_rebalance:36 INFO yesterday AAPL traded 64380213 shares at 93.70 Displayed data results: The displayed data in Daily Positions panel of Backtest shows the adjustment is applied as well 2014-06-05 1000 shares at647.35
2014-06-06 1000 shares at $647.57 2014-06-09 6999 shares at$93.70
2014-06-10 6999 shares at \$94.25
??? What happened to one share? 6999 vs 7000?
Price-based decision results:
2014-06-02 1000 shares purchased
2014-06-11 699 shares sold
If the price data were not split adjusted, then the 6/11/2014 sale would not have been triggered.
For pete's sake (and that was not my first choice of expletives), how Quantopian handles adjustments has been repeatedly explained in great detail. It is more correct than what Yahoo et al do.
The key is that they make no adjustment that was not known at the time. But at each simulation time point, they adjust backwards all historical data requested at that time point as of that time point.
|
2018-10-15 08:08:13
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19686058163642883, "perplexity": 5346.254649826371}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-43/segments/1539583508988.18/warc/CC-MAIN-20181015080248-20181015101748-00248.warc.gz"}
|
http://copper.math.buffalo.edu/418/day21_s17.html
|
# Day 21
Thursday, April 20, 2017
# Sounds and spectra
(from last time in case needed)
This code plots the spectrum of a sound in a wav file: plot_spectrum.py
Example wav files: guitar low E, piano low F, brass bar
Here is the spectrum of the brass bar:
And here is the spectrum of the low piano note:
(Notice that the "fundamental" (the lowest frequency) is almost not present.)
Online tone generator in case it's useful
## Synthesis
This code produces a sound with components we specify: replicate_sound.py
Let us see if we can synthesize the recorded brass bar sound above using frequencies and amplitudes read from the spectrum of the recording:
synthesized_brass.wav
# Drum (circular stretched membrane)
Let's record the sound a drum, and analyze and synthesize it.
[In case of difficulties, I prepared these in advance: small tom closed, small tom open]
# Solving the wave equation on a disk
Let's solve this problem by separation of variables, find the spectrum, and see if it matches the recording of the real drum.
Relationship between F and f:
Code that made the picture above: circular_membrane_modes.py
# Synthesis of drum sound from solution of wave equation on disk
# enter frequency components here
freqs = 120.*array([1,1.593,2.136,2.295,2.653,2.917])
amps = array([1.,1,1,1,1,1])
synthesized_drum.wav (first effort, just guessing frequency of fundamental)
actual drum recording
|
2017-11-23 18:33:14
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43749573826789856, "perplexity": 5938.2072182160055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-47/segments/1510934806856.86/warc/CC-MAIN-20171123180631-20171123200631-00336.warc.gz"}
|
http://thinkinator.com/category/statistics/bayesian-statistics-statistics/
|
# Electricity Usage in a High-rise Condo Complex pt 6 (discussion of model)
In Part 4 of this series, I created a Bayesian model in Stan. A member of the Stan team, Bob Carpenter, has been so kind as to send me some comments via email, and has given permission to post them on the blog. This is a great opportunity to get some expert insights! I’ll put his comments as block quotes:
That’s a lot of iterations! Does the data fit the model well?
I hadn’t thought about this. Bob noticed in both the call and the summary results that I’d run Stan for 300,000 iterations, and it’s natural to wonder, “Why so many iterations? Was it having trouble converging? Is something wrong?” The stan command defaults to four chains, each with 2,000 iterations, and one of the strengths of Stan‘s HMC algorithm is that the iterations are a bit slower than other methods but it mixes much faster so you don’t need nearly as many iterations. So 300,000 is a bit excessive. It turns out that if I run for 2,000 iterations, it takes 28 seconds on my laptop and mixes well. Most of the 28 seconds is taken up by compiling the model, since I can get 10,000 iterations in about 40 seconds.
So why 300,000 iterations? For the silly reason that I wanted lots of samples to make the CI’s in my plot as smooth as possible. Stan is pretty fast, so it only took a few minutes, but I hadn’t thought of implication of appearing to need that many iterations to converge.
# Electricity Usage in a High-rise Condo Complex pt 5
Last time, we modeled the Association’s electricity expenditure using Bayesian Analysis. Besides the fact that MCMC and Bayesian are sexy and resume-worthy, what have we gained by using Stan? MCMC runs more slowly than alternatives, so it had better be superior in other ways, and in this posting, we’ll look at an example of how. I’d recommend pulling the previous posting up in another browser window or tab, and position the “Inference for Stan model” table so that you can quickly consult it in the following discussion.
If you look closely at the numbers, you may notice that the high season (warmer-high, ratetemp 3, beta[3]) appears to have a lower slope than the mid season (warmer-low, ratetemp 2, beta[2]), as was the case in an earlier model. This seems backwards: the high season should cost more per additional kWh, and thus should have a higher slope. This raises two questions: 1) is the apparent slope difference real, and 2) if it is real, is there some real-world basis for this counter-intuitive result?
# Electricity Usage in a High-rise Condo Complex pt 4
This is the fourth article in the series, where the techiness builds to a crescendo. If this is too statistical/programming geeky for you, the next posting will return to a more investigative and analytical flavor. Last time, we looked at a fixed-effects model:
m.fe <- lm (dollars ~ 1 + regime + ratetemp * I(dca - 55))
which looks like a plausible model and whose parameters are all statistically significant. A question that might arise is: why not use a hierarchical (AKA multilevel, mixed-effects) model instead? While we’re at it, why not go full-on Bayesian as well? It just so happens that there is a great new tool called Stan which fits the bill and which also has an rstan package for R.
# Bayesian Data Analysis 3
In the first posting of this series, I simply applied Bayes Rule repeatedly: Posterior $\propto$ Prior $\times$ Likelihood. I didn’t have to know anything about conjugate priors, hyperparameters, sufficient statistics, parametric forms, or anything beyond the basics. I got a reasonable posterior for $\theta$ and used that to find the correct answer. Why go beyond this?
Well, first, there’s really no good way for me to communicate my $\theta$ distribution to anyone else. It’s a (long) vector of values, and that’s the weakness of a non-parametric system: there is no well-known function and no sufficient statistics to easily describe what I’ve discovered. (Of course, it’s possible that there is no well-know function that’s appropriate, in which case the simulation method is actually the only option.)
Second, my posterior density is discrete. This works reasonably well for the exercise I attempted, but it’s still a discrete approximation which is less precise and can suffer from simulation-related issues (number of samples, etc) that have nothing to do with my proposed model.
Third, if an analytical method can be used, it may be possible to directly calculate a final posterior without repeated applications of Bayes Rule. As I mentioned in the previous posting, Gelman got an answer of Gamma(238, 10) analytically, not through approximations and simulation. If we look in Wikipedia, we can find that the conjugate prior for $\theta$, the parameter of a Poisson distribution, is the Gamma ($\alpha, \beta$) distribution, and given our series of $n$ accident counts and an initial (posterior) $\alpha$ and $\beta$, the posterior density is Gamma ($\alpha - \sum_{i=1}^{n} \mathrm {accident}_i, \beta + n$)
# Bayesian Data Analysis 2
In the last post (Bayesian Data Analysis 1), I ran a Bayesian data analysis using a simple, first-principles approach. Armed with only the fact that a Poisson distribution is appropriate for modeling airplane accidents, Bayes Rule, and R, we got the correct answer to the problem through non-parametric simulation.
Before we get into precision and the other topics in the list of issues to explore, let’s tie up one loose end. I could have gotten the right answer for the wrong reason, so let’s look at the posterior distribution of $\theta$ as compared to the provided answer to make sure we’re close. In the answer, Gelman looks at the parametrical form for $\theta$ and considers conjugate priors and analytically determines that the correct parametric posterior distribution is a Gamma function with $\alpha=238$ and $\beta=10$. (More on this in the next posting.) So I simulated out 20,000 samples from that distribution:
rgam <- rgamma (20000, 238, 10)
And then we can compare our posterior distribution with the official, parametric distribution:
Looks reasonably close (though biased by about 0.25) and it didn’t take a lot of fancy machinery to pull off. So why not use this method as our default? Why talk of things like conjugate priors and hyper parameters? And did we just get lucky with numeric precision because we only had 10 accidents and hence only 10 applications of Bayes Rule? Let’s cover that in the next posting, and finish this one with a graph of our answer (not $\theta$ but the predictive distribution of accidents) with the 95% CI :
# Bayesian Data Analysis 1
I’ve read quite a few presentations on Bayesian data analysis, but I always seem to fall into the crack between the first, one-step problem (the usual being how likely you are to have a disease), and more advanced problems that involve conjugate priors and quite a few other concepts. So, when I was recently reading Bayesian Data Analysis[1], I decided to tackle Exercise 13 from Chapter 2 using only Bayes rule updates and simulation. I think it’s been illuminating, so decided to write it up here, using my favorite tool R.
Exercise 13 involves annual airline accidents from 1976 to 1985, modeled as a Poisson$(\theta)$ distribution. The data is :
accidents <- c(24, 25, 31, 31, 22, 21, 26, 20, 16, 22)
A little playing around with graphs and R‘s dpois gives me the impression that $\theta$ is probably around 24, but I’ll make three priors (for $\theta$):
r <- seq (10, 45, 0.2) theta1 <- dnorm (r, 15, 5) theta1 <- theta1 / sum (theta1) theta2 <- dnorm (r, 35, 6) theta2 <- theta2 / sum (theta2) theta3 <- 20 - abs (r - mean (r)) theta3 <- theta3 / sum (theta3)
Where theta1 is probably low, theta2 is probably high, and theta3 is not even parametric. (More on this later.) I normalized them to create proper priors so that they graph well together, but that wasn’t necessary. Remember, I’m doing all of my calculations over the discretized range, $r$. Plotting the theta’s together:
So let’s run the numbers (accidents) through a repeated set of Bayes Rule updates to get a posterior distribution for $\theta$ based on the prior distribution, theta1:
|
2015-11-28 18:23:42
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 20, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6523532867431641, "perplexity": 1007.1243729614743}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398453656.76/warc/CC-MAIN-20151124205413-00248-ip-10-71-132-137.ec2.internal.warc.gz"}
|
https://cs.stackexchange.com/questions/47398/matrix-whose-eigenvectors-are-hermite-polynomials
|
# Matrix whose eigenvectors are Hermite polynomials [closed]
I first constructed a symmetric matrix as the Laplacian operator, and its eigenvectors are a series of harmonics functions as expected. I programmed it and convinced myself. The matrix looks like: $$\left(\begin{array}{ccc} 1& -1 & 0 & 0 & 0 \\ -1& 2 & -1 & 0 & 0 \\ 0& -1 & 2 & -1 & 0 \\ 0& 0 & -1 & 2 & -1 \\ 0& 0 & 0 & -1 & 1\end{array}\right)$$ multiplying this matrix with $\{x_0, x_1, x_2, x_3, x_4\}$ leads to $\{x_0-x_1, -x_0+2x_1-x_2,-x_1+2x_2-x_3,..\}$; The term $-x_{i-1}+2x_i-x_{i+1}$ is equivalent to the second order derivative (in the Laplacian) on a discrete 1-dimensional domain: $(x_i-x_{i-1})-(x_{i+1}-x_i)$.
Everything works fine until I try to construct the matrix for Hermite polynomials. Wiki says the operator is $$L[u]=u''-xu'=-\lambda u.\tag1$$ I think the matrix for the first derivative should be $$\left( \begin{array}{ccc} 1& -1 & 0 & 0 & 0 \\ 0& 1 & -1 & 0 & 0 \\ 0& 0 & 1 & -1 & 0 \\ 0& 0 & 0 & 1 & -1 \\ 0& 0 & 0 & 0 & 0\end{array}\right)$$ however, combining this new matrix with the Laplacian matrix (according to $(1)$) does not produce the expected eigenvectors.
Another source derives polynomial from the operator $$H=-\frac12\frac{d^2}{dx^2}+\frac12x^2,$$ but what are the matrix entries for $x^2$?
Any help? Thanks!
• Isn't this a pure mathematics question? Please make a connection to a compute science topic! (Also, please come up with a more descriptive title.) – Raphael Sep 21 '15 at 7:45
• Alternatively, if this should have been on Math.SE, you can click "flag" to flag it for moderator attention and ask the moderators to migrate it to Math.SE. – D.W. Sep 22 '15 at 7:31
• Ahh, I see: This was also posted on Math.SE. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Sep 22 '15 at 7:32
|
2020-08-06 13:35:18
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.690142035484314, "perplexity": 539.2810498207073}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00554.warc.gz"}
|
https://www.vedantu.com/ncert-solutions/ncert-solutions-class-2-maths-chapter-9
|
Courses
Courses for Kids
Free study material
Free LIVE classes
More
# NCERT Solutions for Class 2 Maths Chapter 9
## NCERT Solutions for Class 2 Maths Chapter 9 - My Funday
Last updated date: 23rd Mar 2023
Total views: 406.2k
Views today: 8.84k
NCERT Solutions for Class 2 Maths Chapter 9 provide stepwise error-free solutions to the exercise sums for the practice of students. The solutions are prepared by our highly experienced teachers as per the CBSE guidelines. The NCERT Solutions of Class 2 Maths Chapter 9, My Funday provide accurate solutions with detailed explanations. With these NCERT Solutions, our highly experienced teachers induce a comprehensive approach towards problem-solving. To get specific solutions to each sum, and secure good marks in the exam, download and refer to My Funday Class 2 NCERT Solutions for free from Vedantu.
Do you need help with your Homework? Are you preparing for Exams?
Study without Internet (Offline)
## NCERT Solutions Class 2 Maths Chapter 9 – Free PDF Download
1. Is Sunday a Funday for you?
Ans: Yes, I have fun on Sunday because I don't have school on this day. I sometimes go out with my parents or call friends at home to play with my toys.
2. Monday is happy to be the first day of the week. Now you tell
1. The third day of the week is_________.
Ans: The third day of the week is Wednesday.
1. The fifth day of the week is ______________.
Ans: The fifth day of the week is Friday.
1. The second day of the week is__________.
Ans: The second day of the week is Tuesday.
1. The last day of the week is____________.
Ans: The last day of the week is Sunday.
3. Which day will come?
1. After Sunday? ______________
Ans: After Sunday, it will be Monday.
1. Before Sunday? _____________
Ans: Before Sunday, It will be Saturday.
1. After Wednesday? ____________
Ans: After Wednesday, It will be Thursday.
1. Before Wednesday? ____________
Ans: Before Wednesday, it will be Tuesday.
1. $2$ days after Sunday? ____________
Ans: $2$ days after Sunday, it will be Tuesday.
1. $4$ days after Wednesday? ___________
Ans: $4$ days after Wednesday, it will be Sunday.
1. $7$ days after Monday? _____________
Ans: $7$ days after Monday, it will be Saturday.
4. Which day do you like most? Why?
Ans: I like Sunday because there is no school on that day and I can have fun the whole day.
1. What is the day today? ___________
Ans: Monday
1. Which day was it yesterday? ____________
Ans: Sunday
1. Which day will it be tomorrow? __________
Ans: Tuesday
1. Which day will it be the day after tomorrow? ___________
Ans: Wednesday
1. Which day was it the day before yesterday? ____________
Ans: Saturday
5. Some children of Class II-A love to play “Teacher-Teacher”. They have decided to take turns in playing the teacher’s role.
Day Who Will Play Teacher’s Role Monday Vaibhav Tuesday Alpana Wednesday Gaurav Thursday Gurpreet Friday Deepak Saturday Rehnuma
Now fill in the blanks.
1. ________ will be the teacher the day after Friday.
Ans: According to the table above, Rehnuma will be the teacher the day after Friday.
1. ___________ will play the teacher’s role on the day before Tuesday.
Ans: Vaibhav will play the teacher’s role on the day before Tuesday which is Monday.
1. Gaurav will play the teacher’s role on the day after ___________
Ans: Gaurav will play the teacher’s role on the day after Tuesday i.e. Wednesday.
1. Deepak will play the teacher’s role on the day before________
Ans: Deepak will play the teacher’s role on the day before Saturday which means he will play the teacher's role on Friday.
6.
Time Table of || - A
Period / Day 1 2 3 4 5 6 Monday Hindi Maths Games English E.V.S Music Tuesday Hindi Maths Drawing English E.V.S Library Wednesday Hindi Maths Games English E.V.S Library Thursday Hindi Maths Drawing English E.V.S Music Friday Hindi Maths Games English E.V.S Music Saturday Hindi Maths Drawing English E.V.S Library
Look at the time table of class $||$- A and fill in the blanks:
Period On Which Days? Drawing Music Games Library
Ans:
Period On Which Days? Drawing Tuesday, Thursday and saturday Music Monday, Thursday and Friday Games Monday, Wednesday and Friday Library Tuesday, Wednesday and saturday
7.
1. On which days do you have games period?____________
Ans: Tuesday and Thursday
1. How many children would like to have a game period everyday? _______
Ans: All Children would like to have games period every day.
1. what games do you play in your games period? _______
Ans: Kho-Kho, dog and the wool, Running
1. On which days do you have a drawing period? ______________
Ans: Monday, Wednesday and Friday.
1. Do you have a music period? _______ If yes, on which days? __________
Ans: Yes, Thursday
1. Which day of the week do you like best at school? ___________
Ans: Thursday because we have Games as well as Music class.
8. Which month do you like best? _____________ Why? _____________
Ans: December is my favourite month because I love winter. Moreover, celebrating Christmas and New Year in the cold is Fun.
9. Draw what you like to do in your favourite month.
Ans:
10. Fill in the table.
Ans:
These are the following things that I like best in the following months.
1. Lichies - December
2. Potato - all $12$ months
3. Rose – January
11. Write the names of five festivals that you or your friends celebrate. Also write the months in which these festivals come.
Name of the Festival Month in Which it Comes
Ans:
Name of the Festival Month in Which it Comes Holi March Onam September Diwali October Gurupurab November Christmas December
12. Name of some of the months that are missing in the list given below. Fill in names of those months.
January, February, _________, April, ___________, June, ___________, August, September, October, _____________, December.
Ans:
January, February, March, April, May, June, July, August, September, October, November and December.
13. Look at the calendar to find out:
1. Which is the first month of the year?
Ans: January
1. Which month comes after March? ___________
Ans: April
1. Which month comes before August? _________
Ans: July
1. Which is the last month of the year? ____________
Ans: December
14. Look at a calendar to find out:
1. Which months have $30$ days? _______________
Ans: April, June, September and November
1. How many months have $31$ days? _______________
Ans: $7$
1. How many days are there in February? _____________
Ans: $28$ or $29$
1. How many days together are there in May and June? _____________
Ans: $61$
1. How many Sundays are there in July? ______________
Ans: $4$
1. What is the day on your birthday? _______________
Ans: $DD - MM - YYYY$
15. Find Out:
1. How many days do you get for your summer holidays? __________
Ans: $20$ days
1. How many winter holidays do you have? ____________
Ans: $15$ days.
16. Have you ever felt :
1. Your teeth chattering with cold? Yes/No
Ans: Yes
1. Your feet burning on walking barefoot? Yes/No
Ans: Yes
1. Name one month when you can easily walk barefoot. ___________
Ans: February
17.
1. Have you seen it raining for many days? Yes/No
Ans: Yes
1. In which month does it rain the mat in your area?____________
Ans: July and August.
18.
1. Which are the hottest months in your area? ____________________
Ans: May
1. Which is the coldest month in your area? ______________-
Ans: December
### NCERT Solutions Class 2 Maths Chapter 9 – Free PDF Download
Mathematics is crucial in almost all aspects of our life be it academics to simple grocery shopping. The NCERT Class 2 Mathematics syllabus comprises a selection of intricate chapters to trigger an extensive learning experience for the students. The solutions for Class 2 Maths Chapter 9 My Funday have been formulated by our teachers in a simple and stepwise manner for the easy understanding of all students. Download a free PDF for Class 2 Chapter 9 Solutions for Maths to have a great learning experience on the fundamental concepts of Mathematics.
The Class 2 Maths Chapter 9 My Funday focuses on teaching the days, months, and composition of a year. An easy download for NCERT Solutions for Class 2 Maths Chapter 9 My Funday is just a click away. By referring to these stepwise NCERT Solutions, students can identify and rectify their doubts easily.
### NCERT Solutions Class 2 Maths My Funday: An Overview of the Exercises
My Funday is a chapter on the days of the week, and months of the year. NCERT Chapter 9 My Funday also discusses seasons and festivals celebrated in a year. By practicing the detailed exercises in this chapter, a student can know the composition of a day, month, or year. Class 2 Maths Ch 9 NCERT Solutions guides the student on the names of the days and months in a chronological manner. The following exercises are solved in the PDF files of CBSE Class 2 Maths Chapter 9 Solutions.
Maths Class 2 Chapter 9: Exercises Page 66-67: One day Monday Page 67-68: Teacher – teacher Page 68-69: Games Every Day? Page 70-72: Favourite Month Page 73: February is Different Page 73-74: Blow Hot, Blow Cold
### Class 2 Maths Chapter-wise Marks Weightage
The CBSE Class 2 syllabus follows a curriculum that stirs a positive problem-solving attitude among students. The chapter-wise weightage of the NCERT Class 2 Maths Syllabus is explained below.
Each of the three chapters- Counting in Groups, Tens and Ones and Patters carries 10 marks. ‘How Much Can You Carry’, Jugs and Mugs, and Add Our Points carry 8 marks each. Footprints and Lines and Lines chapters have been allocated 7 marks each. Five chapters namely- Counting in Tens, My Funday, Give and Take, The Longest Step and Birds Come, Birds Go carry 5 marks each. The chapter What is Long, What is Round has 4 marks. And finally, How Many Ponytails chapter has 3 marks.
Irrespective of marks weightage, all the chapters are equally important for a student of Class 2 to learn. This marks division should only be glanced at before preparing for the terms.
### Benefits of NCERT Solutions for Class 9 Maths
Experienced teachers well adept at the CBSE curriculum have applied an innovative and simple problem-solving methodology to come up with the NCERT Solutions For Class 2 Maths for Chapter 9.
• Quality controlled solutions
• An error-free and precise reasoning for every sum
• Provides strong concepts with skill enhancing tips and chapter insights
• Highlights important definitions and formulas for easy revision
• Step by step approach towards easy understanding
These NCERT Solutions for Class 2 Maths Chapter 9 My Funday is specially curated to clear all shortcomings and doubts of the students.
### Q1. What is the Importance of Maths Class 2 Chapter 9 in the Curriculum?
Ans. The chapter teaches the number of days in a week, month, and a year. It gives a comprehensive idea about time and calendar. The four seasons and the temperature changes associated with them are also discussed.
Q2. Which Days Come Before and After Sunday?
Ans. Saturday comes before Sunday, and Monday comes after Sunday.
Q3. Your Grandma Said She will Visit Your Place Two Days after Wednesday. When will She Come?
Ans. Grandma will come to my house on Saturday.
Q4. How Many Months and Weeks Make a Year?
Ans: Twelve months make a year or 52 weeks make a year.
Q5. Why is February Different?
Ans. February unlike the other months has 28 days. In leap years, it has 29 days.
Interesting right? If you want to practise the lesson with more such fun questions, download NCERT Solutions for Class 2 Maths Chapter 9 My Funday from Vedantu. The study material has many solved questions from this chapter and other chapters of Class 2 Maths.
## FAQs on NCERT Solutions for Class 2 Maths Chapter 9
1. What is Unique about Vedantu’s NCERT Class 2 Maths Solutions?
Vedantu’s uniqueness lies in the fact that our subject experts provide logical reasoning for every answer. The solutions have a precise explanation fit for the revisions before the exam. As it is in the form of a free PDF, it can be accessed by all students. The stepwise solutions will help students to secure good marks in the examination.
2. What is the Concept Behind NCERT Class 2 Maths Chapter 9?
Class 2 Maths Chapter 9 My Funday has the following ideas in store:
• The days of the week and their sequence
• The months of the year and their order
• The number of days in each month
• Significant seasonal changes and associated temperature change
• Festivals in a year
3. How Important are the NCERT Solutions For Class 2 Maths Chapter 9?
The solutions to Class 2 Chapter 9 My Funday are important for concept building and last-minute revisions. On Vedantu you can get free PDF files of solutions for Ch 9 Class 2 Maths and you can download them on any device. There are other study resources like study materials and solutions for doubts, FREE LIVE MASTER-Classes, and FREE Conceptual Videos offered by Vedantu that can make you reach the peak of success.
4. What is the basic idea of Class 2 Maths Chapter 9?
The basic idea of Class 2 Maths Chapter 9 is to teach students about the weekdays. Students need to understand the concept of weekdays in Class 2. Students can understand the week's days by practicing the NCERT Solutions given in the textbook. PDF files can be downloaded free of cost from the internet for practicing NCERT questions. All questions are given in simple language for a clear understanding of the students.
5. In your class, how many students agree that Sunday is a Funday?
In Class 2 Maths Chapter 9, different questions are given based on the days of the week. The main concept of the chapter is to teach students about weekdays. This chapter tells them that Sunday is a holiday and it is a fun day. All students in the class agree that Sunday is a fun day because they can stay at home and can do any type of activity that they like.
1. Which is the first day of the week after funday?
2. What comes before the first day of the week?
3. Which is the best day of the week?
Students of Class 2 can learn about the concept of weekdays in Maths Chapter 9. They should know the right order of the days of the week. They should learn the days of the week in chronological order.
1. The first day of the week after funday is Monday.
2. Sunday comes before the first day of the week.
3. Sunday is the best day of the week because it is a holiday and Funday.
7. Why do students like Sunday the most?
Students like Sunday the most because it is a holiday from school. They do not have to go to school on Sunday. They can stay at home and can do any activity of their choice. All members of the family are at home on Sunday. Students can spend time with their brothers and sisters and parents. They can eat anything they prefer at home on Sunday. They can also watch their favorite TV shows on Sunday.
8. Where can I find NCERT Solutions Class 2 Maths Chapter 9?
Vedantu offers NCERT Solutions Class 2 Maths Chapter 9 free of cost. Parents can download the NCERT Solutions. Vedantu is a learning hub for the students of all classes. Parents can easily download the NCERT Solutions for class 2 from the link given. They can help class 2 students to understand the basic concept of the chapter and practice more questions. Solutions are easy and simple to understand to help students prepare better for exams. The solutions can also be viewed on the Vedantu app.
|
2023-03-27 19:36:16
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.1718626171350479, "perplexity": 4099.58539627829}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00617.warc.gz"}
|
http://math.stackexchange.com/questions/303699/v-1-v-2-subspaces-of-v-show-that-there-is-a-basis-mathcal-b-for-v-such
|
# $V_1, V_2$ subspaces of $V$. Show that there is a basis $\mathcal B$ for V such that $\mathcal B \cap (V_1 \cup V_2)= \emptyset$
Let $V$ be a vector space over a field $F$. Suppose $V_1,V_2$ are subspaces of $V$ that lies strictly between $\{ 0_V \}$ and $V$. Show that there is a basis $\mathcal B$ for $V$ such that $\mathcal B\cap ( V_1 \cup V_2)=\emptyset$.
My work so far: My guess is that $V \setminus (V_1 \cup V_2)$ will make a linearly independent set and therefore can be extended to a basis for $V$. I showed that this set is non empty. So we can consider $S:=V\setminus (V_1 \cup V_2)$ How do I show that $S$ is linearly independent? Or is that even the right way to proceed?.
Any hints, suggestions? Thanks for your time.
-
$V\setminus (V_1 \cup V_2)$ is not a linearly independent set. Have you tried any examples to get a feel for what you're trying to do? Maybe do the exercise for $\mathbb R^2$ and let $V_1$ be the subspace generated by $(1,0)$ and $V_2$ the subspace generated by $(0,1)$. – JSchlather Feb 14 '13 at 3:39
You should in fact focus on the opposite aspect: show that (with additional hypothesis on $F$ which my answers shows is necessary) $V \setminus (V_1 \cup V_2)$ is a spanning set; once this is established, you can select a basis from it. – Marc van Leeuwen Feb 14 '13 at 5:21
Does the characteristic of the field $F$ matter here?. What if I assume that $char (F) \neq 0$, does that matter at all here? – Jack Dawkins Feb 14 '13 at 5:22
This result does not hold in general. If $F=\Bbb F_2$ and $\dim V=2$ (whence $\dim V_i=1$ for $i=1,2$) and $V_1\neq V_2$, then $V\setminus(V_1\cup V_2)$ is a singleton, so you cannot choose a basis from its element.
The example is somewhat marginal, and the result ought to hold for any field with at least three elements, but this counterexample shows one will need to use the fact of having three distinct elements in your fields in a proof of that restricted statement. For $F=\Bbb F_2$ you can get counterexamples in any dimension by taking $V_1,V_2$ to be distinct hyperplanes (subspace of dimension $\dim V-1$); reduction modulo $V_1\cap V_2$ proves this (it reduces the question to the initial counterexample, since a basis must project to a spanning set of the quotient).
@Mark van Leeuwen: Thanks for your answer.What do you mean by $\mathbb{F}_2$. Field of characteristic 2? – Jack Dawkins Feb 14 '13 at 5:23
Yes. ${}{}{}{}$ – Gerry Myerson Feb 14 '13 at 5:33
@user54755: In fact, it denotes the unique field with $2$ elements, which has characterisitic $2$, but is not the only field of characteristic $2$. For other fields of characteristic $2$, even finite ones, the result should be OK. – Marc van Leeuwen Feb 14 '13 at 5:49
|
2015-09-04 10:58:08
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9394671320915222, "perplexity": 162.49287328288347}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440645340161.79/warc/CC-MAIN-20150827031540-00058-ip-10-171-96-226.ec2.internal.warc.gz"}
|
https://docs.eyesopen.com/toolkits/python/oefftk/OEFFFunctions/OESetForceFieldDummyAtom.html
|
# OESetForceFieldDummyAtom¶
void OESetForceFieldDummyAtom(OEChem::OEAtomBase *atom)
Modify information associated with an atom to indicate that if the atom has atomic number OEElemNo_Du it can be used for certain force field calculations where it would not normally be valid.
Note
Currently, only MMFF and Amber support support dummy atoms at all, and even then only for calculations involving non-bonded terms and when this special indication has been set.
Note
This function is defined in the OEChem TK, but in the OEFF main namespace.
|
2021-06-19 06:44:58
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5264030694961548, "perplexity": 2765.124441152217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487643703.56/warc/CC-MAIN-20210619051239-20210619081239-00341.warc.gz"}
|
http://www.ualberta.ca/~bhan/abstracts/2004hanriesz.abs.html
|
## Abstract
Let $\phi$ be a compactly supported refinable function in $L_2(\RR)$ such that the shifts of $\phi$ are stable and $\hat\phi(2\xi)=\hat a(\xi)\hat \phi(\xi)$ for a $2\pi$-periodic trigonometric polynomial $\hat a$. A wavelet function $\psi$ can be derived from $\phi$ by $\hat \psi(2\xi):=e^{-i\xi}\ol{\hat a(\xi+\pi)} \hat \phi(\xi)$. If $\phi$ is an orthogonal refinable function, then it is well known that $\psi$ generates an orthonormal wavelet basis in $L_2(\RR)$. Recently, it has been shown in the literature (\cite{DS, HS:rw1d}) that if $\phi$ is a $B$-spline or pseudo-spline refinable function, then $\psi$ always generates a Riesz wavelet basis in $L_2(\RR)$. It was an open problem whether $\psi$ can always generate a Riesz wavelet basis in $L_2(\RR)$ for any compactly supported refinable function in $L_2(\RR)$ with stable shifts. In this paper, we settle this problem by proving that for a family of arbitrarily smooth refinable functions with stable shifts, the derived wavelet function $\psi$ does not generate a Riesz wavelet basis in $L_2(\RR)$. Our proof is based on some necessary and sufficient conditions on the $2\pi$-periodic functions $\hat a$ and $\hat b$ in $C^{\infty}(\RR)$ such that the wavelet function $\psi$, defined by $\hat \psi(2\xi):=\hat b(\xi)\hat \phi(\xi)$, generates a Riesz wavelet basis in $L_2(\RR)$.
Back to Preprints and Publications
|
2014-10-26 05:57:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6444782018661499, "perplexity": 176.83369367664466}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414119658026.51/warc/CC-MAIN-20141024030058-00056-ip-10-16-133-185.ec2.internal.warc.gz"}
|
https://www.ncatlab.org/nlab/show/local+BRST+cohomology
|
Contents
# Contents
## Idea
Given a Lagrangian field theory $(E,\mathbf{L})$ with field bundle $E \overset{fb}{\to} \Sigma$ over some spacetime $\Sigma$ and local Lagrangian density $\mathbf{L}$, then its local BV-BRST complex (or local BRST complex, for short) is the realization of the BV-BRST complex not on local observables $A = \tau_{\Sigma} \alpha$ given by functionals on the space of field histories $\Gamma_{\Sigma}(E)_{\delta_{EL} = 0}$ which are transgressions $\tau_{\Sigma}$ of variational differential forms $\alpha \in \Omega^{\bullet, \bullet}_\Sigma(E)$ on the jet bundle, but on these variational differential forms themselves (whence “local”, i.e. before transgression).
If $s$ denotes the BV-BRST differential in a BV-resolution $\Omega^{\bullet,\bullet}_\Sigma(E)\vert_{\mathcal{E}_{BV}}$ of the restriction to the shell $\mathcal{E} \hookrightarrow J^\infty_\Sigma(E)$ of the variational bicomplex $\Omega^{\bullet,\bullet}_\Sigma(E)$ with its total spacetime derivative $d$ (horizontal derivative), then the local BV-BRST cohomology is the cochain cohomology of $s + d$, hence of the total complex of the double complex given by $s$ and $d$.
Generally, considering variational differential forms up to $d$-exact terms is the “local” incarnation of what under the integration involved in the transgression is integration by parts and it is in this way that “local BV-BRST cohomology” knows about the actual BV-BRST cohomology on local observables.
## Example
Consider local coordinates $(\phi^a)$ on the fibers of the field bundle. The corresponding antifield coordinates are to be denoted $\overline{\phi}_a$ and the BV-BRST differential takes them to the corresponding component
$s(\overline{\phi}_a) = \frac{\delta_{El} L}{\delta \phi^a}$
of the Euler-Lagrange form.
In degree $(p+1,0)$ the $s+d$-closed elements in vanishing ghost degree are pairs $(v,J_v)$ consisting of an infinitesimal symmetry of the Lagrangian $v$, regarded as an antifield density $v^a \overline{\phi}_a dvol_\Sigma$, together with a corresponding conserved Noether current $J_v$:
$\array{ \{J_v\} &\overset{d}{\longrightarrow}& \{ \overset{= 0}{\overbrace{ d J_v - \iota_v \delta_{EL}\mathbf{L} }} \} \\ && \uparrow\mathrlap{-s} \\ && \{ v^a \overline{\phi}_a dvol_\Sigma\} }$
Such pairs are $(s+d)$-exact if on-shell the infintiesimal symmetry coincides with an infinitesimal gauge symmetry. To see this, recall:
An infinitesimal gauge symmetry $v_\epsilon$ of gauge parameter $(\epsilon^\alpha)$ is a vector field on the jet bundle with components of the form
$\mathcal{L}_{v_\epsilon} \phi^a \;\coloneqq\; R^a_\alpha \epsilon^\alpha + R^{a \mu}_\alpha \frac{d \epsilon^\alpha}{d x^\mu}$
such that this is an infinitesimal symmetry of the Lagrangian in that
\begin{aligned} \iota_{v_\epsilon} \delta_{EL} \mathbf{L} & = v^a \frac{\delta_{EL} L}{\delta \phi^a} dvol_\Sigma \\ & = \epsilon^\alpha \left( R^a_\alpha \frac{\delta_{EL} L}{ \delta \phi^a} - \frac{d}{d x^\mu} \left( R^{a \mu}_\alpha \frac{\delta_{EL} L}{\delta \phi^a} \right) \right) dvol_\Sigma + d\left( \epsilon^\alpha R^{a \mu}_\alpha \frac{\delta_{EL} L}{\delta \phi^a} \right) \iota_{\partial_\mu} dvol_\Sigma \\ & = 0 + d(...) \end{aligned}
for all $(\epsilon^\alpha)$.
The corresponding antighosts $\overline{c}_\alpha$ are taken by the BV-BRST differential to the antifield-preimage of the term on the left:
$s\left(\overline{c}_\alpha\right) \;=\; R^a_\alpha \overline{\phi}_a - \frac{d}{d x^\mu} \left( R^{a \mu}_\alpha \overline{\phi}_a \right) \,.$
Moreover, an on-shell vanishing infinitesimal symmetry of the Lagrangian is a vector field with components of the form
$\kappa^{a b} \frac{\delta_{EL} L}{\delta \phi^a}$
for $\kappa^{a b} = - \kappa^{b a}$ a skew-symmetric system of smooth functions on the jet bundle.
The linear combination of such an infinitesimal gauge symmetry and an on-shell vanishing infinitesimal symmetry is $(s+d)$-exact:
\begin{aligned} v^a dvol_\Sigma & = \left( R^a_\alpha \epsilon^\alpha + R^{a \mu}_\alpha \frac{d \epsilon^\alpha}{d x^\mu} + \kappa^{a b} \frac{\delta_{EL} L }{ \delta \phi^a } \right) dvol_\Sigma \\ & = s \left( \epsilon^\alpha \overline{c}_\alpha - \tfrac{1}{2}\kappa^{a b} \overline{\phi}_a \overline{\phi}_b \right) dvol_\sigma + d\left( \epsilon^\alpha R^{a \mu}_\alpha \right) \iota_{\partial_\mu} dvol_\Sigma \end{aligned}
It may be useful to organize this expression into the $s+d$-bicomplex like so:
$\array{ \{K\} &\overset{d}{\longrightarrow}& \{ d K + \epsilon^\alpha R^{a \mu}_\alpha \frac{\delta_{EL}\mathbf{L}}{ \delta \phi^a} \} &\overset{d}{\longrightarrow}& \{ \overset{= 0}{\overbrace{ d J_v - \iota_v \delta_{EL}\mathbf{L} }} \} \\ && \mathllap{s}\uparrow && \uparrow\mathrlap{-s} \\ && \epsilon^\alpha R^{a \mu}_\alpha \overline{\phi}_a \iota_{\partial_\mu} dvol_\Sigma &\underset{d}{\longrightarrow}& \left\{ d\left( \epsilon^\alpha R^{a \mu}_\alpha \overline{\phi}_a \right) \iota_{\partial_\mu} dvol_\Sigma + \left( R^a_\alpha \epsilon^\alpha + R^{a \mu}_\alpha \frac{d \epsilon^\alpha}{d x^\mu} + \kappa^{a b} \frac{\delta_{EL} L }{ \delta \phi^a } \right) \overline{\phi}_a \, dvol_\Sigma \right\} \\ && && \uparrow\mathrlap{-s} \\ && && \left( - \epsilon^\alpha \overline{c}_\alpha + \tfrac{1}{2}\kappa^{a b } \overline{\phi}_a \overline{\phi}_b \right) dvol_\Sigma }$
## References
Review includes
The general theory is discussed in
Details of the local antibracket are discussed in
Application to gravity and/or Yang-Mills theory (Einstein-Yang-Mills theory) is discussed in
Last revised on April 18, 2019 at 04:41:01. See the history of this page for a list of all contributions to it.
|
2021-03-09 01:20:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 41, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9989332556724548, "perplexity": 2101.7711913796825}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178385534.85/warc/CC-MAIN-20210308235748-20210309025748-00199.warc.gz"}
|
http://www.lastfm.fr/user/vurtomatic/library/music/Hammock/_/I+Can+Almost+See+You?setlang=fr
|
# Bibliothèque
Musique » Hammock »
## I Can Almost See You
37 écoutes | Se rendre sur la page du titre
Titres (37)
Titre Album Durée Date
I Can Almost See You 4:12 27 nov. 2010, 2h38m
I Can Almost See You 4:12 31 mai 2010, 18h55m
I Can Almost See You 4:12 17 avr. 2010, 17h08m
I Can Almost See You 4:12 17 avr. 2010, 10h36m
I Can Almost See You 4:12 8 juin 2009, 10h17m
I Can Almost See You 4:12 20 mars 2009, 13h09m
I Can Almost See You 4:12 11 fév. 2009, 12h30m
I Can Almost See You 4:12 26 jan. 2009, 11h25m
I Can Almost See You 4:12 26 jan. 2009, 10h10m
I Can Almost See You 4:12 26 jan. 2009, 8h54m
I Can Almost See You 4:12 26 jan. 2009, 7h38m
I Can Almost See You 4:12 26 jan. 2009, 6h23m
I Can Almost See You 4:12 26 jan. 2009, 5h07m
I Can Almost See You 4:12 26 jan. 2009, 3h52m
I Can Almost See You 4:12 26 jan. 2009, 2h36m
I Can Almost See You 4:12 26 jan. 2009, 1h21m
I Can Almost See You 4:12 26 jan. 2009, 0h05m
I Can Almost See You 4:12 25 jan. 2009, 22h50m
I Can Almost See You 4:12 25 jan. 2009, 21h34m
I Can Almost See You 4:12 25 jan. 2009, 20h19m
I Can Almost See You 4:12 22 jan. 2009, 20h19m
I Can Almost See You 4:12 21 jan. 2009, 20h50m
I Can Almost See You 4:12 19 jan. 2009, 16h00m
I Can Almost See You 4:12 19 jan. 2009, 9h10m
I Can Almost See You 4:12 18 jan. 2009, 20h19m
I Can Almost See You 4:12 18 jan. 2009, 11h26m
I Can Almost See You 4:12 18 jan. 2009, 11h22m
I Can Almost See You 4:12 18 jan. 2009, 11h19m
I Can Almost See You 4:12 18 jan. 2009, 11h14m
I Can Almost See You 4:12 18 jan. 2009, 11h10m
I Can Almost See You 4:12 18 jan. 2009, 11h06m
I Can Almost See You 4:12 18 jan. 2009, 11h02m
I Can Almost See You 4:12 18 jan. 2009, 10h58m
I Can Almost See You 4:12 18 jan. 2009, 9h17m
I Can Almost See You 4:12 18 jan. 2009, 9h13m
I Can Almost See You 4:12 18 jan. 2009, 9h09m
I Can Almost See You 4:12 18 jan. 2009, 9h05m
|
2014-04-18 17:54:30
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9181492924690247, "perplexity": 10355.850032679342}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00017-ip-10-147-4-33.ec2.internal.warc.gz"}
|
https://deepai.org/publication/analysis-for-the-slow-convergence-in-arimoto-algorithm
|
Analysis for the Slow Convergence in Arimoto Algorithm
In this paper, we investigate the convergence speed of the Arimoto algorithm. By analyzing the Taylor expansion of the defining function of the Arimoto algorithm, we will clarify the conditions for the exponential or 1/N order convergence and calculate the convergence speed. We show that the convergence speed of the 1/N order is evaluated by the derivatives of the Kullback-Leibler divergence with respect to the input probabilities. The analysis for the convergence of the 1/N order is new in this paper. Based on the analysis, we will compare the convergence speed of the Arimoto algorithm with the theoretical values obtained in our theorems for several channel matrices.
Authors
• 3 publications
• 3 publications
• 3 publications
• Analysis of the Convergence Speed of the Arimoto-Blahut Algorithm by the Second Order Recurrence Formula
In this paper, we investigate the convergence speed of the Arimoto-Blahu...
09/17/2020 ∙ by Kenji Nakagawa, et al. ∙ 0
• Elimination of ISI Using Improved LMS Based Decision Feedback Equalizer
This paper deals with the implementation of Least Mean Square (LMS) algo...
08/10/2012 ∙ by Mohammad Havaei, et al. ∙ 0
• On the Effectiveness of Fekete's Lemma
Fekete's lemma is a well known combinatorial result pertaining to number...
10/19/2020 ∙ by Holger Boche, et al. ∙ 0
• Proximal Langevin Algorithm: Rapid Convergence Under Isoperimetry
We study the Proximal Langevin Algorithm (PLA) for sampling from a proba...
11/04/2019 ∙ by Andre Wibisono, et al. ∙ 0
• The Global Convergence Analysis of the Bat Algorithm Using a Markovian Framework and Dynamical System Theory
The bat algorithm (BA) has been shown to be effective to solve a wider r...
03/27/2019 ∙ by Si Chen, et al. ∙ 12
• On the Theoretical Properties of the Exchange Algorithm
Exchange algorithm is one of the most popular extensions of Metropolis-H...
05/19/2020 ∙ by Guanyang Wang, et al. ∙ 0
• Zeroth-order Stochastic Compositional Algorithms for Risk-Aware Learning
We present Free-MESSAGEp, the first zeroth-order algorithm for convex me...
12/19/2019 ∙ by Dionysios S. Kalogerias, et al. ∙ 0
This week in AI
Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.
1 Introduction
Arimoto [4] proposed a sequential algorithm for calculating the channel capacity of a discrete memoryless channel. Based on the Bayes probability, the algorithm is given by the alternating minimization between the input probabilities and the reverse channel matrices. For arbitrary channel matrix the convergence of the Arimoto algorithm is proved and the convergence speed is evaluated. In the worst case, the convergence speed is the order, and if the input distribution that achieves the channel capacity is in the interior of the set of input distributions, the convergence is exponential.
In this paper, we first consider the exponential convergence and evaluate the convergence speed. We show that there exist cases of exponential convergence even if is on the boundary of . Moreover, we also consider the convergence of the order, which is not dealt with in the previous studies. Especially, when the input alphabet size , we will analyze the convergence of the order in detail and the convergence speed is evaluated by the derivatives of the Kullback-Leibler divergence with respect to the input probabilities.
As a basic idea for evaluating the convergence speed, we consider that the function which defines the Arimoto algorithm is a differentiable mapping from to , and notice that the capacity achieving input distribution is the fixed point of . Then, the convergence speed is evaluated by analyzing the Taylor expansion of about the fixed point .
2 Related works
There have been many related works on the Arimoto algorithm. For example, extension to different types of channels [11], [15], [17], acceleration of the Arimoto algorithm [10], [18], characterization of Arimoto algorithm by divergence geometry [8], [10], [12], etc. If we focus on the analysis for the convergence speed of the Arimoto algorithm, we see in [4],[10],[18]
that the eigenvalues of the Jacobian matrix are calculated and the convergence speed is investigated in the case that
is in the interior of .
In this paper, we consider the Taylor expansion of the defining function of the Arimoto algorithm. We will calculate not only the Jacobian matrix of the first order term of the Taylor expansion, but also the Hessian matrix of the second order term, and examine the convergence speed of the exponential or order based on the Jacobian and Hessian matrices. Because our approach for the evaluation of the convergence speed is very fundamental, we hope that our results will be applied to all the existing works.
3 Channel matrix and channel capacity
Consider a discrete memoryless channel with the input source and the output source . Let be the input alphabet and be the output alphabet.
The conditional probability that the output symbol is received when the input symbol was transmitted is denoted by
and the row vector
is defined by . The channel matrix is defined by
Φ=⎛⎜ ⎜⎝P1⋮Pm⎞⎟ ⎟⎠=⎛⎜ ⎜⎝P11⋯P1n⋮⋮Pm1⋯Pmn⎞⎟ ⎟⎠. (1)
We assume that for any there exist at least one with . This means that there are no useless output symbols.
The set of input probability distributions on the input alphabet
is denoted by . The interior of is denoted by . Similarly, the set of output probability distributions on the output alphabet is denoted by .
Let be the output distribution for the input distribution , where the representation by components is , then the mutual information is defined by . The channel capacity is defined by
C=maxλ∈Δ(X)I(λ,Φ). (2)
The Kullback-Leibler divergence for two output distributions is defined by
D(Q∥Q′)=n∑j=1QjlogQjQ′j. (3)
The Kullback-Leibler divergence satisfies , and if and only if [7].
An important proposition for investigating the convergence speed of the Arimoto algorithm is the Kuhn-Tucker condition on the input distribution to achieve the maximum of (2).
Theorem (Kuhn-Tucker condition) In the maximization problem (2), a necessary and sufficient condition for the input distribution to achieve the maximum is that there is a certain constant with
D(Pi∥λ∗Φ){=~C,\rm for i \rm with λ∗i>0,≤~C,\rm for i \rm with λ∗i=0. (4)
In (4), is equal to the channel capacity .
Since this Kuhn-Tucker condition is a necessary and sufficient condition, all the information about the capacity achieving input distribution can be derived from this condition.
4 Arimoto algorithm for calculating channel capacity
4.1 Arimoto algorithm [4]
A sequence of input distributions
{λN=(λN1,⋯,λNm)}N=0,1,⋯⊂Δ(X) (5)
is defined by the Arimoto algorithm as follows. First, let be an initial distribution taken in , i.e., . Then, the Arimoto algorithm is given by the following recurrence formula;
λN+1i=λNiexpD(Pi∥λNΦ)m∑k=1λNkexpD(Pk∥λNΦ),i=1,⋯,m,N=0,1,⋯. (6)
On the convergence of this Arimoto algorithm, the following results are obtained in Arimoto [4];
By defining
C(N+1,N)≡−m∑i=1λN+1ilogλN+1i+m∑i=1n∑j=1λN+1iPijlogλNiPijm∑k=1λNkPkj, (7)
they obtained the following theorems;
Theorem A1: If the initial input distribution is in , then
limN→∞C(N+1,N)=C. (8)
Theorem A2: If , then
0≤C−C(N+1,N)≤logm−h(λ0)N, (9)
where is the entropy of .
Theorem A3: If the capacity achieving input distribution is in , then
0≤C−C(N+1,N)
where and is a constant.
In [4], they consider the Taylor expansion of by , and the Taylor expansion of by , however they do not consider the Taylor expansion of the mapping , which will be considered in this paper. Further, in the above Theorem A3, they consider only the case , where the convergence is exponential.
In Yu [18], they consider the mapping and the Taylor expansion of about . They calculate the eigenvalues of the Jacobian matrix , however they do not consider the Hessian matrix. Further, they consider only the case as in [4].
4.2 Mapping from Δ(X) to Δ(X)
Let be the defining function of the Arimoto algorithm (6), i.e.,
Fi(λ)=λiexpD(Pi∥λΦ)m∑k=1λkexpD(Pk∥λΦ),i=1,⋯,m. (11)
Define , then we can consider that is a differentiable mapping from to , and (6) is represented by
λN+1=F(λN). (12)
In this paper, for the analysis of the convergence speed, we assume
rankΦ=m. (13)
Lemma 1
The capacity achieving input distribution is unique.
Proof: By Csiszàr[7], p.137, eq.(37), for arbitrary ,
m∑i=1λiD(Pi∥Q)=I(λ,Φ)+D(λΦ∥Q). (14)
By the assumption (13), we see that there exists [14] with
D(P1∥Q0)=⋯=D(Pm∥Q0)≡C0. (15)
Substituting into (14), we have . Because is a constant,
maxλ∈Δ(X)I(λ,Φ)⟺minλ∈Δ(X)D(λΦ∥Q0). (16)
Define , then is a closed convex set, thus by Cover [6], p.297, Theorem 12.6.1, that achieves exists and is unique. By the assumption (13), the mapping is one to one, therefore, with is unique.
Remark 1
Due to the equivalence (16), the Arimoto algorithm can be obtained by Csiszàr [8], Chapter 4, “Minimizing information distance from a single measure”, Theorem 5.
Lemma 2
The capacity achieving input distribution is the fixed point of the mapping in . That is, .
Proof: In the Kuhn-Tucker condition (4), let us define as the number of indices with , i.e.,
λ∗i{>0,i=1,⋯,m1,=0,i=m1+1,⋯,m, (17)
then
D(Pi∥λ∗Φ){=C,i=1,⋯,m1,≤C,i=m1+1,⋯,m. (18)
We have
m∑k=1λ∗kexpD(Pk∥λ∗Φ)=m1∑k=1λ∗keC=eC, (19)
hence by (11), (17), (19),
Fi(λ∗) ={e−Cλ∗ieC,i=1,⋯,m1,0,i=m1+1,⋯,m, (20) =λ∗i,i=1,⋯,m, (21)
which shows .
The sequence of the Arimoto algorithm converges to the fixed point , i.e.,
λN→λ∗,N→∞. (22)
We will investigate the convergence speed by using the Taylor expansion of about .
4.3 Type of index
Now, we classify the indices
in the Kuhn-Tucker condition (4) in more detail into the following 3 types;
D(Pi∥λ∗Φ)⎧⎪ ⎪⎨⎪ ⎪⎩=C,\rm for i \rm with λ∗i>0 (type I),=C,\rm for i \rm with λ∗i=0 (type II),
Let us define the sets of indices as follows;
all the indices: I≡{1,⋯,m}, (24) type I indices: II≡{1,⋯,m1}, (25) type II indices: III≡{m1+1,⋯,m1+m2}, (26) type III indices: IIII≡{m1+m2+1,⋯,m}. (27)
, , , . We have and .
is not empty and for any channel matrix, but and may be empty for some channel matrix.
4.4 Examples of convergence speed
Let us consider the difference of convergence speed of the Arimoto algorithm depending on the channel matrices.
For many channel matrices , the convergence is exponential, but for some special the convergence is very slow. Let us consider the following examples taking types I, II, III into account, where the input alphabet size and the output alphabet size .
Example 1
(only type I) If only type I indices exist, then , hence is in the interior of . As a concrete channel matrix of this example, let us consider
Φ(1)=⎛⎜⎝0.8000.1000.1000.1000.8000.1000.2500.2500.500⎞⎟⎠. (28)
For this , we have and . See Fig.1. The vertices of the large triangle in Fig.1 are the output probability distributions . We have , then considering the analogy to Euclidean geometry, can be regarded as an “acute triangle”.
Example 2
(types I and II) If there are type I and type II indices, we can assume without loss of generality, hence is on the side and . As a concrete channel matrix of this example, let us consider
Φ(2)=⎛⎜⎝0.8000.1000.1000.1000.8000.1000.3000.3000.400⎞⎟⎠. (29)
For this , we have and . See Fig.2. Considering the analogy to Euclidean geometry, can be regarded as a “right triangle”.
Example 3
(types I and III) If there are type I and type III indices, we can assume without loss of generality, hence is on the side and . As a concrete channel matrix of this example, let us consider
Φ(3)=⎛⎜⎝0.8000.1000.1000.1000.8000.1000.3500.3500.300⎞⎟⎠. (30)
For this , we have and . See Fig.3. Considering the analogy to Euclidean geometry, can be regarded as an “obtuse triangle”.
For the above , Fig.4 shows the state of convergence of . By this Figure, we see that in Examples 1 and 3 the convergence is exponential, while in Example 2 the convergence is slower than exponential.
From the above three examples, it is inferred that the Arimoto algorithm converges very slowly when type II index exists, and converges exponentially when type II index does not exist. We will analyze this phenomenon in the following.
5 Taylor expansion of F(λ) about λ=λ∗
We will examine the convergence speed of the Arimoto algorithm by the Taylor expansion of about the fixed point . Taylor expansion of the function about is
F(λ)=F(λ∗)+(λ−λ∗)J(λ∗)+12!(λ−λ∗)H(λ∗)t(λ−λ∗)+o(∥λ−λ∗∥2), (31)
where denotes the transpose of and denotes the Euclidean norm .
In (31), is the Jacobian matrix at , i.e.,
J(λ∗) =(∂Fi∂λi′∣∣∣λ=λ∗)i′,i=1,⋯,m. (32)
We consider in this paper that the input probability distribution is a row vector, thus the Jacobian matrix is such as
←i→ J(λ∗) =↑i′↓⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝∂F1∂λ1∣∣∣λ=λ∗⋯∂Fm∂λ1∣∣∣λ=λ∗⋮⋮∂F1∂λm∣∣∣λ=λ∗⋯∂Fm∂λm∣∣∣λ=λ∗⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠∈Rm×m, (33)
i.e., is the component. Note that our is the transpose of a usual Jacobian matrix corresponding to column vector.
Because by (11), we have by (33),
Lemma 3
Every row sum of is equal to .
In (31), , where is the Hessian matrix of at , i.e.,
Hi(λ∗)=(∂2Fi∂λi′∂λi′′∣∣∣λ=λ∗)i′,i′′=1,⋯,m, (34)
and is an abbreviated expression of the dimensional vector
Remark 2
satisfy the constraint , but in (31), (32), (34) we consider as independent variables to have the Taylor series approximation (31). This approximation is justified as follows. By the Kuhn-Tucker condition (4), , hence by the assumption put below (1), we have . See [4]. For , define , i.e., is an open ball in centered at with radius . Note that is free from the constraint . Taking sufficiently small, we can have , for any . The function is defined for with , even if some . Therefore, the domain of definition of can be extended to , where is the inverse image of by the mapping . is an open neighborhood of in . Then is a function of as independent variables (free from the constraint ). We can consider (31) to be the Taylor expansion by independent variables , then substituting into (31) to obtain the approximation for about .
Now, substituting into (31), then by and , we have
λN+1=λ∗+(λN−λ∗)J(λ∗)+12!(λN−λ∗)H(λ∗)t(λN−λ∗)+o(∥λN−λ∗∥2). (35)
Then, by putting , (35) becomes
μN+1=μNJ(λ∗)+12!μNH(λ∗)tμN+o(∥μN∥2). (36)
By (22), we will investigate the convergence
μN→0,N→∞, (37)
based on the Taylor expansion (36). Let
μNi≡λNi−λ∗i,i=1,⋯,m, (38)
denote the components of , and write by components as , then we have
m∑i=1μNi=0,N=0,1,⋯, (39)
because .
5.1 Basic analysis for fast and slow convergence
For the investigation of the convergence speed, we consider the following simple case.
Let us define a real sequence by the recurrence formula;
μN+1 =θμN−ρ(μN)2,N=0,1,⋯, (40) 0 <θ≤1,ρ>0,0<μ0<θ/ρ. (41)
If , then we have , hence decays exponentially.
While, if , (40) becomes . This recurrence formula cannot be solved explicitly, however, we see the state of convergence by Fig.5.
Because the differential coefficient of the function at is 1, the convergence speed is very slow. In fact, this convergence is slower than exponential. From Lemma 7 in section 7 below, we will see that the convergence speed is the order and .
5.2 On Jacobian matrix J(λ∗)
Let us consider the Jacobian matrix for any . We are assuming in (13), hence .
We will calculate the components (32) of .
Defining
Di≡D(Pi∥λΦ),i=1,⋯,m, (42) Fi≡Fi(λ),i=1,⋯,m, (43)
we can write (11) as
Fi=λieDim∑k=1λkeDk,i=1,⋯,m. (44)
From (44),
Fim∑k=1λkeDk=λieDi, (45)
then differentiating the both sides of (45) by , we have
∂Fi∂λi′m∑k=1λkeDk+Fi∂∂λi′m∑k=1λkeDk=δi′ieDi+λieDi∂Di∂λi′, (46)
where is the Kronecker delta.
Before substituting into the both sides of (46), we define the following symbols. Remember that the integer was defined in (17). See also (25).
Let us define
Q∗ ≡Q(λ∗)=λ∗Φ, (47) Q∗j ≡Q(λ∗)j=m∑i=1λ∗iPij=m1∑i=1λ∗iPij,j=1,⋯,n, (48) D∗i ≡D(Pi∥Q∗),i=1,⋯,m, (49) D∗i′,i ≡∂Di∂λi′∣∣∣λ=λ∗,i′,i=1,⋯,m, (50) F∗i ≡Fi(λ∗),i=1,⋯,m. (51)
Lemma 4
m∑k=1λkeDk∣∣ ∣∣λ=λ∗=eC, (52) ∂Di∂λi′=−n∑j=1Pi′jPijQj,i′,i=1,⋯,m, (53) ∂∂λi′m∑k=1λkeDk∣∣ ∣∣λ=λ∗=eD∗i′−eC,i′=1,⋯,m, (54) F∗i=λ∗i,i=1,⋯,m. (55)
Proof: We have (52), (53) by simple calculation. See (19). (55) is the result of Lemma 2. (54) is proved as follows;
∂∂λi′m∑k=1λkeDk∣∣ ∣∣λ=λ∗ =m∑k=1(δi′keDk+λkeDk∂Dk∂λi′)∣∣ ∣∣λ=λ∗ =eD∗i′+m1∑k=1λ∗keC⎛⎝−n∑j=1PkjPi′jQ∗j⎞⎠ =eD∗i′−eCn∑j=1Pi′j1Q∗jm1∑k=1λ∗kPkj =eD∗i′−eC.
Note that , from Remark 2.
Substituting the results of Lemma 4 into (46), we have
∂Fi∂λi′
|
2021-10-26 14:52:05
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9195939898490906, "perplexity": 735.6709762212801}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587908.20/warc/CC-MAIN-20211026134839-20211026164839-00590.warc.gz"}
|
https://vuhavan.wordpress.com/2013/08/25/how-many-real-zeroes-does-a-random-polynomial-have/
|
Finding real zeroes of a polynomial (with real coefficients) is perhaps one of the oldest and most important problems in mathematics. There are many theorems which give estimates for the number of real roots. A famous example is Descartes’ rule of sign changes:
“The number of positive zeroes is at most the number of sign changes in the coefficients sequence. ”
For example, $x^3 +x^2-x-1$ has at most one positive real zero. As a matter of fact, it has exactly one positive zero. On the other hand, I do not know of a good way to give a lower bound on the number of real zeroes.
In general, the number of real zeroes can be seen as a function of the coefficients. If one chooses these coefficients randomly, then this number becomes a random variable. Waring was the first to study the distribution of this random variable (he considered cubic polynomials), but a systematic studies only started in the 1940s with a series of works of Littlewood and Offord, following some earlier consideration of Bloch and Polya. This series of papers has become the defining work of Offord’s career.
The precise setting of the problem is as follows. Let $c_0, \dots, c_n$ be deterministic numbers which may depend on $n$, and $\xi_0, \dots, \xi_n$ be iid copies of a real random variable $\latex \xi$ with mean zero and variance one. Let $N_n$ be the number of real zeroes of the random polynomial $f_n (z) = \sum_{i=0}^n c_i \xi_i z^i.$
One would like to study the distribution of $N_n$. Another, more difficult question is to study the distribution of the real zeroes themselves.
In the 40s to 60s many leading researchers (including Bloch, Polya, Littlewood, Offord, Erdos, Kac etc) have obtained significant results in bounding $N_n$ (either in probability or expectation), with respect to various atom variable $\xi$. In particular, Kac obtained an asymptotic sharp bound on $E (N_n)$ for the case $\xi$ is real gaussian and $c_0=c_1= \cdots =c_n=1$. In this case, the expectation of $N_n$ isapproximately C log n with $C =2/\pi$. His approach has led to the general Kac-Rice formula which is an important tool in the field. Several decades later, in 1998, Edelman and Kostlan came up with a nice geometric approach that allows one to compute the expectation of $N_n$ for a general sequence of $c_i$, given that the atom variable $\xi$ is still gaussian.
Motivated by the universality phenomenon, one would conjecture that given a “reasonable” coefficient sequence, any result for $N_n$ in the gaussian case should hold for a general variable $\xi$ with the same mean and variance. However, Edelman-Kostlan proof relies heavily on the geometric fact that a gaussian vector is rotationally symmetric. Even for the case when all $c_i=1$, extending Kac’s result to other random variables is highly trivial task and required a considerable amount of effort. In 1947, Kac extended his famous result in the gaussian case to the case when $\xi$ has the uniform distribution on the interval $[-1,1]$. In 1956, Erdos and Offord extended the result to the Bernoulli case ($\xi$ is uniform on the set $\{-1,+1\}$). Stevens in 1969 pushed the extension to a quite wide class of distributions, and around the same time Ibragimov and Maslova extended Kac’s result to all mean-zero distributions in the domain of attraction of the normal law, under the extra assumption that $Pr(\xi=0 )=0$. Few years later, Maslova computed the variance of $N_n$ and furthermore established a central limit theorem.
Let us now turn to the general case when $c_i$ can be rather arbitrary (and can depend on both $i$ and $n$ in particular). A deeper level of understanding $N_n$ is to determine the correlation functions of the real zeroes. These functions are essential tools in the study of random point processes. In simple cases, such as when the process is Poisson (one simply puts $n$ independent random points on $[-1,1]$, say), the correlation between the points is trivial (there is none). However, when the points come from
a complex system (such as zeroes of a random polynomial or eigenvalues of a random matrix), the situation is highly sophisticated, as the points strongly correlate and nearby points exercise some repelling force. (This is the reason why researchers like to use such processes for modeling in physics.) In the polynomial case, the correlation functions provide information about the interaction between real zeroes in a very short interval, where one expects to see only $O(1)$ zeroes. In principle, from there one can determine all moments of the number of zeroes in that short interval. Given these local information, in most cases one can determine all moments of the global variable $N_n$. For example, by cutting the real line into small intervals, one obtains the expectation of $N_n$ by the additivity of expectations. For more details, including the precise definition of correlation functions, we refer to Hough et. al. Thus, a more ambitious goal is to go for the universality of the correlation functions directly, instead of proving universality for $N_n$.
In the gaussian case, the correlation functions can be computed using Kac-Rice formula. (As a matter of fact, this formula is available in an abstract form for any atom variable $\xi$; but its explicit calculation is extremelly hard and one needs strong properties of gaussian vectors to see it through.) Universality for correlation functions has been established by Bleher and Di in 2004 for random binomial polynomials $latex c_i := \sqrt {n \choose i}$, given that the atom variable $\xi$ has a continuous distribution function which is sufficiently smooth. Their method relies on Kac-Rice formula. This formula gives an abstract form for correlation functions of all orders. As mentioned above, it is very hard to compute this formula in non-gaussian cases. Using analytic arguments, Bleher and Di showed that the difference between the value of the formula in the gaussian case and a general case is negligible. The authors mentioned that a similar argument would also work for Weyl polynomial ($c_i:= \sqrt {1/ i!}$), another class of well studied random polynomials. On the other hand, their proof required the analytic assumption on the distribution of $\xi$ in an essential way and so does not apply to discrete cases, such as Bernoulli. Furthermore, the exact formula for the coefficients $c_i$ (${n \choose i}^{1/2}$ and $\sqrt{1/i!}$, respectively) also plays a role and it is not clear if one can extend the result to other sequences.
In a recent attempt, Tao and I proposed another way to prove universality. What we did was to prove a general replacement principle which shows universality of correlations given some assumptions about the (joint) distribution of the values of $f_n (z)$ ((for a general sequence $c_0, \cdots, c_n$) and some easy bounds of the correlation functions in the gaussian case. It is not to difficult to show that these requirements are satisfied in the Weyl and binomial case. For Kac’s polynomial ($\latex c_i =1$) it requires a little more work, especially the application of Inverse Littlewood-Offord theory and a quantitatively version of Gromov’s growth theorem. The method also works for many other distributions; in fact we believe it works for most “natural” sequences. Our result can be seen as a two moment theorem for random polynomials (in order to compare to our earlier four moment theorem for random Hermitian matrices). However, the approach is quite different and more general. As a matter of fact, the polynomials considered in the paper does not need to have independent coefficients, thus our method can be applied to the characteristic polynomial of a random matrix as well.
While the above discussion sounds rather analytical, our proof, somewhat surprisingly, is essentially combinatorial in spirit. Here are two new, simple, (take home) arguments that I like a lot. Our starting idea is to use sampling to show that two quantities are approximately the same. In particular, we can approximate an integral by a random Riemann sum. As well known, this approximation is good if we have sufficiently many sampling points and the “variance” of the function in question is small, so Chebysev inequality can be used. The first twist in the proof here is that our function has a large variance. To fight this problem, we apply a smoothening process, which replaces a function $f$ by $f-g$ where $g$ is chosen carefully so that $\int g =0$ and $f-g$ is much smoother than $f$, which then reduces the “variance” significantly. Another new, useful idea, is a new kind of small ball probability (Littlewood-Offord type) using the lacunary structure of the coefficient sequence.
From → Không phân loại
7 phản hồi
Một vấn đề thú vị!
Thưa Thầy em đang là sinh viên mới ngành toán ở VN. Em cũng rất yêu thích các vấn đề về tổ hợp. Qua tìm hiểu em tìm được cuốn Additive Combinatorics của Thầy và GS Tao. Em có đọc sơ qua về các chương thì thấy nhiều kiến thức chưa biết, em xin hỏi là để đọc hiểu cuốn sách của Thầy em cần đọc những cuốn nào trước đó ạ. Cảm ơn Thầy rất nhiều về bài viết này.
• Em co the doc truoc cac sach co ban ve: Probability (Feller), Harmonic Analysis (Stein), Number Theory (Hardy), Graph Theory (West or Bollobas).
That ra quyen Additive Combinatorics la self-contained, tuc la cac khai niem dung doc dinh nghia tu dau, chi co speed move tuong doi nhanh, nen khong doc luot luot duoc.
Em xin cảm ơn Thầy về những hướng dẫn và lời khuyên. Chúc Thầy và gia đình luôn khỏe.
Dear Prof. Vu,
is there any generalization of Descartes’ rule of sign changes for random polynomial?
In other words, I am interested in the following question.
How many real positive zeroes does a random polynomial have?
Could you please give me any reference on this.
Thank you!
Yours,
Hong
• If you care only about polynomials with gaussian coefficients, the density function for real roots are know, using Kac’s formula. Thus, in principle, you can compute the expectation of number of real roots in any interval. See, for instance, Edelman-Kostlan paper in the Bullentin of the AMS.
• My recent paper with Tao (Universality of Random polynomials) extend Kac’s result to general ensembles.
|
2017-06-26 15:37:33
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 50, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8864197731018066, "perplexity": 864.7549729456492}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320823.40/warc/CC-MAIN-20170626152050-20170626172050-00083.warc.gz"}
|
https://mathhelpboards.com/threads/advanced-integration-techniques.3233/
|
### Welcome to our community
#### Be a part of something great, join today!
Status
Not open for further replies.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Hello
Integration is one of the most interesting topics in Mathematics. It has a wide area of applications in very different aspects of engineering and science. There are numerous ways to tackle integration problems. For elementary ones please refer to the following http://mathhelpboards.com/calculus-10/integral-calculus-tutorial-1711.html for basic introduction to integral calculus .
For the most part, I don't want you to be afraid by the word ''advanced'' the most important factor is practice and practice as long as you have basic knowledge of elementary integral calculus you should not get afraid. Some exercises here will require basic knowledge of certain properties of special functions which I will introduce before tackling certain problems (I will not focus on the proof) .Furthermore , basic background of complex variables will be such a great help here. I will try to leave certain problems for the reader with a final answer to try in your leisure time .
Finally , I will try to post an exercise every day so if you find any mistake don't hesitate to inform me . Furthermore , if you have any comments , face any problem understanding something or you want to show me your work just send me a pm or post it here and I will try to respond as soon as possible .
[new] I created a pdf for my tutorials here.
Last edited by a moderator:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
1. Differentiation under the integral sign : (Leibniz_integral)
This is one of the most commonly used techniques to solve numerous numbers of questions I will be using this technique to solve many other exercises in the coming posts
[HR][/HR]
Assume that we have the following function of two variables :
$$\displaystyle \int ^{b}_a \, f(x,y) \, dx$$
Then we can differentiate it with respect to y provided that f has partial continuous derivative on a chosen interval.
$$\displaystyle F'(y) = \int^{b}_{a} f_{y}(x,y)\,dx$$
[HR][/HR]
Now using this in many problems is not that clear you have to think a lot to get the required answer because many integral questions are just in one variable so you add the second variable and assume it is a function of two variables .
Assume we want to solve the following integral :
$$\displaystyle \int ^{1}_{0} \, \frac{x^2-1}{\ln (x) }$$
Now that seems very difficult to solve but using this technique we can solve it easily no matter how much power is x raised to . So the crux move is to decide ,where to put the second variable ! So the problem with the integral is that we have a logarithm in the denominator which makes the problem so difficult to tackle !
Remember that we can get a natural logarithm if we differentiate exponential functions i.e $F(a) = 2^a \Rightarrow \,\, F'(a) = \ln(a) \cdot 2^a$
Applying this to our problem
$$\displaystyle F(a)=\int ^{1}_{0} \, \frac{x^a-1}{\ln (x) }$$
Differentiate with respect to $a$
$$F'(b)=\frac{d}{d a}\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx$$
We see that the integral is a function with two variables $f(x,a)$
So we take the partial derivative with respect to a ..
$$F'(a)=\int^{1}_{0} \frac{\partial }{\partial a}\left(\frac{x^a-1}{\ln(x)}\right) \,\,dx$$
$$F'(a)=\int^{1}_{0} x^a\,\,dx$$
$$F'(a)=\frac{x^{a+1}}{a+1} \bigl]^1_0$$
$$F'(a)=\frac{1}{a+1}$$
Now integrate wrt to a to get $$\displaystyle F(a)$$
$$F(a)=\ln{(a+1)}+C$$
$$F(0)=\ln{1}+C\,\,\text{hence : } C=0$$
$$\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx=\ln{(a+1)}$$
By this powerful rule we were not only able to solve the integral we also found a general formula for some $$\displaystyle a$$ .
Now to solve our original integral put $$\displaystyle a=2$$
$$\int^{1}_{0} \frac{x^2-1}{\ln(x)}\,\,dx=\ln{(2+1)}=\ln(3)$$
[HW]
$$\int^{\infty}_{0}\frac{\sin (x) }{x}\,dx$$
To Be Continued ...
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
1.Differentiation under the integral sign (continued)
It is not always easy to find the function of two variables in which to differentiate. So that requires insight and ability to foresee the function.
[HR][/HR]
Here is an example , find the following integral
$$\displaystyle \int_0^{\frac{\pi}{2}} \frac{x}{\tan x}dx$$
[HR][/HR]
So where do we put the variable $$\displaystyle a$$ here , that doesn't seem to be straight forward , how do we proceed ?
Well, take a deep breath , as I said it is not always easy to deduce such a function , it might not be so straight forward .
Let us try the following
$$F(a)= \int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}dx$$
Now differentiate with respect to $$\displaystyle a$$ :
$$F'(a)= \int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx$$
We will show later that
$$\int_0^{\frac{\pi}{2}} \frac{1}{1+(a\tan(x) )^2}dx= \frac{\pi}{2(1+a)}$$
$$F'(a) = \frac{\pi}{2(1+a)}$$
Now Integrate both sides
$$F(a)= \frac{\pi}{2}\ln(1+a) +C$$
Now we need to substitute $a=0$ in order to find $C = 0$
$$\int_0^{\frac{\pi}{2}} \frac{\arctan(a\tan(x))}{\tan (x)}\,dx = \frac{\pi}{2}\ln(1+a)$$
Put $a =1$ in order to get our original integral
$$\int_0^{\frac{\pi}{2}} \frac{x}{\tan (x)}\,dx = \frac{\pi}{2}\ln(2)$$
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Differentiation under the integral sign (continued):
I already gave this as a practice problem I will solve it now to check your solution :
We are given the following :
$$\int^{\infty}_0 \frac{\sin(x) }{x}\, dx$$
This problem can be solved by many ways , but here we will try to solve it by differentiation .
So as I described earlier in the previous examples it is generally not so easy to find the function with two variables .
Actually this step might require trial and error techniques until we get the desired result , so don't just give up if an approach merely doesn't work !.
$$F(a)=\int^{\infty}_0 \frac{\sin(ax) }{x}\, dx$$
Let us try this one :
If we differentiated with respect to a we get the following :
$$F'(a)=\int^{\infty}_0 \cos(ax) \, dx$$
But unfortunately this integral doesn't converge , so this is not the correct one .
Well, that seemed hopeless , but you should benefit from mistakes . The previous integral will converge if there is an exponential (This is merely the Laplace transform which I will illustrate later ... ).
So let us try the following :
$$F(a)=\int^{\infty}_0 \frac{\sin(x) e^{-ax}}{x}\, dx$$
Take the derivative to get :
$$F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx$$
Now this is easy to solve we can use integration by parts twice to get the following :
$$F'(a)=-\int^{\infty}_0 \sin(x) e^{-ax}\, dx=\frac{-1}{a^2+1}$$
Now integrate both sides :
$$F(a)=-\arctan(a)+C$$
Now to find the value of the constant will take the limit as a grows very large :
$$C=\lim_{a\to \infty }F(a) +\arctan(a)= \frac{\pi}{2}$$
So we get our F(a) as the following:
$$F(a)=-\arctan(a)+ \frac{\pi}{2}$$
Now for the particular value of a = 0 we have :
$$\int^{\infty}_0 \frac{\sin(x) }{x}=\frac{\pi}{2}$$
Please if anything not clear you can send me a pm ...
In the next post I will start Hyperbolic Integration.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
2.Hyperbolic Integration : hyperbolic functions
Hyperbolic functions are very interesting , they are used to solve several integration problems. The most interesting thing about them is their relation to geometric functions through the see here.
It is really power full that you can switch between trigonometric functions and hyperbolic functions and vice versa , this requires a basic knowledge of complex numbers.
We know from the general definition of hyperbolic functions that :
$$\sinh(x) = \frac{e^x-e^{-x}}{2}$$
$$\cosh(x) = \frac{e^x+e^{-x}}{2}$$
Well, that seems interesting as it seems very close to the Euler's formula for sine and cosine :
$$\sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$
$$\cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$
It can be deduced easily that we can use the following to convert from trigonometric to hyperbolic functions :
$$\cos(ix)= \cosh(x)$$
$$\sin(ix)= i\sinh(x)$$
Woo , that is really fabulous so we can apply this to several problems to get rid of the complex i which is really annoying .
Here is an interesting problem :
$$\int - \left( \frac{\pi \csc\left(\pi (\frac{-1-ix}{2})\right)}{4}+\frac{\pi \csc\left(\pi (\frac{-1+ix}{2})\right)}{4}\right)\, dx$$
That seems very very very .... complicated , but wait don't be cheated by the complex view of a problem . Think about our hyperbolic friend.
Here we can apply basic trigonometric simplifications :
$$\int \frac{-\pi}{4}\left(\frac{1}{\sin\left(\frac{-\pi}{2})\cos(\frac{i\pi x}{2}\right)}+\frac{1}{\sin\left(\frac{-\pi}{2}\right)\cos\left(\frac{i\pi x}{2}\right)} \right) \, dx$$
Now we see how to apply our magic trick
$$\int \frac{\pi}{4}\left(\frac{1}{\cosh\left(\frac{\pi x}{2}\right)}+\frac{1}{\cosh\left(\frac{\pi x}{2}\right)} \right) \, dx$$
This can be further simplified to get the following :
$$\frac{\pi}{2}\int \text{sech}\left(\frac{\pi x}{2}\right)\,dx=2\arctan \left( \tanh\left(\frac{\pi x}{4}\right) \right)+C$$
Ooh , wait , hang on How is that ?!!
Well, differentiate the right side to see what is going on :
$$2\frac{\frac{\pi}{4} \text{sech}^2 \left(\frac{\pi x}{4}\right)}{1+\tanh^2 \left(\frac{\pi x}{4}\right)}$$
$$\frac{\pi}{2}\left(\frac{1}{\cosh^2\left(\frac{\pi x}{4}\right) +\sinh^2\left(\frac{\pi x}{4}\right)} \right)$$
Further simplification will do the task :
$$\frac{\pi}{2}\left(\frac{1}{\cosh^2(\frac{\pi x}{4}) +\cosh^2(\frac{\pi x}{4})-1}\right)=\frac{\pi}{2}\left(\frac{1}{\cosh(\frac{\pi x}{2})}\right)= \frac{\pi}{2}\text{sech}\left(\frac{\pi x}{2}\right)$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Hyperbolic Integration (continued)
As we have seen in the previous post hyperbolic functions can be so helpful as they help us simplify a lot of operations in a way that is really neat. Also, they help us handle complex variables to get a result that is completely free of complex numbers. This is basically because of the ability to convert from trigonometric functions though Euler formula.
Working with hyperbolic functions for the most part requires a lot of practice. It might not be so clear that the function can be integrated , this will help us in complex integration in the future .
Let us have the following example :
prove that :
$$I=\int -\frac{1}{2} \left( \csc(-\alpha-ix)+\csc(-\alpha+ix)\right)= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$$
$$\frac{1}{2}\int \csc(\alpha+ix)+\csc(\alpha-ix) \,dx$$
$$\frac{1}{2}\int \frac{1}{\sin(\alpha+ix)}+\frac{1}{\sin(\alpha-ix)}\, dx$$
Applying basic geometry identities :
$$=\frac{1}{2}\int \left(\frac{1}{\sin(\alpha)\cosh(x) +i\sinh(x) \cos(\alpha)}+\frac{1}{\sin(\alpha)\cosh(x) -i\sinh(x) \cos(\alpha)}\right)\, dx$$
Notice that the denominators are the complex conjugate of each other :
$$\frac{1}{2}\int \left(\frac{2\sin(\alpha)\cosh(x)}{\sin^2(\alpha) \cosh ^2(x) +\sinh^2(x) \cos^2(\alpha)}\right)\, dx=\, \int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha)(1+\sinh^2(x)) +\sinh^2(x) (1-\sin^2(\alpha))}\right)\, dx$$
$$\int \left(\frac{\sin(\alpha)\cosh(x)}{\sin^2(\alpha) +\sinh^2(x)}\right)\, dx= \int \frac{\csc(\alpha)\cosh(x)}{ 1+\csc^2(\alpha)\sinh^2(x)}\, dx$$
$$I= \arctan\left(\csc(\alpha) \sinh(x) \right) +C$$
I will start Laplace transform in the next post ...
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
3.Integration by Laplace Transform (Laplace)
3.1.Basic Introduction :
Laplace transform is a very powerful transform. It can be used in many applications . For example, it can be used to solve Differential Equations and its rules can be used to solve integration problems .
The basic definition of Laplace transform :
$$F(s) = \mathcal{L}(f(t)) = \int ^{\infty}_0 e^{-st}f(t) \, dt\, \text{ ----(1)}$$
This integral will converge when $$\text{Re}(s)>a \text{ for } |f(t)| \leq Me^{at}$$ (Exponential Type )
Let us see the Laplace transform for some functions f(t) :
1.f(t) =1 Let us find F(s) :
This is elementary $$F(s) =\int ^{\infty}_0 e^{-st} \, dt= \frac{1}{s}$$
2.Also in general $f(t) = t^n \text{ where }n \geq 0$
We can prove by parts that $$F(s) =\int ^{\infty}_0 e^{-st} t^n \, dt=\frac{n!}{s^{n+1}}\text{ ----(2)}$$ ( Try to prove it )
3.What about $f(t)= \cos(at)$
Also by parts we can prove that $$F(s) =\int ^{\infty}_0 e^{-st} \cos(at) \, dt=\frac{s}{s^2+a^2}\text{ ----(3)}$$
You can assistant a table for Laplace transform ...
One of the most interesting results of the Laplace :
$$(f * g)(t)= \int^{t}_0 f(s)g(t-s)\,ds\text{ ----(4)}$$ (Convolution)
$$\mathcal{L}\left((f * g)(t)\right)= \mathcal{L}(f(t)) \mathcal{L}(g(t)) \text{ ----(5)}$$
You can see the proof here.
Let us see some examples on integration:
Find the following integral :
$$\int^{\infty}_{0}e^{-2t}t^3\,dt$$
We can directly use the above formula
$$\int ^{\infty}_0 e^{-st}t^n \, dt=\frac{n!}{s^{n+1}}$$
here we have s= 2 , and n =3 , so directly we get :
$$\int^{\infty}_{0}e^{-2t}t^3\,dt= \frac{3!}{2^{3+1}}=\frac{3}{8}$$
So , we see that is becoming easy to find ...
Now let us think about the Laplace Inverse :
So, basically you are given F(s) and we want to get f(t) this is denoted by :
$$\mathcal{L}^{-1}(F(s))= f(t)\text{ or }\mathcal{L}^{-1}(\mathcal{L}(f(t)))= f(t)$$
We are given a function in the variable s and we want to transform it into another in the variable t .
Suppose we have the following examples :
Find the Laplace inverse of the following :
1) $$\frac{1}{s^3}$$
2) $$\frac{s}{s^2+4}$$
1- we can use the result in (2) so we have
$$\mathcal{L}(t^2)= \frac{2!}{t^3}\,\, \Rightarrow \,\, \frac{1}{2}\mathcal{L}(t^2)= \frac{1}{s^3}$$
Now take the inverse to both sides :
$$\frac{t^2}{2}= \mathcal{L}^{-1}\left(\frac{1}{s^3}\right)$$
2- we can use the result in (3) so we have
$$\cos(2t)=\mathcal{L}^{-1}\left(\frac{s}{s^2+4}\right)$$
[HW.1] Find the Laplace of $\sin(at)$
[HW.2] Find the inverse Laplace of $$\frac{1}{s^{n+1}}$$
To be continued ...
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration by Laplace Transform (continued)
3.2.1.Interesting results :
Prove the following :
$$B(x+1,y+1)=\int^{1}_{0}t^{x}\, (1-t)^{y}\,dt= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$ B is Beta function and $\Gamma$ is Gamma function
Note : Just as a basic introduction to special functions, you can think of gamma function as the following : $n!=\Gamma{(n+1)}$ we can extend n to exist in the whole complex plane {except n is a negative integer }.
The beta function is so interesting , we will explain some of its basic properties later .It can be used to solve many integrals.
For the proof :
We have the defnition (4) in the previous post about convolution .
Let us choose some functions f , and g :
$f(t) = t^{x} \,\, , \, g(t) = t^y$
By substituting in (4) we get the following :
$$(t^x*t^y)= \int^{t}_0 s^{x}(t-s)^{y}\,ds \text{ ---- (*)}$$
We have from definition (5) that :
$\mathcal{L}\left(t^x*t^y\right)= \mathcal{L}(t^x) \mathcal{L}(t^y )$
Now we can use the result (2) to deduce
$$\mathcal{L}\left(t^x*t^y\right)= \frac{x!\cdot y!}{s^{x+y+2}}$$
Notice that we need to find the inverse of Laplace $\mathcal{L}^{-1}$
$$\mathcal{L}^{-1}\left(\mathcal{L}(t^x*t^y)\right)=\mathcal{L}^{-1}\left( \frac{x!\cdot y!}{s^{x+y+2}}\right)=t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$
So we have the following :
$$(t^x*t^y) =t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!}$$
Now substitute in --(*) we get :
$$t^{x+y+1}\frac{x!\cdot y!}{(x+y+1)!} = \int^{t}_0 s^{x}(t-s)^{y}\,ds$$
This looks good , put t=1 we get :
$$\frac{x!\cdot y!}{(x+y+1)!} = \int^{1}_0 s^{x}(1-s)^{y}\,ds$$
By using that $n! = \Gamma{(n+1)}$
we arrive happily to our formula :
$$\int^{1}_0 s^{x}(1-s)^{y}\,ds= \frac{\Gamma(x+1)\Gamma{(y+1)}}{\Gamma{(x+y+2)}}$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration by Laplace Transform (continued)
3.2.2.Interesting results :
Prove the following :
$$\int^{\infty}_0 \frac{f(t)}{t}\,dt=\, \int^{\infty}_0 \, \mathcal{L}(f(t))\, ds\text{ ----(6)}$$
This is a very powerful rule it is basically saying that if we have a function divided by its independent variable and we integrated it in the half positive plane then we can transform it into an integral wrt to the Laplace transform of the function .
So Let us prove it :
we will start from the right hand side of (6)
we know from the definition --(1) that :
$$\int^{\infty}_0 \, \mathcal{L}(f(t))\, ds= \int^{\infty}_0 \,\left( \int^{\infty}_0e^{-st}\, f(t) \,dt\right)\, ds$$
Now by the Fubini's theorem we can rearrange* the double integral :
$$\int^{\infty}_0 \, f(t) \left(\int^{\infty}_0 e^{-st}\,ds\right)\,dt \text{ ----(**)}$$
The integral inside the parenthesis :
$$\int^{\infty}_0 e^{-st}\,ds$$
I know this is elementary ,but you may realize this is the Laplace transform but now the function is f(s) so basically we are just evaluating F(t) when f(s) =1 .
$$\int^{\infty}_0 e^{-st}\,ds = \mathcal{L}(1)= \frac{1}{t}$$
Now substitute this value in the integral in (**)
$$\int^{\infty}_0 \, \frac{f(t)}{t}\,dt$$
which is the left hand side of ----(6) as just required ...
This is just a fabulous rule , it can be used to simplify many computations ...
Let us find the following integral :
$$\int^{\infty}_0 \frac{\sin(t)}{t}\,dt$$
This is not the first time we see this integral and not the last . We have seen that we can find it using differentiation under the integral sign .
But you are just about to see the power of definition (6) [Doe that look like a movie trailer ] , never mind .
Back to our integral , so the integral is of the type $\frac{f(t)}{t}$ so we can easily find it .
$$\int^{\infty}_0 \frac{sin(t)}{t}\,dt= \int^{\infty}_{0}\mathcal{L}(\sin(t))\,ds$$
Now what is the Laplace transform of $\sin(t)$ , did you do your homework ?
$$\mathcal{L}(\sin(at)) = \frac{a^2}{s^2+a^2}$$
So putting a =1 we get :
$$\mathcal{L}(\sin(t)) = \frac{1}{s^2+1}$$
Just substitute in our integral :
$$\int^{\infty}_0 \frac{ds}{1+s^2}= \tan^{-1}(s)|_{s=\infty}-\tan^{-1}(s)|_{s=0}=\frac{\pi}{2}$$
Just as expected ...
In the next post , I will explain the Gamma function .
[HR][/HR]* We must prove that we can rearrange the double integral , I didn't go into the details.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions
4.1.Gamma Function :
Gamma function is really interesting it is used to solve many interesting integrals, here we try to define some basic properties , prove some of them and take some examples.
Definition :
$$\Gamma{(x+1)} = \int^{\infty}_0 e^{-t}\, t^{x}\, dt \text{ ---(1)}$$
For the first glance that just looks like the Laplace Transform , actually they are closely related .
So let us for simplicity assume that x=n where $$n\geq 0 \text{ , n }$$ (is an integer )and substitute in (1) we get :
$$\Gamma{(n+1)} = \int^{\infty}_0 e^{-t}\, t^{n}\, dt$$
Well, we can use the Laplace transform - you might revise if you forgot -
$$\int^{\infty}_0 e^{-t}\, t^{n}= \frac{n!}{s^n+1}|_{s=1}= n!$$
So we see that there is a relation between the gamma function and the factorial ..
We will assume for the time being that the gamma function is defined as the following
$$n!= \Gamma{(n+1)}$$.
By this definition n\geq 0 where n is any positive integer which is pretty limited but surely this definition will be soon replaced by a stronger one.
Let us have some Examples :
Find the following integrals :
$$\int^{\infty}_0 \, e^{-t}t^4\, dt$$
By definition (1) this can be replaced by
$$\int^{\infty}_0 \, e^{-t}t^4\, dt =\Gamma(4+1)= 4! = 24$$
Pretty good let us continue with some more examples :
1) $$\int^{\infty}_0 \, e^{-t^2}t \, dt$$
2) $$\int^{1}_{0}\ln (t) \, t^2 \, dt$$
1) For the first one clearly we need a substitution before we go ahead :
so let us start by putting $$x=t^2$$ so the integral becomes :
$$\frac{1}{2}\int^{\infty}_0 \, e^{-x}x^{\frac{1}{2}}\cdot x^{-\frac{1}{2}}\, dt =\frac{1}{2}\Gamma(1+0)=\frac{1}{2}$$
2) For the second we use the substitution $$t=e^{-\frac{x}{2}}$$
$$\int^{1}_{0}\ln (t) \, t^2 \, dt = -\frac{1}{4}\int^{\infty}_{0}e^{-\frac{3x}{2}}\cdot x\,dx$$
Using another substitution $$t=\frac{3x}{2}$$ (Back to my favourite symbol )
$$\frac{-1}{9}\int^{\infty}_0 e^{-x}\, x\, dx =\frac{- \Gamma(2)}{9}= \frac{-1}{9}$$
It is an important thing to get used to the symbol $\Gamma$. I am sure that you are saying (boring...) that this seems elementary , but my main aim here is to let you practice the new symbol and get used to solving some problems using it .
[HW] prove $$\frac{\Gamma(5)\cdot \Gamma(2)}{\Gamma(7)}=\frac{1}{30}$$
[HW] Find the integral : $$\int^{\infty}_{0}e^{-\frac{1}{60}t}\, t^{20}\, dt$$ in terms of the gamma function .
In the next post I should extend the idea of a factorial ... you know this is only get exciting ...
To be continued.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.1.Gamma Function (continued) :
For simplicity we assumed int the gamma definition (1) (look the previous post) that gamma only works for positive integers . This definition was so helpful as we assumed the relation between gamma and factorials . Actually ,this restricts the gamma function so much , we want to exploit the real strength of this very nice function .
Hence, we must extend the gamma function to work for all real numbers except for some values. Actually we will see soon that we can extend it to work for all complex numbers except where the function has poles .
There are many representations for the gamma function :
$$\Gamma(z) = \lim_{n=\infty}\frac{n! \, \, n^z}{z(z+1)(z+2) \, . \, . \, . \, (z+n)}\text{ ---- (2) }$$
$$\Gamma(z)\,= \frac{e^{-\gamma z}}{z} \prod_{n=1}^\infty \left(1 + \frac{z}{n}\right)^{-1} e^{z/n}\text{ ----(3)}$$
Let us focus for the time being on the first representation :
This tells us a lot about the analyticity of gamma function , we notice that the gamma function is analytic everywhere except for z={0,-1,-2,-3, ... } , so zero and the negative integers are excluded and the function is not defined in this set .
We know have extended the gamma function and we want to evaluate interesting values .
Let us start by the following :
Find the integral :
$$\int^{\infty}_{0}\, \frac{e^{-t}}{\sqrt{t}}\, dt$$
Now according to definition (1) this is equal to $$\Gamma\left(\frac{1}{2}\right)$$ but we want to find a value for this if it exists ?
Laplace transform will not help here since we don't know what is the Laplace of a root !
Let us first make a substitution $$\sqrt{t}=x$$
so we have the integral as : $$2\int^{\infty}_{0}\, e^{-x^2}\, dx$$
Now to find this integral we need to do a simple trick ....
$$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx \right)^2$$
we will first try to find the upper integral ...
$$\left( \int^{\infty}_{0}\, e^{-x^2}\, dx\right)^2=\left( \int^{\infty}_{0}\, e^{-x^2}\, dx\right)\cdot \left(\int^{\infty}_{0}\, e^{-x^2}\, dx\right)$$
Now this is the crux movement (stay seated , don't shout, it is all ok )
$$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx\right)\cdot \left(\int^{\infty}_{0}\, e^{-y^2}\, dy\right)$$
So we have done nothing just changed the second x by sub
.
Now since they are two independent variables we can do the following :
$$\int^{\infty}_0\int^{\infty}_{0}\, e^{-(x^2+y^2)}\, dy \, dx$$
Now by polar substitution we can do the following :
$$\int^{\frac{\pi}{2} }_0\int^{\infty}_{0}\, e^{-r^2}\, r\,dr \, d\theta$$
so this becomes elementary :
$$\int^{\frac{\pi}{2}}_0\,\frac{1}{2}\, d\theta= \frac{\pi}{4}$$
So we have $$\left(\int^{\infty}_{0}\, e^{-x^2}\, dx \right)^2=\frac{\pi}{4}$$
Take the square root to both sides :
$$\int^{\infty}_{0}\, e^{-x^2}\, dx =\frac{\sqrt{\pi}}{2}$$
$$2\int^{\infty}_{0}\, e^{-x^2}\, dx =\sqrt{\pi}$$
So we have our result $$\fbox{[tex] \Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$$ }[/tex]
This result is very interesting and will use it to solve many integrals .
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.1.Gamma Function (continued) :
There are still lots of properties of gamma function one of which is related to the
factorial function . we know that $$n!=n(n-1)(n-2)\, ... \, 4\cdot 3\cdot 2\cdot 1$$.
For gamma function we have the following property :
$$\Gamma(x+1) = x\Gamma(x)$$
This is interesting because we can find some values for certain inputs for gamma .
Assume that we want to find
$$\Gamma\left(\frac{3}{2}\right)$$:
If we used this property we get :
$$\Gamma\left(1+\frac{1}{2}\right)= \frac{1}{2}\cdot\Gamma\left(\frac{1}{2}\right)=$$$$\frac{\sqrt{\pi}}{2}$$
Not all the time the result will be reduced to a simpler form as the previous example. For example we dont know how to express $$\Gamma(\frac{1}{4})$$ in a simpler form but we can approximate its value :
$$\Gamma(\frac{1}{4})\approx 3.6256 \, ...$$
so sometimes we just solve some integrals in terms of gamma function since we don't know a simpler form ...
For example solve the integral :
$$\int^{\infty}_0 e^{-t}t^{\frac{1}{4}}$$
we know by definition --- (1) of gamma function that this reduces to:
$$\int^{\infty}_0 e^{-t}t^{\frac{1}{4}}=\Gamma\left(\frac{5}{4}\right)=$$$$\frac{\Gamma\left(\frac{1}{4}\right)}{4}$$
We have seen that $$\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$$ but what about $$\Gamma\left(\frac{-1}{2}\right)$$ ?
The question is elementary since we know that $$\Gamma\left(1-\frac{1}{2}\right)=\frac{-1}{2}\Gamma\left(\frac{-1}{2}\right)$$
so we have that $$\Gamma\left(-\frac{1}{2}\right)=-2\sqrt{\pi}$$
Then we can prove that any fraction with the denomenator =2 and the numerator is odd can be reduced into
$$\Gamma\left(\frac{2n+1}{2}\right)= C \,\Gamma\left(\frac{1}{2}\right)$$ where $$C\in \mathbb {Q}\text{ and n}\in \mathbb{Z}$$
ِAs a practice for Gamma find :
$$\int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt$$
We have learned that gamma function has an exponent muliplied by a polynomial but here we have a hyperoblic function !
Hange on we know that we can expand cosh using power series :
$$\cosh(x) = \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$
put $x=a\sqrt{t}$ we have
:
$$\cosh(a\sqrt{t}) = \sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!}$$
subsituting back in the integral we have :
$$\int_{0}^{\infty}e^{-t}\sum_{n=0}^{\infty}\frac{a^{2n}\cdot t^n}{(2n)!\, \sqrt{t}}dt$$
Now since the series is always positive we can swap the integral and the series :
$$\sum_{n=0}^{\infty}\frac{a^{2n}}{(2n)!}\left[ \int_{0}^{\infty}e^{-t} t^{n-\frac{1}{2}}dt\right]$$
so we have using gamma :
$$\sum_{n=0}^{\infty}\frac{a^{2n}\cdot \Gamma(n+\frac{1}{2})}{2n!}$$
But we don't know how to simplify further because we need another property which will help us find the solution .
To be continued ...
Last edited:
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.1.Gamma Function (continued) :
In the previous post we were stuck trying to simplify $$\Gamma(n+\frac{1}{2})$$ but ,you know, the following property will make our life just easier.
Legendre Duplication Formula: (LDF)*
$$\Gamma\left(\frac{1}{2}+n\right) = {(2n)! \over 4^n n!} \sqrt{\pi}$$
The proof of this identity requires some manipulation of beta function .
So , we have just got our free gift to continue our example , so we were stuck at :
$$\sum^{\infty}_{n=0} \frac{a^{2n}\,\Gamma\left(\frac{1}{2}+n\right)}{(2n!)}$$
so using LDF we get the following :
$$\sum^{\infty}_{n=0}\frac{a^{2n}\,\left({(2n)! \over 4^n n!} \sqrt{\pi}\right)}{(2n!)}$$
Further simplification we get :
$$\sqrt{\pi}\,\sum^{\infty}_{n=0} \frac{{a^{2n}\over 4^n }}{n!}$$
Now that looks familiar since we know that :
$$\sum^{\infty}_{n=0}\frac{z^n}{n!}=e^z$$
Putting $$z={a^2\over 4 }$$ and multiplying by $\sqrt{\pi}\,$ we get :
$$\sqrt{\pi}\sum^{\infty}_{n=0} \frac{\left({a^2\over 4 }\right)^n}{n!}=\sqrt{\pi}\,e^{{a^2\over 4 }}$$
So we have finally that :
$$\int_{0}^{\infty}\frac{e^{-t}\cosh(a\sqrt{t})}{\sqrt{t}}dt\,=\,\sqrt{\pi}e^{{a^2 \over 4}}$$
Finally we have the interesting property Euler's Reflection Formula : (ERF)*
$$\Gamma \left({z}\right) \Gamma \left({1 - z}\right) = \frac \pi {\sin \left({\pi z}\right)} \,\,\forall z \notin \mathbb{Z}$$
Let us have some examples on that :
Simplify the following expressions :
1. $$\Gamma\left(\frac{3}{4}\right)\,\Gamma\left(\frac{1}{4}\right)$$
2. $$\Gamma{\left(\frac{1+i}{2}\right)} \, \Gamma{\left(\frac{1-i}{2}\right)}$$
Solution :
1. we can rewrite as $$\Gamma\left(1-\frac{1}{4}\right)\,\Gamma\left(\frac{1}{4}\right)$$
Now by ERF we have the following :
$$\Gamma\left(1-\frac{1}{4}\right)\,\Gamma\left(\frac{1}{4}\right)=\frac \pi {\sin \left({\frac{\pi}{4} }\right)}=\sqrt{2}\,\pi$$
2. We can rewrite as $$\Gamma{\left(\frac{1+i}{2}\right)} \, \Gamma{\left(1-\frac{1+i}{2}\right)}$$
This expression simplifies to:
$$\frac{\pi}{\sin\left(\frac{\pi(1+i)}{2}\right)}= \frac{\pi}{\cos \left( \frac{i \pi}{2}\right)}$$
By geometry to hyperbolic conversions we get :
$$\frac{\pi}{\cosh\left(\frac{\pi}{2}\right)}=\pi \, \text{sech}\left(\frac{\pi}{2}\right)$$
LDF and ERF will be so much beneficial when we discuss the beta function in the next post .
[HR][/HR]* I will use these shortcuts when ever calling these formulas.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.2.Beta Function :
General definition :
$$\int^1_0 t^{x-1}(1-t)^{y-1} \, dt= B(x,y)$$
This is actually my favourite function , and what is so nice about it , that it has many representations. Also it is related to Gamma function :
$$B(x,y)\,=\,{\Gamma(x)\Gamma(y) \over \Gamma(x+y)}$$
We have proved this identity earlier when we discussed convolution.
We shall realize the symmetry of beta function that is to say $$B(x,y) = B(y,x)$$.
Beta function has many other representations all can be deduced through substitution :
$$B(x,y) \,=\, \int ^{\infty}_0 \frac{t^{x-1}}{(1+t)^{x+y}}\, dt$$
$$B(x,y) \,=\,2\,\int^{\frac{\pi}{2}}_0 \cos^{2x-1}(t) \sin^{2y-1}(t)\,dt$$
The proofs are left to the reader as practice ...
Let us discuss some examples :
Find the following integral :
$$\int^{\infty}_0 \frac{1}{(x^2+1)}\, dx$$
we know that $$\int^{\infty}_0 \frac{1}{(x^2+1)}=\frac{\pi}{2}$$
Let us try to solve it by beta ,First put $x=\sqrt{t}$
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)}\,dt$$
Now we know the second representation of beta so we have :
$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$
$$x+y={1} \,\, \Rightarrow \,\, y={1\over 2}$$
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)}\,dt=\frac{B({1 \over 2},{1 \over 2})}{2}=\frac{\Gamma({1 \over 2})\, \Gamma({1\over 2})}{2}= \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{2}=\frac{\pi}{2}$$
Now let us try to solve another similar integral :
$$\int^{\infty}_0 \frac{1}{(x^2+1)^2}\, dx$$
Using the same substitution as the previous example we get :
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^2}\,dt$$
$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$
$$x+y={2} \,\, \Rightarrow \,\, y={3\over 2}$$
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^2}\,dt=\frac{B({1 \over 2},{3 \over 2})}{2}=\frac{\Gamma({1 \over 2})\, \Gamma({3\over 2})}{2}= \frac{\Gamma({1 \over 2})\, \Gamma({1\over 2})}{4}=\frac{\pi}{4}$$
We might try to generalize :
$$\int^{\infty}_0 \frac{1}{(x^2+1)^n}\, dx \,\,\, \forall\,\, n> \frac{1} {2}$$
Using the same substitution again and again ..(boring) ..
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt$$
$$x-1={-1 \over 2} \,\, \Rightarrow \,\, x={1\over 2}$$
$$x+y={n} \,\, \Rightarrow \,\, y=n-{1\over 2}$$
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt=\frac{\Gamma ({1\over 2})\cdot \Gamma\left(n-{1\over 2}\right)}{2\Gamma(n)}$$
Now how to simplify $$\Gamma\left(n-{1\over 2}\right)$$ ?
But remember that by LDF that $$\Gamma\left(k+{1\over2}\right) =\frac{(2k)!\sqrt{\pi}}{4^k\,k!}$$
Now let $$k=n-1$$ so we have $$\Gamma\left(n-{1\over2}\right) =\frac{(2n-2)!\sqrt{\pi}}{4^{n-1}\,(n-1)!}=\frac{\Gamma\left(2n-1\right)\sqrt{\pi}}{4^{n-1}\,\Gamma(n)}$$
Substituting in our integral we have the following :
$$\frac{1}{2}\int^{\infty}_0 \frac{t^{\frac{-1}{2}}}{(t+1)^n}\,dt=\frac{2\pi \,\cdot \Gamma\left(2n-1 \right)}{4^n \cdot \Gamma^2(n)}$$
$$\int^{\infty}_0 \frac{1}{(x^2+1)^n}\, dx \,=\frac{\pi \,\cdot \Gamma\left(2n-1 \right)}{2^{2n-1}\cdot \Gamma^2(n)}$$
It is easy to say that for $n\in \mathbb{Z}^+ \,$ we get a $\pi$ multiplied by some rational number ...
[HW] Verify the previous exercises using the formula ...
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.2.Beta Function (continued):
Prove that :
$$\int_0^1\frac{x^n}{\sqrt{1-x}}\,dx=2\,\frac{(2n)!!}{(2n+1)!!}$$
Where the double factorial !! is defined as the following :
$$n!! = \begin{cases}n\cdot(n-2)... 5\cdot 3\cdot 1&; \mbox{if }n>0 \, \text{odd} \\n\cdot(n-2)... 6\cdot 4\cdot 2&; \mbox{if } n>0 \, \text{even}\\1 &; \mbox{if } n=0,-1\end{cases}$$
Solution :
Now , we shall be familiar by just looking that this is the beta function :
$$\int_0^1 x^n\cdot (1-x)^{-\frac{1}{2}}\, dx$$
$$\fbox{[tex]x-1 = n \,\, \Rightarrow \,\, x=n+1\,\,\,\,$$
$$y-1= \frac{-1}{2}\,\, \Rightarrow \,\, y=\frac{1}{2}$$}[/tex]
$$\int_0^1 x^n\cdot (1-x)^{-\frac{1}{2}}\, dx=B(n+1,\frac{1}{2})$$
$$B\left( n+1,\frac{1}{2} \right)=\frac{\Gamma \left( \frac{1}{2} \right)\Gamma(n+1)}{\Gamma \left( n+\frac{3}{2}\right)}=\frac{2 \sqrt{\pi}\, \Gamma(n+1)}{(n+\frac{1}{2})\Gamma \left( n+\frac{1}{2} \right)}$$
Now you shall realize that we must use LDF:
$$\frac{\sqrt{\pi}\Gamma(n+1)}{(n+\frac{1}{2})\Gamma\left(n+\frac{1}{2}\right)}=\frac{2\sqrt{\pi}\, n!}{(2n+1)\sqrt{\pi}\, \frac{(2n)!}{4^n\, n!}}$$
$$\frac{2\, 2^{2n} (n!)^2}{(2n)!}=2\frac{2^{2n}(n\cdot(n-1)\,...\,3\cdot 2\cdot 1)^2}{(2n+1)\cdot 2n\cdot(2n-1)\, ... \, 3\cdot 2\cdot 1}$$
Now we should separate odd and even terms in the denomenator :
$$2\frac{2^{2n}(n\cdot(n-1)\,...\,3\cdot 2\cdot 1)^2}{(2n\cdot(2n-2)\, ... \, 4\cdot 2)((2n+1)\cdot (2n-1)\, ... \, 3\cdot 1)}$$
$$2\frac{(2n\cdot(2n-2)\,...\,6\cdot 4\cdot 2)^2}{(2n \cdot (2n-2)\, ... \, 4\cdot 2)((2n+1)\cdot (2n-1)\, ... \, 3\cdot 1)}=2\frac{(2n)!!}{(2n+1)!!}$$
[HR][/HR]
Find the following integral:
$$\int^{\infty}_{-\infty}\, \left(1+\frac{x^2}{n-1}\right)^{-\frac{n}{2}}\,dx$$
Solution :
First we shall realize the evenness of the integral :
$$2\int^{\infty}_{0}\, \left(1+\frac{x^2}{n-1}\right)^{-\frac{n}{2}}\,dx$$
We should think of a substitution to get the second beta integral ...
Let $$t=\frac{x^2}{n-1}\,\,\, \Rightarrow \,\,\, dt =\frac{2x}{n-1}\,dx$$
$$\sqrt{n-1}\int^{\infty}_{0}\frac{t^{-\frac{1}{2}}}{(1+t)^{\frac{n}{2}}}\, \,dx$$
Now we see that our integral becomes so familiar :
$$\sqrt{n-1}\, B\left(\frac{1}{2},\frac{n-1}{2}\right)= \frac{\sqrt{\pi(n-1)}\Gamma\left(\frac{n-1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}$$
[HW] Prove $$\int^{\infty}_{0}\frac{x^{2m+1}}{(ax^2+c)^n} dx\,=\frac{m!\,(n -m -2)!}{2(n-1)!\, a^{m+1}\,c^{n-m-1}}$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.2.Beta Function (continued):
Find the following integral:
$$\int^{\infty}_0\, \frac{x^{-p}}{x^3+1}\,dx$$
Solution :
We should foresee that this is the second beta integral.
Let us do the substitution :
$$x^3\,=t\,$$
$$\frac{1}{3}\, \int^{\infty}_0\, \frac{t^{-\frac{p+2}{3}}}{t+1}\,dt$$
Now we should find x, y :
$$x =\frac{1-p}{3}$$
$$y+x =1\,\,\, \Rightarrow \,\, y = 1-\frac{1-p}{3}$$
so we have our beta representation of the integral :
$$\frac{B\left(\frac{1-p}{3}, \frac{1-p}{3}\right)}{3}= \frac{\Gamma\left(\frac{1-p}{3}\right) \Gamma\left(1-\frac{1-p}{3}\right)}{3}$$
Now we should use the
ERF :
$$\frac{\Gamma \left( \frac{1-p}{3}\right) \Gamma\left(1-\frac{1-p}{3}\right)}{3}= \frac{\pi}{3\, \sin\left(\frac{\pi(1-p)}{3}\right)}= \, \frac{\pi}{3}\,\csc\left(\frac{\pi-\pi p}{3}\right)$$
[HR][/HR]
Now let us try to find :
$$\int^{\frac{\pi}{2}}_0 \, \sqrt[3]{\sin^{2}x} \,\, dx$$
Solution:
Rewrite as :
$$\int^{\frac{\pi}{2}}_0 \, \sin^{\frac{3}{2}}x \,\cos^{0}x\, dx$$
$$2x-1= \frac{3}{2}\,\, \Rightarrow \,\,\, x = \frac{5}{4}$$
$$2y-1= 0\,\, \Rightarrow \,\,\, y = \frac{1}{2}$$
$$\int^{\frac{\pi}{2}}_0 \, \sin^{\frac{3}{2}}x \, dx= \frac{B\left(\frac{5}{4}, \frac{1}{2}\right)}{2}=\frac{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{7}{4}\right)}$$
simplification is left to the reader as practice ...
[HR][/HR]
Find the following integral :
$$\int^{\frac{\pi}{2}}_{0} (\sin x)^i \cdot (\cos x)^{-i}\, dx$$
Solution :
Don't be intimidated by the complex looking of the integral , we have the third beta integral :
$$2x-1= i \,\, \Rightarrow \,\,\, x = \frac{1+i}{2}$$
$$2y-1= -i \,\, \Rightarrow \,\,\, y = \frac{1-i}{2}$$
$$\,\frac{1}{2} \Gamma\left(\frac{1+i}{2}\right) \, \Gamma\left(\frac{1-i}{2}\right)$$
Now we see that we have to use ERF
$$\,\frac{1}{2} \Gamma\left(\frac{1+i}{2}\right) \, \Gamma\left(1-\frac{1+i}{2}\right)= \frac{\pi}{2\sin\left(\frac{\pi(1+i)}{2}\right)}= \frac{\pi}{2}\text{sech}\,\left( {\pi \over 2}\right)$$
Digamma function is explored in the next post
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.3.Digamma (psi) function :
General definition :
$$\psi(x) \, = \, \frac{\Gamma'(x)}{\Gamma(x)}$$
Assume that we have $$F(x)= \ln \left(\Gamma(x)\right)$$ then if we differentiate both sides we get $$f(x)=\frac{\Gamma'(x)}{\Gamma(x)}$$ then $f(x)= \psi(x)$ .
So clearly we see the strong relation between gamma and digamma through differentiation .
One of the most used results in solving integrals is the fact that can be obtained from the general definition $$\Gamma'(x)=\psi(x)\, \Gamma(x)$$.
Differentiating the gamma function is pretty unique since we can get the same function multiplied by digamma.
Assume we want to differentiate the following:
$$\frac{\Gamma(2x+1) }{\Gamma(x)}$$
We can use differentiating rule for quotients as the following :
$$\frac{2\Gamma'(2x+1)\Gamma(x)-\Gamma'(x)\Gamma(2x+1) }{\Gamma^2(x)}$$
Now we can use the result $$\Gamma'(x)=\psi(x)\, \Gamma(x)$$ :
$$\frac{2\Gamma(2x+1)\psi(2x+1)\Gamma(x)-\psi(x)\Gamma(x)\Gamma(2x+1) }{\Gamma^2(x)}$$
Now we can separate the numerator and simplify :
$$\frac{\Gamma(2x+1)}{\Gamma(x)} \left( 2\psi(2x+1)-\psi(x) \right)$$
Now that is interesting it is like we have just multilplied by the difference of psi if we neglect 2 .
[HR][/HR]
A very interesting result is the following :
$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$
Poof :
we know by ERF that:
$$\Gamma(x)\Gamma(1-x)=\pi \csc(\pi x)$$
Now differentiate both sides :
$$\psi(x)\Gamma(x)\Gamma(1-x)-\psi(1-x)\Gamma(x)\Gamma(1-x)=-\pi^2 \csc(\pi x)\,\cot(\pi x)$$
Take $$-\Gamma(x)\Gamma(1-x)$$ as a common factor :
$$\Gamma(x)\Gamma(1-x)\left(\psi(1-x)-\psi(x)\right)=-\pi^2 \csc(\pi x)\,\cot(\pi x)$$
Now we see the ERF again so we can simplify to get :
$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$
[HR][/HR]
That was basically a differentiating tutorial in the next post we will solve some integrals ...
[HW] $$\psi(1+x)-\psi(x)= \frac{1}{x}$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.3.Digamma function(continued) :
I gave this as HW :
$$\psi(1+x)-\psi(x) = \frac{1}{x}$$
Proof:
Let us start by the following :
$$\frac{\Gamma(1+x)}{\Gamma(x)}= x\text{ ---- (1) }$$
Now differentiate both sides :
$$\frac{\Gamma(1+x)}{\Gamma(x)}\left(\psi(1+x)-\psi(x)\right)= 1$$
$$\psi(1+x)-\psi(x)= \frac{\Gamma(x)}{\Gamma(1+x)}$$
The right hand side is the reciprocal of (1)
$$\psi(1+x)-\psi(x) = \frac{1}{x}$$
OK, Now that we have proved this identity we will use it in the next example.
[HR][/HR]
Find the following integral :
$$\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx\text{ *}$$
Solution:
Let us try by finding the following :
$$\int^{\infty}_{0} \frac{x^a}{(1+x^2)^2}\, dx$$
Now use the following substitution : $$x^2= t$$
$$\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt$$
By the beta function this is equivalent to :
$$\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt = \frac{1}{2}B\left(\frac{a+1}{2},2-\frac{a+1}{2}\right)=\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(2-\frac{a+1}{2}\right)$$
Now let the following :
$$F(a) =\frac{1}{2}\int^{\infty}_{0}\frac{t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(2-\frac{a+1}{2}\right)$$
Differentiate with respect to a
$$F'(a) =\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{a-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma\left(\frac{a+1}{2}\right)\Gamma \left(2-\frac{a+1}{2}\right)\left[\psi \left(\frac{a+1}{2}\right)-\psi \left(2-\frac{a+1}{2}\right) \right]$$
Now put a =0 so we have :
$$\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{-1}{2}}}{(1+t)^2}\, dt =\frac{1}{4}\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)\left[\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)\right]$$
Now we use our identity :
$$\psi \left(\frac{1}{2}\right)-\psi \left(\frac{3}{2}\right)=-\left(\psi \left(1+\frac{1}{2}\right)-\psi \left(\frac{1}{2}\right)\right)=-2$$
Also by some gamma manipulation we have :
$$\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{3}{2}\right)=\frac{1}{2}\,\Gamma^2 \left(\frac{1}{2}\right)=\frac{\pi}{2}$$
so the integral reduces to :
$$\frac{1}{4}\int^{\infty}_{0}\frac{\log(t) \,t^{\frac{-1}{2}}}{(1+t)^2}\, dt=-\frac{\pi}{4}$$
putting $$x^2= t$$ we have our result :
$$\int^{\infty}_{0}\frac{\log(x) }{(1+x^2)^2}\, dx=-\frac{\pi}{4}$$
To be continued ...
[HR][/HR]* In higher studies we usually donate the natural logarithm as log
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)
4.3.Digamma function(continued) :
I introduced the
Weierstrass representation of gamma function :
$$\Gamma(x)=\frac{e^{-{\gamma}\,x}}{x}\,\prod^{\infty}_{n=1}\left(1+x/n \right)^{-1}e^{\frac{x}{n}}$$
Now to derive the digamma function :
First take the natural logarithm to both sides :
$$\log\left(\Gamma(x)\right)=-\gamma\, x\,-\,\log(x)\,+\sum^{\infty}_{n=1}-\ln\left(1+\frac{x}{n}\right)+{x\over n}$$
Now we shall differentiate with respect to x to get digamma :
$$\psi(x)=-\gamma\,-\frac{1}{x}\,+\sum^{\infty}_{n=1}\frac{\frac{-1}{n}}{1+\frac{x}{n}}+{1 \over n}$$
Further simplification will result in the following :
$$\psi(x)=-\gamma\,-\frac{1}{x}\,+\sum^{\infty}_{n=1}\frac{x}{n(n+x)}$$
[HR][/HR]
Now let us evaluate some values :
1- $$\psi(1)=-\gamma\,-1 \,+\sum^{\infty}_{n=1}\frac{1}{n(n+1)}$$
$$\sum^{\infty}_{n=1}\frac{1}{n(n+1)}=1$$ (see why ? )
$$\fbox{[tex]\psi(1)\,=\,-\gamma$$}[/tex]
2- $$\psi\left(\frac{1}{2}\right)=-\gamma\,-2\,+\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}$$
We need to find : $$\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}$$
This requires the result $$\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n}=-\log(2)$$
So we can prove easily that :
$$\sum^{\infty}_{n=1}\frac{1}{n(2n+1)}=2-2\log(2)$$
hence , $$\fbox{ [tex]\psi \left( \frac{1}{2} \right)=-\gamma \,-\, 2\log (2)$$} [/tex]
[HR][/HR]
Let us know have some examples on integrals :
Find the following integral :
$$\int^{\infty}_{0}e^{-a t}\, \log(t) \, dt$$
Let us start by noting that seems like a gamma function , but differentiated so :
$$F(b)= \int^{\infty}_{0}e^{-at}\, t^b\,dt$$
Now use the substitution $x=at$ we get
$$F(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\, \left(\frac{x}{a}\right)^b\,dx$$
Now that looks exactly as the gamma definition so :
$$F(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\, \left(\frac{x}{a}\right)^b\,dx=\frac{\Gamma(b+1)}{a^{b+1}}$$
Now differentiate with respect to b :
$$F'(b)= \frac{1}{a}\int^{\infty}_{0}e^{-x}\,\log\left(\frac{x}{a}\right) \left(\frac{x}{a}\right)^b\,dx=\frac{\Gamma(b+1) \psi (b+1)}{a^{b+1}}-\frac{\log(a)\Gamma(b+1)}{a}$$
Now put $b=0$ and $at=x$
$$\int^{\infty}_{0}e^{-at}\,\ln(t) \,dx=\frac{\psi(1)-\log(a)}{a}=-\frac{\gamma+\log(a)}{a}$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration lessons continued ....
4.Integration using special functions (continued)
4.3.Digamma (psi) function :
Prove the following identity :
$\displaystyle \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\,dz = \psi(a)$
[HR][/HR]
Solution :
We begin with the double integral :
$$\displaystyle \int^{\infty}_0 \int^t_1 \,e^{-xz}\,dx\,dz=\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz$$
Using fubini theorem we also have :
$$\displaystyle \int^t_1 \int^{\infty}_0\,e^{-xz}\,dz \,dx = \int^t_1 \frac{1}{x}\,dx = \ln t$$
Hence we have the following :
$$\displaystyle \int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz = \ln(t) \text{ --------(1)}$$
We also know that :
$$\displaystyle \Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\ln t \, dt \text{ --------(2)}$$
Substituting (1) in (2) we have :
$$\displaystyle \Gamma'(a) = \int^{\infty}_0 t^{a-1}e^{-t}\,\left(\int^{\infty}_{0}\frac{e^{-z}-e^{-tz}}{z}\, dz \right)\, dt$$
$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dz \, dt$$
Now we can use the fubini theorem :
$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \int^{\infty}_{0}\frac{t^{a-1}e^{-t}e^{-z}-t^{a-1}e^{-t(z+1)}}{z}\, dt \, dz$$
$$\displaystyle \Gamma'(a) = \int^{\infty}_0 \frac{1}{z} \left( e^{-z}\int^{\infty}_{0}t^{a-1}e^{-t}\, dt-\int^{\infty}_0 t^{a-1}e^{-t(z+1)}\, dt \right)\, dz$$
But we can easily deduce using Laplace that :
$$\displaystyle \int^{\infty}_0 t^{a-1}e^{-t(z+1)}\,dt= \Gamma(a) \,(z+1)^{-a}$$
Aslo we have :
$$\displaystyle \int^{\infty}_{0}t^{a-1}e^{-t}\, dt=\Gamma(a)$$
Hence we can simplify our integral to the following :
$$\displaystyle \Gamma'(a) = \Gamma(a)\, \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z} dz$$
$$\displaystyle \frac{\Gamma'(a)}{\Gamma(a)} = \int^{\infty}_0 \frac{e^{-z}-(1+z)^{-a}}{z}\, dz = \psi(a)$$
[HR][/HR]
[HW] Prove that $$\displaystyle \int^{\infty}_{0}\, \frac{e^{-ax}-e^{-bx}}{x}\, dx=\ln\left({b\over a}\right)$$
you may use differentiation under the integral sign or double integration
[HW] Find $$\displaystyle \int^{\infty}_0 \left(e^{-x} -\frac{1}{x+1}\right) \frac{dx}{x}$$
[HR][/HR]
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration lessons continued
4.Integration using special functions (continued)
4.3.Digamma function(continued) :
We have the following digamma property :
$\displaystyle \psi(s+1)\,=\, -\gamma \,+\, \int^{1}_{0}\frac{1-x^s}{1-x}$
[HR][/HR][HR][/HR]
Prove the following integral :
$$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$
[HR][/HR]
Solution :
As crazy as it looks , it becomes very easy to solve if we know how to start !
First note that since there is a log in the denominator that gives as an idea to differentiate ...
$$\displaystyle \text{Let : }F(c)=\int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$$
Differentiate with respect to c :
$$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a)(1-x^b)x^c}{(1-x)}dx$$
$$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a-x^b+x^{a+b})x^c}{(1-x)}dx$$
$$\displaystyle F'(c)=\int_0^1 \frac{x^c\,-\,x^{a+c}\,-\,x^{b+c}\,+\,x^{a+b+c}}{(1-x)}dx$$
$$\displaystyle F'(c)=\int_0^1 \frac{(x^c\,-1)\,+\,(1-\,x^{a+c})\,+\,(1-\,x^{b+c})\,+\,(x^{a+b+c}-1)}{(1-x)}dx$$
$$\displaystyle F'(c)=-\int_0^1 \frac{1-x^c\,}{1-x}\,dx+\int_0^1\frac{1-x^{a+c}}{1-x}\,dx+\int_0^1\frac{1-x^{b+c}}{1-x}\,dx-\int_0^1\frac{1-x^{a+b+c}}{1-x}dx$$
Now use the following result :
$$\displaystyle \psi(s+1)=-\gamma+\int^{1}_0\frac{1-x^s}{1-x}\,dx$$
$$\displaystyle F'(c)=-\psi(c+1)-\gamma+\psi(a+c+1)+\gamma+\psi(b+c+1)+\gamma-\psi(a+b+c+1)-\gamma$$
$$\displaystyle F'(c)=-\psi(c+1)+\psi(a+c+1)+\psi(b+c+1)-\psi(a+b+c+1)$$
Integrate with respect to c we have :
$$\displaystyle F(c)=-\log\left[\Gamma(c+1)\right]+\log \left[\Gamma(a+c+1)\right] +\log\left[\Gamma(b+c+1)\right]-\log \left[\Gamma(a+b+c+1)\right] +C_1$$
Which reduces to :
$$\displaystyle \log\left[\frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma (c+1) \Gamma (a+b+c+1)} \right] +C_1$$
Now put c= 0 we have :
$$\displaystyle 0=\log \left[ \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right]+C_1$$
$$\displaystyle C_1=-\log \left[ \frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$$
So we have the following :
$$\displaystyle F(c)=\log\left[ \frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma(c+1) \Gamma (a+b+c+1)}\right] -\log\left[\frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$$
Hence we have the result :
$$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$$
To be Continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration lessons continued ....
4.Integration using special functions (continued)
4.3.Digamma function(continued) :
We continue with some more exercises :
[HR][/HR]
Find the following integral :
$\displaystyle \int^{\infty}_0\left(e^{-bx}-\frac{1}{1+ax}\right)\,\frac{dx}{x}$
[HR][/HR]
Solution :
Let us first use the substitution t = ax so we get the following :
$$\displaystyle \int^{\infty}_0\left(e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$$
Now we must realize the result we get earlier in this series that :
$$\displaystyle \int^{\infty}_0 \frac{e^{-x}-(1+x)^{-a}}{x}\,dx = \psi(a)$$
But there we don't have $$\displaystyle e^{-x}$$ so let us add and subtract it :
$$\displaystyle \int^{\infty}_0\left(e^{-t}-e^{-t}+e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$$
$$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\,\frac{dt}{t}+\int^{\infty}_0 \frac{e^{-\frac{bt}{a}}-e^{-t}}{t}\, dt$$
$$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\frac{dt}{t}=-\gamma$$
We already also proved several posts ago that :
$$\displaystyle \int^{\infty}_0 \frac{e^{-\frac{bt}{a}}- e^{-t}}{t}\, dt= -\log\left(\frac{b}{a}\right) = \log\left(\frac{a}{b}\right)$$
Hence the result :
$$\displaystyle \int^{\infty}_0 \left(e^{-bx}-\frac{1}{1+ax} \right)\, \frac{dx}{x} = \log\left(\frac{a}{b} \right)-\gamma$$
[HR][/HR]
Prove the following integral :
$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$
[HR][/HR]
Solution :
We know the following hyperoblic identity : (hopefully )
$$\displaystyle \text{coth}(x)= \frac{1+e^{-2x}}{1-e^{-2x}}$$
$$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\frac{1+e^{-2x}}{1-e^{-2x}} \right) \, dx$$
Now let 2x =t so we have :
$$\displaystyle \int^{\infty}_0 \, e^{-\left(\frac{at}{2}\right)} \left(\frac{1}{t}-\frac{1+e^{-t}}{2(1-e^{-t})} \right) \, dt$$
$$\displaystyle \int^{\infty}_0 \, \frac{e^{-\left(\frac{at}{2}\right)}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}+e^{\left(-\frac{at}{2}-t\right)}}{2(1-e^{-t})} \, dt$$
I will add and subtract some terms :
$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}+e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t}\,-\,\frac{e^{-\left(\frac{at}{2}\right)}+e^{-\left(\frac{at}{2}\right)}-e^{-\frac{at}{2}}+e^{-\left(\frac{at}{2}\right)-t}}{2(1-e^{-t})} \, dt$$
$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\,dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{\left(-\frac{at}{2}-t\right)}} {2(1-e^{-t})}\, dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt$$
First : We use the identity :
$$\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$$
$$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\, dt=\psi \left(\frac{a}{2}\right)$$
The second integral can easily be proven :
$$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-\left(\frac{at}{2}-t\right)}}{2(1-e^{-t})}=\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}}{2}\,dt=\frac{1}{a}$$
We also proved earlier that :
$$\displaystyle \int^{\infty}_0\frac{e^{-t}\,-\,e^{-st}}{t} \, dt = \ln \left( s \right)$$
$$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt = -\ln \left(\frac{a}{2}\right)$$
Hence we have the full integral :
$$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$$
[HR][/HR]
[HW] Prove : $$\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$$
[HR][/HR]
Zeta function ,,, is on the way
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration lessons (continued )
4.Integration using special functions (continued)
4.4.Riemann Zeta function :
You surely had heard about this interesting function, not only has it resulted in many interesting founding but also the yet to be solved Riemann Hypothesis makes it one of the most celebrated functions in history of mathematics. Its interesting aspect is the relation to prime numbers.
In this tutorial we will not go so deep in the proofs and analyticity of it, we will try to describe its properties and mainly focus on the integral representation and its relation to other functions.
Let us first start by defining the zeta function as the following :
$\displaystyle \zeta(s) \,=\, \sum_{n=1}^{\infty}\,\frac{1}{n^s}\,\,\, \,\, \text{Re}(s)>1$
There are more general representation of the zeta function but we will stick to the one we defined above. For the first glance it seems that zeta has no relations to other functions but it turned out that it has a strong relation to other functions such as gamma and digamma functions.
4.4.1 zeta and gamma representation :
$\displaystyle \Gamma(s) \, \zeta(s) \, = \int^{\infty}_{0}\, \frac{t^{s-1}}{e^t-1}\, dt$
This relation turns out to be so much interesting since we can evaluate the right-hand integral using known results for both zeta and gamma.
This most famous value for zeta function is when $s=2$ which represents the infinite sum of reciprocals of squares :
$\displaystyle \zeta(2) =\sum^{\infty}_{n=1} \frac{1}{n^2}\,= \frac{\pi^2}{6}$
Actually we can derive all the values of $\displaystyle \zeta(2k) \,\,\,\, k\in \mathbb{Z}$ but unfortunately there is no known way to find the zeta for odd integers $\displaystyle \zeta(2k+1)\,$ .
Now let us use this result to find some integrals :
$\displaystyle \int^{\infty}_0 \frac{t}{e^t-1} \,dt$
It follows directly from the gamma-zeta relation that taking $s=2$ we have :
$\displaystyle \Gamma(2) \, \zeta(2) \, = \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt$
$$\displaystyle \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt=\zeta(2)=\frac{\pi^2}{6}$$
[HR][/HR]
Solve the following integral :
$\displaystyle \int_0^{\frac{\pi}{2}} \,\frac{\log \left(\sec(x) \right)}{\tan(x)} \,dx$
use the substitution : $\displaystyle \sec(x)=e^t$
$$\displaystyle \int_0^{\infty} \, \frac{t}{e^{2t}-1}\,dt\, =\, \frac{1}{4} \cdot \int_0^{\infty}\,\frac{t}{e^t-1}\,dt=\frac{\pi^2}{24}$$
Another similar relation between zeta and gamma is the following identity :
$\displaystyle \Gamma(s)\, \zeta(s) \, (1-2^{1-s})\, = \int^{\infty}_0\, \frac{t^{s-1}}{e^t+1}\,$
It follows from that equation that :
$$\displaystyle \int^{\infty}_0\, \frac{t}{e^t+1}\,dt = \frac{\pi^2}{12}$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Integration lessons continued ....
4.4.2.Zeta function and Bernoulli numbers
We will see in this section some values of the zeta function using an equation due to Euler , but let us first define the Bernoulli numbers .
It is usually defined as $$\displaystyle B_{k}$$ and the easiest way to derive them is find the coefficients of the power series of $$\displaystyle \frac{x}{e^x-1}$$
[HR][/HR]
Definition
$$\displaystyle \frac{x}{e^x-1} = \sum_{k \geq 0} \frac{B_k}{k!}x^k$$
[HR][/HR]
Now let us derive some values for the Bernoulli numbers , rewrite the power series as
$$\displaystyle x = (e^x-1) \, \sum_{k\geq 0} \frac{B_k}{k!}x^k$$
$$\displaystyle x = \left( x+\frac{1}{2!}x^2 + \frac{1}{3!} x^3+ \frac{1}{4!} x^4+\cdots \right) \cdot \left( B_0 +B_1 \, x+\frac{B_2}{2!}\,x^2 +\frac{B_3}{3!} x^3+ \cdots \right)$$
$$\displaystyle x = B_0 x + \left(B_1+\frac{B_0}{2!} \right)x^2 + \left(\frac{B_0}{3!}+ \frac{B_2}{2!}+\frac{B_1}{2!}\right)x^3 +\left( \frac{B_0}{4!}+\frac{B_1}{3!}+\frac{B_2}{2!\, 2!}+\frac{B_3}{3!}\right) x^4+ \cdots$$
By comparing the terms we get the following values
$$\displaystyle {B_0 = 1 \, , \, B_1 =-\frac{1}{2} \, + \, B_2 = \frac{1}{6} \, , \, B_3=0 \, , \, B_4 = -\frac{1}{30} , \cdots }$$
Actually continuing with this we deduce
• $$\displaystyle B_{2k+1}=0 \,\, \, \, \, \,\,\, \, \, \, \, \forall \,\, \, \, \, \, k\in \mathbb{Z}^+$$
• Every Bernoulli number depends on all the numbers before it so it can be defined recursively .
• $$\displaystyle B_{2k} \, , \, k\geq 1$$ are alternating .
[HR][/HR]
Now according to Euler we have the following interesting result :
$$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$
[HR][/HR]
PROOF
Let us start by the following definition due to Euler
$$\displaystyle \frac{\sin(z)}{z} = \prod_{n\geq1} \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$
Take the logarithm to both sides
$$\displaystyle \ln(\sin(z)) - \ln(z) = \sum_{n\geq1} \ln \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$$
By differentiation with respect to z
$$\displaystyle \cot(z) -\frac{1}{z} = \sum_{n\geq1} \frac{-2 \frac{z}{n^2 \, \pi^2 }}{1-\frac{z^2}{n^2 \, \pi^2}}$$
By simple algaberic manipulation we have
$$\displaystyle z\cot(z) = 1+2\sum_{n\geq1} \frac{z^2}{z^2-n^2 \, \pi^2}$$
$$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \frac{z^2}{ n^2 \, \pi^2} \left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right)$$
Now using the power series expansion
$$\displaystyle \frac{1}{1-\frac{z^2}{\pi^2 \, n^2}}= \sum_{k\geq 0} \frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$$ converges for $$\displaystyle |z|< \pi \, n$$
$$\displaystyle \frac{z^2}{ n^2 \, \pi^2}\left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right) = \sum_{k\geq 1}\frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$$ notice the change in the index !
So now the sums becomes
$$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \sum_{k \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$$
Now if we invert the order of summation we have
$$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \sum_{n \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$$ (Wow do you notice !)
$$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \frac{\zeta(2k) }{\pi^{2k}}z^{2k} \text{ }\cdots(1)$$
Euler didn't stop here , he used power series estimation for $$\displaystyle z\cot(z)$$ using the Bernoulli numbers
staring by the equation $$\displaystyle \frac{x}{e^x-1} = \sum_{k\geq 0} \frac{B_k}{k!}x^k$$
by putting $$\displaystyle x=2iz$$ we have
$$\displaystyle \frac{2iz}{e^{2iz}-1} = \sum_{k\geq0} \frac{B_k}{k!}{(2iz)}^k$$
Which can be reduced directly to the following by noticing that $$\displaystyle B_{2k+1}=0, \,\,\, k>0$$
$$\displaystyle z \cot(z) = 1- \sum_{k\geq 1}(-1)^{k-1} B_{2k} \frac{2^{2k}}{(2k)!}z^{2k} \text{ } \cdots(2)$$
The result is immediate by comparing (1) and (2) $$\displaystyle \square$$
[HR][/HR]
[HW] find $$\displaystyle \zeta(4) \, ,\, \zeta(6) \, , \, B_5 \, , \, B_6$$
To be continued ...
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....
4.4.3. Hurwitz zeta and polygamma functions
Hurwitz zeta is a generalization of the zeta function by adding a parameter . This intimate relation between the two functions arises from multiple differentiations of the the digamma function .
[HR][/HR]
Let us first start by defining the Hurwitz zeta function
Definition
$$\displaystyle \zeta(a,z) \, = \, \sum_{n\geq 0} \frac{1}{(n+z)^a}$$
[HR][/HR]
Note : according to this definition we have $$\displaystyle \zeta(a,1) = \zeta(a)$$
Let us define the polygamma function as the function produced by differentiating the digamma function and it is often denoted by $$\displaystyle \psi_{n}(z) \,\,\,\, \forall \,\, n\geq 0$$ . We define the digamma function by setting $n=0$ so it's denoted by $\psi_0(z)$ .
Other values can be found by the following recurrence relation $$\displaystyle \psi'_{n}(z) = \psi_{n+1}(z)$$ , so we have $$\displaystyle \psi_{1}(z) = \psi'_{0}(z)$$
[HR][/HR]
We have the following relation between Hurwitz zeta and the polygamma function
DEFINITION
$$\displaystyle \psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z) \,\,\, \, \forall \,\, n\geq 1$$
[HR][/HR]
PROOF
We have already proved the following relation
$$\displaystyle \psi_{0}(z) = -\gamma -\frac{1}{z}+ \sum_{n\geq 1}\frac{z}{n(n+z)}$$
This can be written as the following
$$\displaystyle \psi_{0}(z) = -\gamma + \sum_{k\geq 0}\frac{1}{k+1}-\frac{1}{k+z}$$
By differentiating with respect to $z$
$$\displaystyle \psi_{1}(z) = \sum_{k\geq 0}\frac{1}{(k+z)^2}$$
$$\displaystyle \psi_{2}(z) = -2\sum_{k\geq 0}\frac{1}{(k+z)^3}$$
$$\displaystyle \psi_{3}(z) = 2 \cdot 3 \, \sum_{k\geq 0}\frac{1}{(k+z)^4}$$
$$\displaystyle \psi_{4}(z) = -2 \cdot 3 \cdot 4 \,\sum_{k\geq 0}\frac{1}{(k+z)^5}$$
$$\displaystyle \text{ }\cdot$$
$$\displaystyle \text{ }\cdot$$
$$\displaystyle \text{ }\cdot$$
$$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\sum_{k\geq 0}\frac{1}{(k+z)^{n+1}}$$
We realize the RHS is just the Hurwitz zeta function
$$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\zeta(n+1,z)$$
As required to prove $$\displaystyle \square$$.
[HR][/HR]
By setting $z=1$ we have an equation in terms of the ordinary zeta function
$$\displaystyle \psi_{n}(1) = (-1)^{n+1}n!\,\zeta(n+1)$$
Now since we already proved in the preceding section that
$$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$
we can easily verify the following
$$\displaystyle \psi_{2k-1}(1)= (2k-1)!\, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$$
$$\displaystyle \psi_{2k-1}(1)= (-1)^{k-1} B_{2k} \frac{2^{2k-2}}{k}{\pi}^{2k} \,\,\, \, , \,\, k\geq 1$$
This can be used to evaluate some values for the polygamma function
$$\displaystyle \psi_{1}(1)= \frac{\pi^2}{6} \,\, , \,\, \psi_{3}(1) = \frac{\pi^4}{15}$$
Other values can be evaluated in terms of the zeta function
$$\displaystyle \psi_{2}(1)= -2 \zeta(3) \,\, , \,\, \psi_{4}(1) = -24 \zeta(5)$$
To Be Continued ...
Status
Not open for further replies.
|
2021-06-18 18:40:30
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "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, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9486082792282104, "perplexity": 520.2697495091273}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00621.warc.gz"}
|
https://apache.googlesource.com/xerces-c/+/6da3df209543499899f022d03e4d47bf63dd111f/obj/Makefile.in
|
blob: fbb380c326f7ee8c005edf480cf93294bd7b7954 [file] [log] [blame]
# # The Apache Software License, Version 1.1 # # Copyright (c) 1999 The Apache Software Foundation. All rights # reserved. # # Redistribution and use in source and binary forms, with or without # modification, are permitted provided that the following conditions # are met: # # 1. Redistributions of source code must retain the above copyright # notice, this list of conditions and the following disclaimer. # # 2. Redistributions in binary form must reproduce the above copyright # notice, this list of conditions and the following disclaimer in # the documentation and/or other materials provided with the # distribution. # # 3. The end-user documentation included with the redistribution, # if any, must include the following acknowledgment: # "This product includes software developed by the # Apache Software Foundation (http://www.apache.org/)." # Alternately, this acknowledgment may appear in the software itself, # if and wherever such third-party acknowledgments normally appear. # # 4. The names "Xerces" and "Apache Software Foundation" must # not be used to endorse or promote products derived from this # software without prior written permission. For written # permission, please contact apache\@apache.org. # # 5. Products derived from this software may not be called "Apache", # nor may "Apache" appear in their name, without prior written # permission of the Apache Software Foundation. # # THIS SOFTWARE IS PROVIDED AS IS'' AND ANY EXPRESSED OR IMPLIED # WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES # OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE # DISCLAIMED. IN NO EVENT SHALL THE APACHE SOFTWARE FOUNDATION OR # ITS CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, # SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT # LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF # USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND # ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, # OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT # OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF # SUCH DAMAGE. # ==================================================================== # # This software consists of voluntary contributions made by many # individuals on behalf of the Apache Software Foundation, and was # originally based on software copyright (c) 1999, International # Business Machines, Inc., http://www.ibm.com . For more information # on the Apache Software Foundation, please see # . # # # $Log$ # Revision 1.4 1999/11/24 23:21:42 rahulj # Now works under HPUX 10.20 with CC and aCC, with ICU and native distclean target now remove and remaining Makefile(s) clean target now removes the template repository directory # # Revision 1.3 1999/11/23 01:59:16 rahulj # Code now works under HPUX 11. Tested inmemory message loader. Revamped makefiles. Builds with both DCE threads as well as pthread libraries. # # Revision 1.2 1999/11/12 20:38:43 rahulj # Replaced XML4CROOT with XERCESCROOT. # # Revision 1.1.1.1 1999/11/09 01:10:49 twl # Initial checkin # # Revision 1.2 1999/11/08 20:43:14 rahul # Swat for adding in Product name and CVS comment log variable. # # ################################################################### # IMPORTANT NOTE # ################################################################### # If you are going to do the OS390BATCH build, make sure you have # # the OS390BATCH environment variable set. # # # # export OS390BATCH=1 # # # ################################################################### PLATFORM = @platform@ CC = @cc@ CXX = @cxx@ PREFIX = @prefix@ LDFLAGS = @ldflags@ LIBS = @libs@ OSVER = @osver@ include ../src/Makefile.incl LIB=${XERCESCROOT}/lib ## OS390BATCH ifeq (${OS390BATCH},1) BATCH_TARGET = "//'${LOADMOD}(XML4C30)'" endif LIBNAME=libxerces-c THISLIB=${LIB}/${LIBNAME} VER=1_0 ALL_OBJECTS_DIR=${XERCESCROOT}/obj ALL_OBJECTS=$(wildcard$(ALL_OBJECTS_DIR)/*.o) ####################################################### # HP-UX with CC compiler is awkward since it does # # not instantiate templates automatically. To do so # # you need to invoke the linker. So, just for this # # compiler, we try to create a dummy executable # # before we create the shared library. This will # # force all templates to be instantiated and the # # shared library is created with all template # # instantiations in place. # ####################################################### ifeq (${PLATFORM},HPUX) ifeq (${CXX},CC) ###### Start HP-UX CC compiler specific stuff ####### DUMMYEXE=$(XML_LIB_DIR)/dummyExe TEMPLATESOBJS=$(wildcard $(TEMPLATESREPOSITORY)/*.o) all:$(DUMMYEXE) ${THISLIB}${VER}${SHLIBSUFFIX} \rm -f$(DUMMYEXE) $(DUMMYEXE):$(ALL_OBJECTS) ${MAKE_SHARED} -o${@} $^$(PLATFORM_LIBRARIES) $(ALLLIBS) ###### End HP-UX CC compiler specific stuff ####### else all:${THISLIB}${VER}${SHLIBSUFFIX} endif else ## OS390BATCH ifeq (${OS390BATCH},1) all:${BATCH_TARGET} else all: ${THISLIB}${VER}${SHLIBSUFFIX} endif endif ######################################################## # # # OS/390 works pretty much the same way as Windows NT # # as far as linking goes. The runtime library is # # called .dll, and the link-time libraries have a .x # # extension. You need the .x files to link # # your application at build time and need the .dll # # file to run it. So, we need to copy over the .x # # file to the lib directory as well, so that you can # # link your application. # ########################################################${THISLIB}${VER}${SHLIBSUFFIX}: $(ALL_OBJECTS) @echo Building${THISLIB}${VER}${SHLIBSUFFIX} ${MAKE_SHARED} -o${@} $^$(TEMPLATESOBJS) $(PLATFORM_LIBRARIES)$(EXTRA_LINK_OPTIONS) $(ALLLIBS) ## OS390 ifeq (${PLATFORM},OS390) ## OS390BATCH ifeq (${OS390BATCH},1)${BATCH_TARGET}: $(ALL_OBJECTS) @echo Building${BATCH_TARGET} ${MAKE_SHARED} -o${@} $^$(TEMPLATESOBJS) $(PLATFORM_LIBRARIES)$(EXTRA_LINK_OPTIONS) $(ALLLIBS) else$(CP) ${LIBNAME}${VER}${OS390SIDEDECK}${LIB} endif endif clean: @echo "Making clean in obj ..." -rm -f $(ALL_OBJECTS)${THISLIB}${VER}${SHLIBSUFFIX} ifneq ($strip$(TEMPLATESREPOSITORY)),) -rm -f $(TEMPLATESREPOSITORY)/*.o -rm -f$(TEMPLATESREPOSITORY)/*.c -rm -f $(TEMPLATESREPOSITORY)/*.cs -rm -f$(TEMPLATESREPOSITORY)/*.he endif distclean: clean rm -f Makefile ifneq ($strip$(TEMPLATESREPOSITORY)),) -rm -rf $(TEMPLATESREPOSITORY) endif install: ifeq (${OS390BATCH},1) @echo "Nothing to do for OS390BATCH ..." else -mkdir -p ${PREFIX}/lib$(CP) ${THISLIB}${VER}${SHLIBSUFFIX}$(PREFIX)/lib ifeq (${PLATFORM},OS390)$(CP) ${LIBNAME}${VER}${OS390SIDEDECK}$(PREFIX)/lib endif endif
|
2021-01-16 22:31:38
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2839049696922302, "perplexity": 5913.057204993993}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703507045.10/warc/CC-MAIN-20210116195918-20210116225918-00043.warc.gz"}
|
https://proxieslive.com/why-is-readq-timestamp-is-not-checked-when-a-transaction-issues-a-command-reading-q/
|
# Why is Read(Q) timestamp is not checked when a transaction issues a command reading Q?
Consider a concurrent schedule of two transactions $$T_1,T_2$$:
$$R_x(Q): T_x$$ Reads $$Q$$
$$W_x(Q): T_x$$ Writes $$Q$$
$$S:R_2(A),R_1(A),W_1(A),W_2(A)$$, this schedule is not conflict serializable.However, if we follow timestamping protocol, where $$TS(T_1) S is allowed because $$T_1$$ can read a variable after $$T_2$$ has read it, which is not conflict equivalent to schedule $$T_1 T_2$$ and is contradictory to the fact that timestamping protocol gives a conflict serializable schedule. What am I missing here?
|
2020-09-29 20:27:25
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2730095088481903, "perplexity": 6065.838293305173}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402088830.87/warc/CC-MAIN-20200929190110-20200929220110-00578.warc.gz"}
|
https://almostsuremath.com/tag/math-pr/
|
# Multivariate Normal Distributions
I looked at normal random variables in an earlier post but, what does it mean for a sequence of real-valued random variables ${X_1,X_2,\ldots,X_n}$ to be jointly normal? We could simply require each of them to be normal, but this says very little about their joint distribution and is not much help in handling expressions involving more than one of the ${X_i}$ at once. In case that the random variables are independent, the following result is a very useful property of the normal distribution. All random variables in this post will be real-valued, except where stated otherwise, and we assume that they are defined with respect to some underlying probability space ${(\Omega,\mathcal F,{\mathbb P})}$.
Lemma 1 Linear combinations of independent normal random variables are again normal.
Proof: More precisely, if ${X_1,\ldots,X_n}$ is a sequence of independent normal random variables and ${a_1,\ldots,a_n}$ are real numbers, then ${Y=a_1X_1+\cdots+a_nX_n}$ is normal. Let us suppose that ${X_k}$ has mean ${\mu_k}$ and variance ${\sigma_k^2}$. Then, the characteristic function of Y can be computed using the independence property and the characteristic functions of the individual normals,
\displaystyle \begin{aligned} {\mathbb E}\left[e^{i\lambda Y}\right] &={\mathbb E}\left[\prod_ke^{i\lambda a_k X_k}\right] =\prod_k{\mathbb E}\left[e^{i\lambda a_k X_k}\right]\\ &=\prod_ke^{-\frac12\lambda^2a_k^2\sigma_k^2+i\lambda a_k\mu_k} =e^{-\frac12\lambda^2\sigma^2+i\lambda\mu} \end{aligned}
where we have set ${\mu_k=\sum_ka_k\mu_k}$ and ${\sigma^2=\sum_ka_k^2\sigma_k^2}$. This is the characteristic function of a normal random variable with mean ${\mu}$ and variance ${\sigma^2}$. ⬜
The definition of joint normal random variables will include the case of independent normals, so that any linear combination is also normal. We use use this result as the defining property for the general multivariate normal case.
Definition 2 A collection ${\{X_i\}_{i\in I}}$ of real-valued random variables is multivariate normal (or joint normal) if and only if all of its finite linear combinations are normal.
# The Riemann Zeta Function and Probability Distributions
The famous Riemann zeta function was first introduced by Riemann in order to describe the distribution of the prime numbers. It is defined by the infinite sum
\displaystyle \begin{aligned} \zeta(s) &=1+2^{-s}+3^{-s}+4^{-s}+\cdots\\ &=\sum_{n=1}^\infty n^{-s}, \end{aligned} (1)
which is absolutely convergent for all complex s with real part greater than one. One of the first properties of this is that, as shown by Riemann, it extends to an analytic function on the entire complex plane, other than a simple pole at ${s=1}$. By the theory of analytic continuation this extension is necessarily unique, so the importance of the result lies in showing that an extension exists. One way of doing this is to find an alternative expression for the zeta function which is well defined everywhere. For example, it can be expressed as an absolutely convergent integral, as performed by Riemann himself in his original 1859 paper on the subject. This leads to an explicit expression for the zeta function, scaled by an analytic prefactor, as the integral of ${x^s}$ multiplied by a function of x over the range ${ x > 0}$. In fact, this can be done in a way such that the function of x is a probability density function, and hence expresses the Riemann zeta function over the entire complex plane in terms of the generating function ${{\mathbb E}[X^s]}$ of a positive random variable X. The probability distributions involved here are not the standard ones taught to students of probability theory, so may be new to many people. Although these distributions are intimately related to the Riemann zeta function they also, intriguingly, turn up in seemingly unrelated contexts involving Brownian motion.
In this post, I derive two probability distributions related to the extension of the Riemann zeta function, and describe some of their properties. I also show how they can be constructed as the sum of a sequence of gamma distributed random variables. For motivation, some examples are given of where they show up in apparently unrelated areas of probability theory, although I do not give proofs of these statements here. For more information, see the 2001 paper Probability laws related to the Jacobi theta and Riemann zeta functions, and Brownian excursions by Biane, Pitman, and Yor. Continue reading “The Riemann Zeta Function and Probability Distributions”
# Manipulating the Normal Distribution
The normal (or Gaussian) distribution is ubiquitous throughout probability theory for various reasons, including the central limit theorem, the fact that it is realistic for many practical applications, and because it satisfies nice properties making it amenable to mathematical manipulation. It is, therefore, one of the first continuous distributions that students encounter at school. As such, it is not something that I have spent much time discussing on this blog, which is usually concerned with more advanced topics. However, there are many nice properties and methods that can be performed with normal distributions, greatly simplifying the manipulation of expressions in which it is involved. While it is usually possible to ignore these, and instead just substitute in the density function and manipulate the resulting integrals, that approach can get very messy. So, I will describe some of the basic results and ideas that I use frequently.
Throughout, I assume the existence of an underlying probability space ${(\Omega,\mathcal F,{\mathbb P})}$. Recall that a real-valued random variable X has the standard normal distribution if it has a probability density function given by,
$\displaystyle \varphi(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}.$
For it to function as a probability density, it is necessary that it integrates to one. While it is not obvious that the normalization factor ${1/\sqrt{2\pi}}$ is the correct value for this to be true, it is the one fact that I state here without proof. Wikipedia does list a couple of proofs, which can be referred to. By symmetry, ${-X}$ and ${X}$ have the same distribution, so that they have the same mean and, therefore, ${{\mathbb E}[X]=0}$.
The derivative of the density function satisfies the useful identity
$\displaystyle \varphi^\prime(x)=-x\varphi(x).$ (1)
This allows us to quickly verify that standard normal variables have unit variance, by an application of integration by parts.
\displaystyle \begin{aligned} {\mathbb E}[X^2] &=\int x^2\varphi(x)dx\\ &= -\int x\varphi^\prime(x)dx\\ &=\int\varphi(x)dx-[x\varphi(x)]_{-\infty}^\infty=1 \end{aligned}
Another identity satisfied by the normal density function is,
$\displaystyle \varphi(x+y)=e^{-xy - \frac{y^2}2}\varphi(x)$ (2)
This enables us to prove the following very useful result. In fact, it is difficult to overstate how helpful this result can be. I make use of it frequently when manipulating expressions involving normal variables, as it significantly simplifies the calculations. It is also easy to remember, and simple to derive if needed.
Theorem 1 Let X be standard normal and ${f\colon{\mathbb R}\rightarrow{\mathbb R}_+}$ be measurable. Then, for all ${\lambda\in{\mathbb R}}$,
\displaystyle \begin{aligned} {\mathbb E}[e^{\lambda X}f(X)] &={\mathbb E}[e^{\lambda X}]{\mathbb E}[f(X+\lambda)]\\ &=e^{\frac{\lambda^2}{2}}{\mathbb E}[f(X+\lambda)]. \end{aligned} (3)
# Brownian Drawdowns
Here, I apply the theory outlined in the previous post to fully describe the drawdown point process of a standard Brownian motion. In fact, as I will show, the drawdowns can all be constructed from independent copies of a single ‘Brownian excursion’ stochastic process. Recall that we start with a continuous stochastic process X, assumed here to be Brownian motion, and define its running maximum as ${M_t=\sup_{s\le t}X_s}$ and drawdown process ${D_t=M_t-X_t}$. This is as in figure 1 above.
Next, ${D^a}$ was defined to be the drawdown ‘excursion’ over the interval at which the maximum process is equal to the value ${a \ge 0}$. Precisely, if we let ${\tau_a}$ be the first time at which X hits level ${a}$ and ${\tau_{a+}}$ be its right limit ${\tau_{a+}=\lim_{b\downarrow a}\tau_b}$ then,
$\displaystyle D^a_t=D_{({\tau_a+t})\wedge\tau_{a+}}=a-X_{({\tau_a+t)}\wedge\tau_{a+}}.$
Next, a random set S is defined as the collection of all nonzero drawdown excursions indexed the running maximum,
$\displaystyle S=\left\{(a,D^a)\colon D^a\not=0\right\}.$
The set of drawdown excursions corresponding to the sample path from figure 1 are shown in figure 2 below.
As described in the post on semimartingale local times, the joint distribution of the drawdown and running maximum ${(D,M)}$, of a Brownian motion, is identical to the distribution of its absolute value and local time at zero, ${(\lvert X\rvert,L^0)}$. Hence, the point process consisting of the drawdown excursions indexed by the running maximum, and the absolute value of the excursions from zero indexed by the local time, both have the same distribution. So, the theory described in this post applies equally to the excursions away from zero of a Brownian motion.
Before going further, let’s recap some of the technical details. The excursions lie in the space E of continuous paths ${z\colon{\mathbb R}_+\rightarrow{\mathbb R}}$, on which we define a canonical process Z by sampling the path at each time t, ${Z_t(z)=z_t}$. This space is given the topology of uniform convergence over finite time intervals (compact open topology), which makes it into a Polish space, and whose Borel sigma-algebra ${\mathcal E}$ is equal to the sigma-algebra generated by ${\{Z_t\}_{t\ge0}}$. As shown in the previous post, the counting measure ${\xi(A)=\#(S\cap A)}$ is a random point process on ${({\mathbb R}_+\times E,\mathcal B({\mathbb R}_+)\otimes\mathcal E)}$. In fact, it is a Poisson point process, so its distribution is fully determined by its intensity measure ${\mu={\mathbb E}\xi}$.
Theorem 1 If X is a standard Brownian motion, then the drawdown point process ${\xi}$ is Poisson with intensity measure ${\mu=\lambda\otimes\nu}$ where,
• ${\lambda}$ is the standard Lebesgue measure on ${{\mathbb R}_+}$.
• ${\nu}$ is a sigma-finite measure on E given by
$\displaystyle \nu(f) = \lim_{\epsilon\rightarrow0}\epsilon^{-1}{\mathbb E}_\epsilon[f(Z^{\sigma})]$ (1)
for all bounded continuous continuous maps ${f\colon E\rightarrow{\mathbb R}}$ which vanish on paths of length less than L (some ${L > 0}$). The limit is taken over ${\epsilon > 0}$, ${{\mathbb E}_\epsilon}$ denotes expectation under the measure with respect to which Z is a Brownian motion started at ${\epsilon}$, and ${\sigma}$ is the first time at which Z hits 0. This measure satisfies the following properties,
• ${\nu}$-almost everywhere, there exists a time ${T > 0}$ such that ${Z > 0}$ on ${(0,T)}$ and ${Z=0}$ everywhere else.
• for each ${t > 0}$, the distribution of ${Z_t}$ has density
$\displaystyle p_t(z)=z\sqrt{\frac 2{\pi t^3}}e^{-\frac{z^2}{2t}}$ (2)
over the range ${z > 0}$.
• over ${t > 0}$, ${Z_t}$ is Markov, with transition function of a Brownian motion stopped at zero.
# Drawdown Point Processes
For a continuous real-valued stochastic process ${\{X_t\}_{t\ge0}}$ with running maximum ${M_t=\sup_{s\le t}X_s}$, consider its drawdown. This is just the amount that it has dropped since its maximum so far,
$\displaystyle D_t=M_t-X_t,$
which is a nonnegative process hitting zero whenever the original process visits its running maximum. By looking at each of the individual intervals over which the drawdown is positive, we can break it down into a collection of finite excursions above zero. Furthermore, the running maximum is constant across each of these intervals, so it is natural to index the excursions by this maximum process. By doing so, we obtain a point process. In many cases, it is even a Poisson point process. I look at the drawdown in this post as an example of a point process which is a bit more interesting than the previous example given of the jumps of a cadlag process. By piecing the drawdown excursions back together, it is possible to reconstruct ${D_t}$ from the point process. At least, this can be done so long as the original process does not monotonically increase over any nontrivial intervals, so that there are no intervals with zero drawdown. As the point process indexes the drawdown by the running maximum, we can also reconstruct X as ${X_t=M_t-D_t}$. The drawdown point process therefore gives an alternative description of our original process.
See figure 1 for the drawdown of the bitcoin price valued in US dollars between April and December 2020. As it makes more sense for this example, the drawdown is shown as a percent of the running maximum, rather than in dollars. This is equivalent to the approach taken in this post applied to the logarithm of the price return over the period, so that ${X_t=\log(B_t/B_0)}$. It can be noted that, as the price was mostly increasing, the drawdown consists of a relatively large number of small excursions. If, on the other hand, it had declined, then it would have been dominated by a single large drawdown excursion covering most of the time period.
For simplicity, I will suppose that ${X_0=0}$ and that ${M_t}$ tends to infinity as t goes to infinity. Then, for each ${a\ge0}$, define the random time at which the process first hits level ${a}$,
$\displaystyle \tau_a=\inf\left\{t\ge 0\colon X_t\ge a\right\}.$
By construction, this is finite, increasing, and left-continuous in ${a}$. Consider, also, the right limits ${\tau_{a+}=\lim_{b\downarrow0}\tau_b}$. Each of the excursions on which the drawdown is positive is equal to one of the intervals ${(\tau_a,\tau_{a+})}$. The excursion is defined as a continuous stochastic process ${\{D^a_t\}_{t\ge0}}$ equal to the drawdown starting at time ${\tau_a}$ and stopped at time ${\tau_{a+}}$,
$\displaystyle D^a_t=D_{(\tau_a+t)\wedge\tau_{a+}}=a-X_{(\tau_a+t)\wedge\tau_{a+}}.$
This is a continuous nonnegative real-valued process, which starts at zero and is equal to zero at all times after ${\tau_{a+}-\tau_a}$. Note that there uncountably many values for ${a}$ but, the associated excursion will be identically zero other than for the countably many times at which ${\tau_{a+} > \tau_a}$. We will only be interested in these nonzero excursions.
As usual, we work with respect to an underlying probability space ${(\Omega,\mathcal F,{\mathbb P})}$, so that we have one path of the stochastic process X defined for each ${\omega\in\Omega}$. Associated to this is the collection of drawdown excursions indexed by the running maximum.
$\displaystyle S=\left\{(a,D^a)\colon a\ge0,\ D^a\not=0\right\}.$
As S is defined for each given sample path, it depends on the choice of ${\omega\in\Omega}$, so is a countable random set. The sample paths of the excursions ${D^a}$ lie in the space of continuous functions ${{\mathbb R}_+\rightarrow{\mathbb R}}$, which I denote by E. For each time ${t\ge0}$, I use ${Z_t}$ to denote the value of the path sampled at time t,
\displaystyle \begin{aligned} &E=\left\{z\colon {\mathbb R}_+\rightarrow{\mathbb R}{\rm\ is\ continuous}\right\}.\\ &Z_t\colon E\rightarrow{\mathbb R},\\ & Z_t(z)=z_t. \end{aligned}
Use ${\mathcal E}$ to denote the sigma-algebra on E generated by the collection of maps ${\{Z_t\colon t\ge0\}}$, so that ${(E,\mathcal E)}$ is the measurable space in which the excursion paths lie. It can be seen that ${\mathcal E}$ is the Borel sigma-algebra generated by the open subsets of E, with respect to the topology of compact convergence. That is, the topology of uniform convergence on finite time intervals. As this is a complete separable metric space, it makes ${(E,\mathcal E)}$ into a standard Borel space.
Lemma 1 The set S defines a simple point process ${\xi}$ on ${{\mathbb R}_+\times E}$,
$\displaystyle \xi(A)=\#(S\cap A)$
for all ${A\in\mathcal B({\mathbb R}_+)\otimes\mathcal E}$.
From the definition of point processes, this simply means that ${\xi(A)}$ is a measurable random variable for each ${A\in \mathcal B({\mathbb R}_+)\otimes\mathcal E}$ and that there exists a sequence ${A_n\in \mathcal B({\mathbb R}_+)\otimes\mathcal E}$ covering E such that ${\xi(A_n)}$ are almost surely finite. The set of drawdowns for the point process corresponding to the bitcoin prices in figure 1 are shown in figure 2 below.
# Criteria for Poisson Point Processes
If S is a finite random set in a standard Borel measurable space ${(E,\mathcal E)}$ satisfying the following two properties,
• if ${A,B\in\mathcal E}$ are disjoint, then the sizes of ${S\cap A}$ and ${S\cap B}$ are independent random variables,
• ${{\mathbb P}(x\in S)=0}$ for each ${x\in E}$,
then it is a Poisson point process. That is, the size of ${S\cap A}$ is a Poisson random variable for each ${A\in\mathcal E}$. This justifies the use of Poisson point processes in many different areas of probability and stochastic calculus, and provides a convenient method of showing that point processes are indeed Poisson. If the theorem applies, so that we have a Poisson point process, then we just need to compute the intensity measure to fully determine its distribution. The result above was mentioned in the previous post, but I give a precise statement and proof here. Continue reading “Criteria for Poisson Point Processes”
# Poisson Point Processes
The Poisson distribution models numbers of events that occur in a specific period of time given that, at each instant, whether an event occurs or not is independent of what happens at all other times. Examples which are sometimes cited as candidates for the Poisson distribution include the number of phone calls handled by a telephone exchange on a given day, the number of decays of a radio-active material, and the number of bombs landing in a given area during the London Blitz of 1940-41. The Poisson process counts events which occur according to such distributions.
More generally, the events under consideration need not just happen at specific times, but also at specific locations in a space E. Here, E can represent an actual geometric space in which the events occur, such as the spacial distribution of bombs dropped during the Blitz shown in figure 1, but can also represent other quantities associated with the events. In this example, E could represent the 2-dimensional map of London, or could include both space and time so that ${E=F\times{\mathbb R}}$ where, now, F represents the 2-dimensional map and E is used to record both time and location of the bombs. A Poisson point process is a random set of points in E, such that the number that lie within any measurable subset is Poisson distributed. The aim of this post is to introduce Poisson point processes together with the mathematical machinery to handle such random sets.
The choice of distribution is not arbitrary. Rather, it is a result of the independence of the number of events in each region of the space which leads to the Poisson measure, much like the central limit theorem leads to the ubiquity of the normal distribution for continuous random variables and of Brownian motion for continuous stochastic processes. A random finite subset S of a reasonably ‘nice’ (standard Borel) space E is a Poisson point process so long as it satisfies the properties,
• If ${A_1,\ldots,A_n}$ are pairwise-disjoint measurable subsets of E, then the sizes of ${S\cap A_1,\ldots,S\cap A_n}$ are independent.
• Individual points of the space each have zero probability of being in S. That is, ${{\mathbb P}(x\in S)=0}$ for each ${x\in E}$.
The proof of this important result will be given in a later post.
We have come across Poisson point processes previously in my stochastic calculus notes. Specifically, suppose that X is a cadlag ${{\mathbb R}^d}$-valued stochastic process with independent increments, and which is continuous in probability. Then, the set of points ${(t,\Delta X_t)}$ over times t for which the jump ${\Delta X}$ is nonzero gives a Poisson point process on ${{\mathbb R}_+\times{\mathbb R}^d}$. See lemma 4 of the post on processes with independent increments, which corresponds precisely to definition 5 given below. Continue reading “Poisson Point Processes”
# Quantum Coin Tossing
Let me ask the following very simple question. Suppose that I toss a pair of identical coins at the same time, then what is the probability of them both coming up heads? There is no catch here, both coins are fair. There are three possible outcomes, both tails, one head and one tail, and both heads. Assuming that it is completely random so that all outcomes are equally likely, then we could argue that each possibility has a one in three chance of occurring, so that the answer to the question is that the probability is 1/3.
Of course, this is wrong! A fair coin has a probability of 1/2 of showing heads and, by independence, standard probability theory says that we should multiply these together for each coin to get the correct answer of ${\frac12\times\frac12=\frac14}$, which can be verified by experiment. Alternatively, we can note that the outcome of one tail and one head, in reality, consists of two equally likely possibilities. Either the first coin can be a head and the second a tail, or vice-versa. So, there are actually four equally likely possible outcomes, only one of which has both coins showing heads, again giving a probability of 1/4. Continue reading “Quantum Coin Tossing”
# Local Time Continuity
The local time of a semimartingale at a level x is a continuous increasing process, giving a measure of the amount of time that the process spends at the given level. As the definition involves stochastic integrals, it was only defined up to probability one. This can cause issues if we want to simultaneously consider local times at all levels. As x can be any real number, it can take uncountably many values and, as a union of uncountably many zero probability sets can have positive measure or, even, be unmeasurable, this is not sufficient to determine the entire local time ‘surface’
$\displaystyle (t,x)\mapsto L^x_t(\omega)$
for almost all ${\omega\in\Omega}$. This is the common issue of choosing good versions of processes. In this case, we already have a continuous version in the time index but, as yet, have not constructed a good version jointly in the time and level. This issue arose in the post on the Ito–Tanaka–Meyer formula, for which we needed to choose a version which is jointly measurable. Although that was sufficient there, joint measurability is still not enough to uniquely determine the full set of local times, up to probability one. The ideal situation is when a version exists which is jointly continuous in both time and level, in which case we should work with this choice. This is always possible for continuous local martingales.
Theorem 1 Let X be a continuous local martingale. Then, the local times
$\displaystyle (t,x)\mapsto L^x_t$
have a modification which is jointly continuous in x and t. Furthermore, this is almost surely ${\gamma}$-Hölder continuous w.r.t. x, for all ${\gamma < 1/2}$ and over all bounded regions for t.
# The Kolmogorov Continuity Theorem
One of the common themes throughout the theory of continuous-time stochastic processes, is the importance of choosing good versions of processes. Specifying the finite distributions of a process is not sufficient to determine its sample paths so, if a continuous modification exists, then it makes sense to work with that. A relatively straightforward criterion ensuring the existence of a continuous version is provided by Kolmogorov’s continuity theorem.
For any positive real number ${\gamma}$, a map ${f\colon E\rightarrow F}$ between metric spaces E and F is said to be ${\gamma}$-Hölder continuous if there exists a positive constant C satisfying
$\displaystyle d(f(x),f(y))\le Cd(x,y)^\gamma$
for all ${x,y\in E}$. Hölder continuous functions are always continuous and, at least on bounded spaces, is a stronger property for larger values of the coefficient ${\gamma}$. So, if E is a bounded metric space and ${\alpha\le\beta}$, then every ${\beta}$-Hölder continuous map from E is also ${\alpha}$-Hölder continuous. In particular, 1-Hölder and Lipschitz continuity are equivalent.
Kolmogorov’s theorem gives simple conditions on the pairwise distributions of a process which guarantee the existence of a continuous modification but, also, states that the sample paths ${t\mapsto X_t}$ are almost surely locally Hölder continuous. That is, they are almost surely Hölder continuous on every bounded interval. To start with, we look at real-valued processes. Throughout this post, we work with repect to a probability space ${(\Omega,\mathcal F, {\mathbb P})}$. There is no need to assume the existence of any filtration, since they play no part in the results here
Theorem 1 (Kolmogorov) Let ${\{X_t\}_{t\ge0}}$ be a real-valued stochastic process such that there exists positive constants ${\alpha,\beta,C}$ satisfying
$\displaystyle {\mathbb E}\left[\lvert X_t-X_s\rvert^\alpha\right]\le C\lvert t-s\vert^{1+\beta},$
for all ${s,t\ge0}$. Then, X has a continuous modification which, with probability one, is locally ${\gamma}$-Hölder continuous for all ${0 < \gamma < \beta/\alpha}$.
|
2021-02-25 08:20:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 157, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9257960319519043, "perplexity": 168.79473411774785}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178350846.9/warc/CC-MAIN-20210225065836-20210225095836-00190.warc.gz"}
|
https://economics.stackexchange.com/tags/game/new
|
# Tag Info
The intersection itself is not terribly interesting, what matters here is only that it is not empty- that it contains some state. To know event $E$ means that one does not consider it possible that the event $E$ does not obtain and that $E$ does indeed obtain. Since the true state is always considered possible, that $i$ knows $E$ at $\omega$ means that \$\...
|
2021-01-19 20:42:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.39069756865501404, "perplexity": 355.99401397206844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703519784.35/warc/CC-MAIN-20210119201033-20210119231033-00168.warc.gz"}
|
http://mymathforum.com/pre-calculus/50762-quick-question.html
|
My Math Forum Quick question.
Pre-Calculus Pre-Calculus Math Forum
February 16th, 2015, 11:51 PM #1 Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty Quick question. Just a quick question: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ can the above be solved as such: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$ $\displaystyle (x-3) = (x+1)$ if so, then could the following be solved the same way? $\displaystyle 2(x-1)^5 - x(x-1)^4 = 0$ $\displaystyle 2(x-1)^5 = x(x-1)^4$ $\displaystyle 2(x-1)^5/(x-1)^4 = x$ $\displaystyle 2(x-1) = x$ Thank you in advance for the feedback.
February 16th, 2015, 11:55 PM #2 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs For the first problem: $\displaystyle (x-3)^3-(x+1)(x-3)^2= 0$ I would factor: $\displaystyle (x-3)^2\left((x-3)-(x+1)\right)= 0$ Collect like terms: $\displaystyle 4(x-3)^2= 0$ Hence: $\displaystyle x=3$ Can you apply the same method to the second problem? Thanks from hyperbola
February 17th, 2015, 12:30 AM #3 Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty $\displaystyle 2(x-1)^5 - x(x-1)^4$ common factor is $\displaystyle (x-1)^4$ $\displaystyle (x-1)^4(2(x-1)-x) = 0$ $\displaystyle (x-1)^4(2x-2-x) = 0$ $\displaystyle (x-1)^4(x-2) = 0$ $\displaystyle (x-1)^4 = 0, x=1$ $\displaystyle (x-2) = 0, x=2$ correct? Thanks from MarkFL Last edited by hyperbola; February 17th, 2015 at 12:32 AM.
February 17th, 2015, 01:48 AM #4 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Yes...perfect! Thanks from hyperbola
February 17th, 2015, 04:44 AM #5 Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty MarkFL could you help me out with one more question please? I'm asked to solve the following by method of 'completing the square'. $\displaystyle 2x^2 + 12x + 3 = 0$ $\displaystyle 2x^2+12x=-3$ $\displaystyle (0.5 x 12)^2=36$ $\displaystyle 2x^2+12x+36=-3+36$ $\displaystyle 2x^2+12x+36=33$ $\displaystyle 2[x^2+6x+18]=33$ No common factors here. As far as I can see, this can only be solved using the quadratic formula. Using quadratic formula I get $\displaystyle x1 = -0.26$ $\displaystyle x2 = -5.74$ Last edited by hyperbola; February 17th, 2015 at 05:38 AM.
February 17th, 2015, 10:14 AM #6 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs For future reference, it is really preferable that you begin a new thread for a new question...this way threads do not become possibly convoluted and hard to follow. We are given: $\displaystyle 2x^2+12x+3=0$ Subtract through by 3: $\displaystyle 2x^2+12x=-3$ Divide through by 2: $\displaystyle x^2+6x=-\frac{3}{2}$ Take the coefficient of the linear term, which is 6, divide it by 2 to get 3, then square that to get 9, and so add 9 to both sides: $\displaystyle x^2+6x+9=\frac{15}{2}$ Rewrite the LHS as a square: $\displaystyle (x+3)^2=\frac{15}{2}$ Use the square root property: $\displaystyle x+3=\pm\sqrt{\frac{15}{2}}=\pm\frac{\sqrt{30}}{2}$ Subtract through by 3: $\displaystyle x=\frac{-6\pm\sqrt{30}}{2}$ Thanks from hyperbola
February 17th, 2015, 12:29 PM #7
Math Team
Joined: Dec 2013
From: Colombia
Posts: 6,444
Thanks: 2115
Math Focus: Mainly analysis and algebra
Quote:
Originally Posted by hyperbola $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$
This division is only valid when $x \ne 3$, otherwise you are dividing by zero. In this case you found that 0 = 4 as a result! For this reason it is best to avoid dividing by terms that may be zero - factorise instead.
Quote:
Originally Posted by hyperbola $\displaystyle 2(x-1) = x$
This time you didn't come out with nonsense because there is a solution under which $x-1 \ne 0$, but you did lose four (repeated) solutions to the equation.
Again, factorise it rather than divide.
Last edited by skipjack; February 20th, 2015 at 08:24 AM.
February 20th, 2015, 04:29 AM #8
Math Team
Joined: Jan 2015
From: Alabama
Posts: 2,279
Thanks: 570
Quote:
Originally Posted by hyperbola Just a quick question: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ can the above be solved as such: $\displaystyle (x-3)^3 - (x+1)(x-3)^2 = 0$ $\displaystyle (x-3)^3 = (x+1)(x-3)^2$ $\displaystyle (x-3)^3/(x-3)^2 = (x+1)$
IF x- 3 is not 0, which is the same as x is not 3, we can do this.
Quote:
$\displaystyle (x-3) = (x+1)$
Obviously there is no value of x that satisfies this so "x = 3" is the only solution.
Quote:
if so, then could the following be solved the same way? $\displaystyle 2(x-1)^5 - x(x-1)^4 = 0$ $\displaystyle 2(x-1)^5 = x(x-1)^4$ $\displaystyle 2(x-1)^5/(x-1)^4 = x$
Again, IF x- 1 is not 0, which is the same as "x is not equal to 1" then we can divide by it.
Quote:
$\displaystyle 2(x-1) = x$
This is the same as 2x- 2 = x or x = 2. So the solutions to this equation are x = 1 and x = 2.
Quote:
Thank you in advance for the feedback.
Last edited by skipjack; February 20th, 2015 at 08:26 AM.
Tags question, quick
### square root method 2x^2 12x 3=0
Click on a term to search for related topics.
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post Pumaftw Algebra 6 December 16th, 2014 07:24 AM Bosanac Algebra 2 January 31st, 2013 11:14 AM seecs Algebra 1 November 10th, 2010 02:10 PM mathhelp123 Calculus 1 October 30th, 2009 02:21 PM Kiranpreet Algebra 2 April 24th, 2008 02:02 PM
Contact - Home - Forums - Top
|
2017-03-24 08:06:55
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6047200560569763, "perplexity": 2245.9190185967227}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218187744.59/warc/CC-MAIN-20170322212947-00611-ip-10-233-31-227.ec2.internal.warc.gz"}
|
http://hindi.orgsyn.in/glossary/chemistry_glossary
|
# रसायन विज्ञान की शब्दावली
Chemistry Glossary |
This page is a glossary of chemistry terms. Chemistry has an extensive vocabulary and a significant amount of jargon. This is a list of chemical terms, including laboratory tools, glassware, and equipment. Chemistry itself is a physical science concerned with the composition, structure, and properties of matter, as well as the changes it undergoes during chemical reactions.
परम ताप (absolute temperature) जब तापमान को परम शून्य के सापेक्ष मापा जाता है तो इस प्रकार प्राप्त तापमान को परम ताप कहते हैं। इसे केल्विन इकाइ का प्रयोग करके व्यक्त किया जाता है।
परम शून्य (absolute zero) यह न्यूनतम ताप है जिसे प्राप्त किया जा सकता है। इससे कम ताप होना सम्भव ही नहीं है। किसी ठोस का ताप चूंकि उसके परमाणुओं के कम्पन का परिमाण प्रदर्शित करतअ है, अतः परम शून्य ताप पर परमाणु पूर्णतः कम्पन करना बद कर देते हैं।
शुद्धि (accuracy) When you measure something, the accuracy is how close your measured value is to the real value. For example, if you're actually six feet tall and your brother measures your height as six feet, one inch, he's pretty accurate. However, if your cousin measures your height as twelve feet, 13 inches, he's not accurate at all.
अम्ल (acid) This is anything that gives off H+ ions in water. Acids have a pH less than 7 and are good at dissolving metals. They turn litmus paper red and phenolphthalein colorless.
acid anhydride This is an oxide that forms an acid when you stick it in water. An example is SO3 - when you add water it turns into sulfuric acid, H2SO4.
acid dissociation constant (Ka) This is equal to the ratio of the concentrations of an acid's conjugate base and the acid present when a weak acid dissociates in water. That is, if you have a solution of Acid X where the concentration of the conjugate base is 0.5 M and the concentration of the acid is 10 M, the acid dissociation constant is 0.5/10 = 0.05.
activated complex In a chemical reaction, the reagents have to join together into a great big blob before they can fall back apart into the products. This great big blob is called the activated complex (a.k.a. transition state)
activation energy The minimum amount of energy needed for a chemical reaction to take place. For some reactions this is very small (it only takes a spark to make gasoline burn)। For others, it's very high (when you burn magnesium, you need to hold it over a Bunsen burner for a minute or so)।
सक्रियता श्रेणी (activity series) This is when you arrange elements in the order of how much they tend to react with water and acids.
actual yield When you do a chemical reaction, this is the amount of chemical that you actually make (i.e. The amount of stuff you can weigh)।
योग अभिक्रिया (addition reaction) A reaction where atoms add to a carbon-carbon multiple bond.
adsorption When one substance collects of the surface of another one.
एल्कोहल (alcohol) An organic molecule containing an -OH group
aldehyde An organic molecule containing a -COH group
alkali metals Group I in the periodic table.
alkaline earth metals Group II in the periodic table.
एल्कने (alkane) An organic molecule which contains only single carbon-carbon bonds.
एल्कीन (alkene) An organic molecule containing at least one C=C bond
एल्काइन (alkyne) An organic molecule containing at least one C-C triple bond.
allotropes When you have different forms of an element in the same state. The relationship that white phosphorus and red phosphorus have to each other is that they're allotropes.
मिश्रधातु (alloy) A mixture of two metals. Usually, you add very small amounts of a different element to make the metal stronger and harder.
अल्फा कण (alpha particle) A radioactive particle equivalent to a helium nucleus (2 protons, 2 neutrons)
amine An organic molecule which consists of an ammonia molecule where one or more of the hydrogen atoms has been replaced by organic groups.
amino acid The basic building blocks of proteins. They're called “amino acids” because they're both amines (they contain nitrogen) and acids (carboxylic acids, to be precise)
amphiprotic When something is both an acid and a base. Like amino acids, for example.
amphoteric When something is both an acid and a base. Sounds familiar, huh?
धनाग्र (anode) The electrode where oxidation occurs. In other words, this is where electrons are lost by a substance.
जलीय (aqueous) dissolved in water
परमाणु द्रब्यमान इकाई (atomic mass unit (a.m.u.)) This is the smallest unit of mass we use in chemistry, and is equivalent to 1/12 the mass of carbon-12. To all intents and purposes, protons and neutrons weigh 1 a.m.u.
परमाणु त्रिज्या (atomic radius) This is one half the distance between two bonded nuclei. Why don't we just measure the distance from the nucleus to the outside of the atom - after all, isn't that the same thing as a radius? It is, but atoms are also (theoretically) infinitely large (due to quantum mechanics), making this impossible to measure.
atomic solid A solid where there's a bunch of atoms in the lattice. This is different from an ionic solid, where ions are the things that are sticking together.
आफबाउ का सिद्धान्त (Aufbau principle) When you add protons to the nucleus to build up the elements, electrons are added into orbitals.
अवगाद्रो का नियम (Avogadro's Law) If you've got two gases under the same conditions of temperature, pressure, and volume, they've got the same number of particles (atoms or molecules)। This law only works for ideal gases, none of which actually exist.
base anhydride An oxide that forms a base when water is added. CaO is an example, turning into calcium hydroxide in water.
क्षार (base) A compound that gives off OH- ions in water. They are slippery and bitter and have a pH greater than 7.
बैटरी (battery) This is when a bunch of voltaic cells are stuck together.
बीटा कण (beta particle) A radioactive particle equivalent to an electron.
bidentate ligand A ligand that can attach twice to a metal ion.
binary compound A compound only having two elements
बंधन उर्जा (binding energy) The amount of energy that holds the neutrons and protons together in the nucleus of an atom. It's a lot of energy, which is why you don't see nuclei falling apart all over the place.
बाण्ड उर्जा (bond energy) The amount of energy it takes to break one mole of bonds.
बाण्ड लम्बाई (bond length) The average distance between the nuclei of two bonded atoms.
बायल का नियम (Boyle's Law) The volume of a gas at constant temperature varies inversely with pressure. In other words, if you put big pressure on something, it gets small.
ब्रान्स्टेट-लौरी अम्ल (Bronsted-Lowry acid) Acids donate protons [H+ ions] and bases grab them
बफर (buffer) A liquid that resists change in pH by the addition of acid or base. It consists of a weak acid and it's conjugate base (acetic acid and sodium acetate, for example)।
उष्मामापन (calorimetry) The study of heat flow. Usually you'd do calorimetry to find the heat of combustion of a compound or the heat of reaction of two compounds.
carboxylic acid An organic molecule with a -COOH group on it. Acetic acid is the most famous one.
उत्प्रेरक (catalyst) A substance that speeds up a chemical reaction without being used up by the reaction. Enzymes are catalysts because they allow the reactions that take place in the body to occur fast enough that we can live.
ऋणाग्र (cathode) The electrode in which reduction occurs. Reduction is when a compound gains electrons.
शृंखला अभिक्रिया (chain reaction) A reaction in which the products from one step provide the reagents for the next one. This is frequently referred to in nuclear fission (when large nuclei break apart to form smaller ones) and in free-radical reactions.
चार्ल का नियम (Charles's Law) The volume of a gas at constant pressure is directly proportional to the temperature. In other words, if you heat something up, it gets big.
रासायनिक समीकरण (chemical equation) The recipe that describes what you need to do to make a reaction take place.
रासायनिक गुण (chemical properties) Properties that can only be described by making a chemical change (by making or breaking bonds)। For example, color isn't a chemical property because you don't need to change something chemically to see what color it is. Flammability, on the other hand, is a chemical property, because you can't tell if something burns unless you actually try to burn it.
chirality When a molecule has a nonsuperimposable mirror image. To imagine this, put your hands together. Although they are mirror images, you can't put them right on top of each other so they are interchangable. Well, normal people can't, anyway.
chromatography This is when you use a system containing a mobile phase (usually a liquid in general chemistry classes) and a stationary phase (something dissolved in the liquid) to separate different compounds. This is usually done by exploiting the differing polarities of solutes, though you can do it a whole slew o' ways.
परिपथ (circuit) The closed path in a circuit through which electrons flow.
coagulation When you destroy a colloid by letting the particles settle out.
colligative property Any property of a solution that changes when the concentration changes. Examples are color, flavor, boiling point, melting point, and osmotic pressure.
colloid It's a suspension.
ज्वलन (combustion) When a compound combines with oxygen gas to form water, heat, and carbon dioxide
समान आयन प्रभाव (common ion effect) When the equilibrium position of a process is altered by adding another compound containing one of the same ions that's in the equilibrium.
complex ion An ion in which a central atom (usually a transition metal) is surrounded by a bunch of molecules like water or ammonia (called “ligands”)
सांद्रता (concentration) A measurement of the amount of stuff (solute) dissolved in a liquid (solvent)। The most common concentration unit is molarity (M), which is equal to the number of moles of solute divided by the number of liters of solution.
condensation When a vapor reforms a liquid. This is what happens on your bathroom mirror when you take a shower.
चालकता (conductance) A measurement of how well electricity can flow through an object.
युग्मी अम्ल (conjugate acid) The compound formed when a base gains a proton (hydrogen atom)।
युग्मी क्षार (conjugate base) The compound formed when an acid loses a proton (hydrogen atom)।
सतत स्पेक्ट्रम (continuous spectrum) A spectrum that gives off all the colors of light, like a rainbow. This is caused by blackbody emission.
covalent bond A chemical bond formed when two atoms share two electrons.
क्रान्तिक द्रब्यमान (critical mass) The minimum amount of radioactive material needed to undergo a nuclear chain reaction.
क्रान्तिक बिन्दु (critical point) The end point of the liquid-vapor line in a phase diagram. Past the critical point, you get something called a “supercritical liquid”, which has weird properties.
crystal lattice see “lattice”
क्रिस्टल (crystal) A large chunk of an ionic solid.
डाल्टन का आंशिक दाब का सिद्धान्त (Dalton's law of partial pressures) The total pressure in a mixture of gases is equal to the sums of the partial pressures of all the gases put together.
decomposition When a big molecule falls apart to make two or more little ones.
degenerate Things (usually orbitals) are said to be degenerate if they have the same energy. This term is used a whole lot in quantum mechanics. Also when dealing with kids who steal cars.
delocalization This is when electrons can move around all over a molecule. This happens when you have double bonds on adjacent atoms in a molecule (conjugated hydrocarbon)
denature When the 3-D structure of a protein breaks down due to heat (or pH, etc), it's said to be denatured. This means that it unravels because the intermolecular forces between atoms in the chain aren't strong enough to hold it together anymore.
विसरण (diffusion) When particles move from areas of high concentration to areas of low concentration. For example, if you open a bottle of ammonia on one end of the room, the concentration of ammonia molecules in the air is very high on that side of the room. As a result, they tend to migrate across the room, which explains why you can smell it after a little while. Be careful not to mix this up with effusion (see definition)
तनुकरण (dilution) When you add solvent to a solution to make it less concentrated.
द्विध्रुव आघूर्ण (dipole moment) When a molecule has some charge separation (usually because the molecule is polar), it's said to have a dipole moment.
द्विध्रुव-द्विध्रुव बल (dipole-dipole force) When the positive end of a polar molecule becomes attracted to the negative end of another polar molecule.
बिलयन (dissociation) When water dissolves a compound.
आसवन (distillation) This is when you separate a mixture of liquids by heating it up. The one with the lowest boiling point evaporates first, followed by the one with the next lowest boiling point, etc.
double-displacement reaction (a.k.a. double replacement reaction) When the cations of two ionic compounds switch places.
effusion When a gas moves through an opening into a chamber that contains no pressure. Effusion is much faster than diffusion because there are no other gas molecules to get in the way.
विद्युत अपघटन (electrolysis) When electricity is used to break apart a chemical compound.
विद्युत अपघट्य' (electrolyte) An ionic compound that dissolves in water to conduct electricity. Strong electrolytes break apart completely in water
weak electrolytes only fall apart a little bit.
(Actually, this isn't entirely true, as Raji Heyovska informs me. Apparently strong electrolytes also dissociate partially in water, though much more so than weak ones. For more info, check out his paper at http://www.jh-inst.cas.cz/~rheyrovs. However, it is also true that the usual definition of a strong electrolyte is one that dissociates completely in water, which is why I include that definition above.) electron affinity
The energy change that accompanies the addition of an electron to an atom in the gas phase. electronegativity A measurement of how much an atom tends to steal electrons from atoms that it's bonded to. Elements at the top right of the periodic table (excluding the noble gases) are very electronegative while atoms in the bottom left are not very electronegative (a.k.a. “electropositive”)
electropositive When something is not at all electronegative. In fact, it tends to lose electrons rather than to gain them. Elements that are electropositive are generally to the left and bottom of the periodic table.
empirical formula A reduced molecular formula. If you have a molecular formula and you can reduce all of the subscripts by some constant number, the result is the empirical formula.
emulsion When very small drops of a liquid are suspended in another. An example of an emulsion is salad dressing after you've shaken it up.
enantiomers molecules that are nonsuperimposable mirror images of each other.
endothermic When a process absorbs energy (gets cold)।
endpoint The point where you actually stop a titration, usually because an indicator has changed color. This is different than the “equivalence point” because the indicator might not change colors at the exact instant that the solution is neutral.
energy level A possible level of energy that an electron can have in an atom.
enthalpy A measurement of the energy content of a system.
entropy A measurement of the randomness in a system.
enzyme A biological molecule that catalyzes reactions in living creatures.
साम्य (equilibrium) When the forward rate of a chemical reaction is the same as the reverse rate. This only takes place in reversible reactions because these are the only type of reaction in which the forward and backward reactions can both take place.
equivalence point The point in a titration at which the solution is completely neutral. This is different than the “endpoint” (see above)।
ester An organic molecule with R-CO-OR' functionality.
excess reagent Sometimes when you do a chemical reaction, there's some of one reagent left over. That's called the excess reagent.
exothermic When a process gives off energy (gets hot)।
family The same thing as a “group” (see above)
first law of thermodynamics The energy of the universe is constant. It's the same thing as the Law of conservation of energy.
संलयन (fission) A nuclear reaction where a big atom breaks up into little ones. This is what happens in nuclear power plants.
मुक्त उर्जा (free energy) also called “Gibbs free energy”, it's the capacity of a system to do work.
free radical An atom or molecule with an unpaired electron. They're way reactive.
functional group A generic term for a group of atoms that cause a molecule to react in a specific way. It's really common to talk about this in organic chemistry, where you have “aldehydes, carboxylic acids, amines” and so on.
गामा किरण (gamma ray) High energy light given off during a nuclear process. When a nucleus gives off this light, it goes to a lower energy state, making it more stable.
geometrical isomer isomerism where atoms or groups of atoms can take up different positions around a double bond or a ring. This is also called cis- trans- isomerism.
ground state The lowest energy state possible for an electron.
group A column (the things up and down) in the periodic table. Elements in the same group tend to have the same properties. These are also called “families”.
अर्ध आयु (half-life) The time required for half of the radioactive atoms in a sample to decay. When talking about chemical reactions, it's the amount of time required to make half the reagent react.
half-reaction The oxidation or reduction part of a redox reaction.
हैलोजन (halogen) The elements in group 17. They're really reactive.
heat of reaction The amount of heat absorbed or released in a reaction. Also called the “enthalpy of reaction”
उष्मा (heat) The kinetic energy of the particles in a system. The faster the particles move, the higher the heat.
हेस का नियम (Hess's Law) The enthalpy change for a change is the same whether it takes place in one big step or in many small ones.
heterogeneous mixture A mixture where the substances aren't equally distributed.
homogeneous mixture A mixture that looks really “smooth” because everything is mixed up really well.
हुन्ड का नियम (Hund's rule) The most stable arrangement of electrons occurs when they're all unpaired.
hybrid orbital An orbital caused by the mixing of s, p, d, and f-orbitals.
hydration When a molecule has water molecules attached to it.
hydrocarbon A molecule containing carbon and hydrogen.
hydrogen bond The tendency of the hydrogen atom stuck to an electronegative atom to become attracted to the lone pair electrons on another electronegative atom. It's a pretty strong intermolecular force, which explains why water has such a high melting and boiling point.
hydrogenation When hydrogen is added to a carbon-carbon multiple bond.
hydronium ion The H+ ion, made famous by acids.
hydroxide ion The OH- ion, made famous by bases.
आदर्श गैस नियम (ideal gas law) PV=nRT
आदर्श गैस (ideal gas) A gas in which the particles are infinitely small, have a kinetic energy directly proportional to the temperature, travel in random straight lines, and don't attract or repel each other. Needless to say, there's no such thing as an ideal gas in the real world. However, we use ideal gases anyway because they make the math work out well for equations that describe how gases behave.
आदर्श विलयन (ideal solution) A solution in which the vapor pressure is directly proportional to the mole fraction of solvent present
immiscible When two substances don't dissolve in each other. Think of oil and water. They're immiscible. Organic compounds and water are frequently immiscible.
indicator A compound that turns different colors at different pH values. We generally like to have the color change at a pH of around seven because that's where the equivalence point of a titration is.
inhibitor A substance that slows down a chemical reaction.
अकार्बनिक यौगिक (inorganic compound) Any compound that doesn't contain carbon (except for carbon dioxide, carbon monoxide, and carbonates)।
अविलेय (insoluble) When something doesn't dissolve.
intermediate A molecule which exists for a short time in a chemical reaction before turning into the product.
अन्तर-आणविक बल (intermolecular force) A force that exists between two different molecules. Examples are hydrogen bonding (which is strong), dipole-dipole forces (which are kind of weak), and London dispersion forces (a.k.a. Van der Waal forces), which are very weak.
आयनिक बन्ध (ionic bond) A bond formed when charge particles stick together.
आयनन की उर्जा (ionization energy) The amount of energy required to pull an electron off of a gaseous atom.
अनुत्क्रंअणीय अभिक्रिया (irreversible reaction) A chemical reaction in which the reagents make products but the products can't reform reagents. Most chemical reactions in basic chemistry classes are thought of as being irreversible.
isotonic solutions Solutions containing the same osmotic pressure.
समस्थानिक (isotope) When an element has more than one possibility for the number of neutrons, these are called isotopes. All known elements posess isotopes. For the record, the word “isotope” doesn't imply that something is radioactive. TV told you that, and TV is stupid.
केल्विन (Kelvin) A unit used to measure temperature. One Kelvin is equal in size to one degree Celsius. To convert between degrees Celsius and Kelvins, simply add 273.15 to the temperature in degrees Celsius to get Kelvins.
ketone A molecule containing a R-CO-R' functional group. Acetone (dimethyl ketone) is a common one.
गतिज उर्जा (kinetic energy) The energy due to the movement of an object. The more something moves, the more kinetic energy it has.
Lanthanide contraction The tendency of the lanthanides to get small when you go from left to right in the periodic table.
lattice energy The energy released when one mole of a crystal is formed from gaseous ions.
lattice The three-dimensional arrangement of atoms or ions in a crystal.
law of conservation of energy The amount of energy in the universe never changes, ever. It just changes form.
द्रब्यमान संरक्षण का नियम (law of conservation of mass) The amount of stuff after a chemical reaction takes place is the same as the amount of stuff you started with.
ला शैतालिए का सिद्धान्त (Le Chatlier's Principle) When you disturb an equilibrium (by adding more chemical, by heating it up, etc.), it will eventually go back into equilibrium under a different set of conditions.
Lewis acid An electron-pair acceptor (carbonyl groups are really good ones)
Lewis base An electron-pair donor. Things with lone pairs like water and ammonia are really good ones.
Lewis structure A structural formula that shows all of the atoms and valence electrons in a molecule.
ligand A molecule or ion that sticks to the central atom in a complex. Common examples are ammonia, carbon monoxide, or water.
limiting reagent If you do a chemical reaction and one of the chemicals gets used up before the other one, the one that got used up is called the “limiting reagent” because it limited the amount of product that could be formed. The other one is called the excess reagent.
line spectrum A spectrum showing only certain wavelengths.
London dispersion force The forces between nonpolar atoms or molecules which is caused by momentary induced dipoles. It's real weak.
lone pair two electrons that aren't involved in chemical bonding. Also frequently referred to as an “unshared pair”.
main-block elements Groups 1,2, and 13-18 in the periodic table. They're called main block elements because the outermost electron is in the s- or p- orbitals. What that has to do with the term “main block” is unclear to me, but hey, that's life.
mass defect The difference between the mass of an atom and the sum of the masses of its individual components. Atoms usually weigh a little less than if you added up the weights of all the particles. This is because that extra mass was converted into the energy which holds the atom together (see “binding energy”)
द्रब्यमान (mass) The amount of matter in an object. The more mass, the more stuff is present.
mechanism A step-by-step sequence that shows how the products of a reaction are made from the reagents. Mechanisms are very frequently shown during organic chemistry.
मोललता (molality) The number of moles of solute per kilogram of solvent in a solution. This is a unit of concentration that's not anywhere near as handy or common as molarity.
मोलर द्रब्यमान (molar mass) The mass of one mole of particles.
मोलर आयतन (molar volume) The volume of one mole of a substance at STP. If you believe that everything is an ideal gas, this is always 22.4 liters. Unfortunately, there's no such thing as an ideal gas.
मोलरता (molarity) A unit of concentration equal to moles of solute divided by liters of solution.
mole fraction The number of moles of stuff in a mixture that are due to one of the compouds.
mole ratio The ratio of moles of what you've been given in a reaction to what you want to find. Handy in stoichiometry.
मोल (mole) 6.02 x 1023 things.
आणविक यौगिक (molecular compound) A compound held together by covalent bonds.
आणविक सूत्र (molecular formula) A formula that shows the correct quantity of all of the atoms in a molecule.
monatomic ion An ion that has only one atom, like the chloride ion.
neutralization reaction The reaction of an acid with a base to form water and a salt.
node A location in an orbital where there's no probability of finding an electron.
nonpolar covalent bond A covalent bond where the electrons are shared equally between the two atoms.
normal boiling point The boiling point of a substance at 1.00 atm.
normal melting point The melting point of a substance at 1.00 atm.
नार्मलता (normality) The number of equivalents of a substance dissolved in a liter of solution.
नाभिकीय संलयन (nuclear fusion) When many small atoms combine to form a large one. This occurs during a thermonuclear reaction.
nuclear fission This is when the nucleus of an atom breaks into many parts.
नाभिकीय अभिक्रिया (nuclear reaction) Any reaction that involves a change in the nucleus of an atom. Nuclear reactions take loads of energy, which is why you don't see them much around the lab.
nucleon A particle (such as proton or neutron) that's in the nucleus of an atom.
अष्टक नियम (octet rule) All atoms want to be like the nearest noble gas. (Well, they all want to have the same number of valence electrons, anyway)। To do this, they either gain or lose electrons (to form ionic compounds) or share electrons (to form covalent compounds)।
प्रकाशीय समावयवता (optical isomerism) Isomerism in which the isomers cause plane polarized light to rotate in different directions.
orbital This is where the electrons in an atom live.
कार्बनिक यौगिक (organic compound) A compound that contains carbon (except carbon dioxide, carbon monoxide, and carbonates)
परासरण (osmosis) The flow of a pure liquid into an area of high concentration through a semi-permeable membrane.
आक्सीकरण संख्या (oxidation number) The apparent charge on an atom.
आक्सीकरण (oxidation) When a substance loses electrons.
आंशिक दाब (partial pressure) The pressure of one gas in a mixture. For example, if you had a 50:50 mix of helium and hydrogen gases and the total pressure was 2 atm, the partial pressure of hydrogen would be 1 atm.
पाली का अपवर्जन का सिद्धान्त (Pauli exclusion principle) No two electrons in an atom can have the same quantum numbers.
percent yield The actual yield divided by the theoretical yield, times 100.
आवर्त काल (period) A row (left to right) in the periodic table.
आवर्तता का नियम (periodic law) The properties of elements change with increasing atomic number in a periodic way. That's why you can stick the elements into a big chart and have the elements line up in nice families.
पीएच (pH) -log[H+]
फेज आरेख (phase diagram) A chart which shows how the phase depends on various conditions of temperature and pressure.
phase The state of a compound (solid, liquid, or gas)
भौतिक गुण (physical property) A property which can be determined without changing something chemically. If that doesn't make sense, see the definition of “chemical change”.
पाई बन्ध (pi-bond) A double bond.
polar covalent bond A covalent bond where one atom tries to grab the electrons from the other one. This occurs because the electronegativities of the two atoms aren't the same.
बहुपरमाणवीय (polyatomic) contains more than one atom.
polymer A molecule containing many repeating units. Plastics are polymers and are formed by free radical chain reactions.
polyprotic acid An acid that can give up more than one hydronium ion. Examples are sulfuric acid and phosphoric acid.
स्थितिज उर्जा (potential energy) The energy something has because of where it is. Things that are way up high have more potential energy than things that are way down low because they have farther to fall.
precision A measurement of how repeatable a measurement is. The more significant figures, the more precise the measurement.
दाब (pressure) Force/area
उत्पाद (product) The thing you make in a chemical reaction.
क्वान्टम सिद्धान्त (quantum theory) The branch of physical chemistry that describes how energy can only exist at certain levels and makes generalizations about how atoms behave from this assumption.
रेडियो सक्रिय (radioactive) When a substance has an unstable nucleus that can fall apart, it's referred to as radioactive.
राउल्ट का नियम (Raoult's Law) The vapor pressure of a solution is directly proportional to the mole fraction of the solvent.
rate determining step The slowest step in a chemical reaction.
rate law A mathematical expression for the speed of a reaction as a function of concentration.
A hint It's usually true that things go faster if you have more stuff in the first place.
redox reaction A reaction that has both an oxidation and reduction.
resonance structure When more than one valid Lewis structure can be drawn for a molecule, these structures are said to be resonance structures. Resonance structures arise from the fact that the electrons are delocalized.
उत्क्रमणीय अभिक्रिया (reversible reaction) A reaction in which the products can make reagents, as well as the reagents making products.
वर्ग मूल माध्य वेग (root mean square velocity (RMS velocity)) The square root of the average of the squares of the individual velocities of the gas particles in a mixture. To put it in a way that a normal human can understand, it's the average of how fast the particles in a gas are going (assuming you ignore the direction they're traveling in)।
लवण (salt) An ionic compound.
संतृप्त (saturated) When the maximum amount of solute is dissolved in a liquid
उष्मागतिकी का द्वितीय नियम (Second law of thermodynamics) Whenever you do something, the universe gets more random.
अर्धचालक (semiconductor) A substance that conducts electricity poorly at room temperature, but has increasing conductivity at higher temperatures. Metalloids are usually good semiconductors.
shielding effect The outer electrons aren't pulled very tightly by the nucleus because the inner electrons repel them. This repulsion is called the shielding effect, and can be used to explain lots of neat-o stuff.
सिग्मा बन्ध (sigma bond) A real fancy way of saying “single bond”
significant figure The number of digits in a number that tell you useful information. For example, when you weigh yourself on a bathroom scale, it says something like 150 pounds rather than 150.32843737 pounds. Why? Because the thing can only weigh accurately to the nearest pound. Any other digits that are on this number don't mean anything, because they're probably wrong anyway.
single-displacement reaction (a.k.a. single replacement reaction) When one unbonded element replaces an element in a chemical compound. These are frequently redox reactions.
बिलेयता (solubility) A measurement of how much of a solute can dissolve in a liquid.
बिलेयता गुणनफल (solubility product constant) Abbreviated Ksp, this value indicates the degree to which a compound dissociates in water. The higher the solubility product constant, the more soluble the compound.
विलेय (solute) The solid that gets dissolved in a solution.
विलायक (solvent) The liquid that dissolves the solid in a solution.
विशिष्ट उष्मा धारिता (specific heat capacity) The amount of heat required to increase the temperature of one gram of a substance by one degree.
spectator ions' The ions in a reaction that don't react.
spontaneous change A change that occurs by itself. All exothermic reactions are spontaneous. However, this doesn't mean that all exothermic reactions are fast. The combustion of gasoline is spontaneous, but not very fast unless you add a little energy.
मानक ताप व दाब (standard temperature and pressure) One atmosphere and 273 K.
steric hindrance This is the idea that the functional groups on big molecules get in the way of a chemical reaction, making it go slower. Imagine a fat guy trying to get into a Honda Prelude - that's steric hindrance.
stoichiometry The art of figuring how much stuff you'll make in a chemical reaction from the amount of each reagent you start with.
STP See standard temperature and pressure.
strong acid An acid that fully dissociates in water
strong nuclear force The force that holds the nucleus together. As the name suggests, this force is strong.
संरचना सूत्र (structural formula) See Lewis structure.
उर्ध्वपातन (sublimation) When a solid can change directly into a gas. Dry ice does this.
अतिशीतलन (supercooling) When you cool something below its normal freezing point
अतिसंतृप्त (supersaturated) When more solute is dissolved in a liquid than is theoretically possible. This doesn't happen much, as you might imagine.
पृष्ट तनाव (surface tension) A measurement of how much the molecules on a liquid tend to like to stick to each other. If something has a high surface tension, it likes to bead up.
suspension A mixture that looks homogeneous when you stir it, but where the solids settle out when you stop. Mud is a very short-lived suspension, while peanut butter is a very long-lived suspension.
संश्लेषण (synthesis) When you make a big molecule from two or more smaller ones.
तन्त्र (system) Everything you're talking about at the moment.
ताप (temperature) A measurement of the average kinetic energy of the particles in a system.
theoretical yield The amount of product which should be made in a chemical reaction if everything goes perfectly.
उष्मागतिकी (thermodynamics) The study of energy
उष्मागतिकी का तृतीय नियम (Third law o' thermodynamics) The randomness of a system at 0 K is zero.
titration When the concentration of an acid or base is determined by neutralizing it.
transition state See “activated complex”
त्रिक बिन्दु (triple point) The temperature and pressure at which all three states of a substance can exist in equilibrium.
unit cell The simplest part of a crystal that can be repeated over and over to make the whole thing.
असंतृप्त (unsaturated) When you haven't yet dissolved all of the solute that's possible to dissolve in a liquid.
unshared electron pair two electrons that aren't involved in chemical bonding. Also frequently referred to as a “lone pair”.
संयोजी एलेक्ट्रान (valence electron) The outermost electrons in an atom.
वाष्प दाब (vapor pressure) The pressure of a substance that's present above it's liquid. For example, you can tell that ammonia has a high vapor pressure because the smell of it is very strong above liquid ammonia.
वाष्पन (vaporization) When you boil a liquid.
volatile A substance with a high vapor pressure.
VSEPR A theory for predicting molecular shapes that assumes that electrons like to be as far from each other as possible.
#to_be_edited
Copyright © 2007 - 2018 सर्वाधिकार सुरक्षित. Dr. K. Singh | Organic Synthesis Insignt.
|
2018-08-18 23:43:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5359959602355957, "perplexity": 1669.9335245815155}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221213903.82/warc/CC-MAIN-20180818232623-20180819012623-00285.warc.gz"}
|
http://math.elinkage.net/showthread.php?tid=721&pid=2671
|
• 0 Votes - 0 Average
• 1
• 2
• 3
• 4
• 5
Notes on 【Galois Theory】
02-20-2017, 11:47 AM
Post: #1
elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
Notes on 【Galois Theory】
1 Revision from Groups, Rings and Fields
1.1 Field Extensions
Let $K,\,L$ be fields, ring homomorphism $\varphi: K\to L$ is
non-zero, then $\small\ker\varphi = \{0\},\;\varphi\,$is a homomorphism of fields
$\small\varphi(a/b)=\varphi(a)/\varphi(b).$
1.1.1 Definition If there is a homomorphism of fields
$\varphi:K\hookrightarrow L$, then $L$ is a field extension of $K$, $\varphi$ is an
embedding of $K$ into $L$.
1.1.2 Remark If $\varphi: K\hookrightarrow L$, then $\varphi:K\to \varphi(K)$ is an
isomorphism. $L$ might be an field extension via different
homomorphism thus we simply use $L/K$ to denote that $L$
is a field extension of $K$ and assume that $K\subset_{\small F}L$.
1.1.3 Lemma $(\forall j\in J\;(K_j\subset_{\small F} L))\implies \displaystyle{\bigcap_{j\in J}K_j \subset_{\small F} L}.$
$\qquad\;(A\subset_{\small F} B\,$means $A$ is a subfield of field $B)$
If $K\subset_{\small F} L,\; S\subset L,\,$then $\small\displaystyle{K(S):=\bigcap\{M\mid K\cup S\subset M\subset_{\small F} L\} }$
is the smallest subfield of $L$ containing $K\cup S$.
Write $K(a_1,\ldots, a_n)\,$For $K(\{a_1,\ldots,a_n\})$
1.1.4 Definition $\small L/K\,$is finitely generated if $\small K(\alpha_1,\ldots,\alpha_n) = L$
(where $n\in\mathbb{N}$). If $L=K(\alpha),\,$the extension is simple.
1.1.5 Definition Given $\small L/K$, we say $\small\alpha\in L$ is algebraic over
$K$ if $\small\exists f\in K[X]\;(f(\alpha) = 0).$ Otherwise $\alpha\in L\,$is transcendental
over $K.$ If $\alpha$ is algebraic, $f: \small X^n+a_{n-1}X^{n-1}+\cdots +a_0$ is the
monic polynomial of smallest degree such that $f(\alpha) = 0$ is
called the minimal polynomial of $\alpha$. Clearly such an $f$ is
unique and irreducible.
1.1.6 Definition $L/K$ is algebraic if $\alpha$ is algebraic over $K$
$(\forall \alpha\in L)$; It is pure transcendental if $\alpha$ is transcendental
over $K\,(\forall \alpha\in L-K).$
02-20-2017, 12:08 PM
Post: #2
elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
1.2 Classification of simple algebraic extensions
Let $K$ be a field, $f\in K[X],\; (f) = \{\lambda f\mid \lambda\in K[x]\}$ Then
$\quad(g'_1\,=g_1+\lambda_1 f)\wedge(g'_2\,=g_2+\lambda_2 f)\implies$
$\qquad\quad(g'_1+g'_2=g_1+g_2+(\lambda_1+\lambda_2)f)\wedge$
$\qquad\quad(g'_1g'_2=g_1g_2+(\lambda_1g2+\lambda_2g_1+\lambda_1\lambda_2)f)$
$\therefore k[X]/(f)=\{\bar{h}=\{h+(f)\}\mid h\in K[x]\}$ is a ring
$\quad($with operations $\bar{u}+\bar{v}=\overline{u+v},\;\bar{u}\bar{v}=\overline{uv}).$
Let $f\underset{\,}{\in} K[X]$ is irreducible, then $k[X]/(f)$ is a field since
$\quad\;\bar{h}\ne 0\iff h\not\in(f)\iff (\exists u,g\in k[X]\,(uf +gh =1))$
$\qquad\qquad\iff (\exists g\in K[x]\,(\bar{g}\bar{h}=1))$.
Now let $\varphi: k[x]\to K[X]/(f),\; g\mapsto \bar{g} = g+(f)$ be an
epimorphism of rings, $\alpha = \varphi(X),\,$then$\,K(\alpha)\subset_{\small F}K[X]/(f)$
since field$\,K[X]/(f)\supset K\cup\{\alpha\}.\;$ Conversetly,
$\small\forall g\in K[x]\,\exists r\in K[X]\;(r = 0)\vee (f\mid g-r,\,\deg(r )<\deg(f))$ thus
$\small K[X]/(f) = \{\bar{r}(\alpha) \mid r\in K[x],\,(r = 0)\vee (\deg( r)<\deg(f))\}\subset K(\alpha)$
Again, we identify $K$ with $\varphi(K)$ here. This can also be stated
as: If $\alpha$ is algebriac, $f$ is the minimal polynomial of $\alpha$ over $K$
then we have a commutative diagram
$\qquad\qquad\qquad\qquad\begin{matrix}K & \rightarrow & K[X]\\ & \searrow & \downarrow\\ & & K(\alpha)\end{matrix}$
inducing an isomorphism of fields $\small K[X]/(f)\cong K(\alpha)$.
Thus any simple algebraic extension of $K$ is of the form
$\small K\hookrightarrow K[X]/(f)$ for some$\small\;f\in K[X]$ irreducible up to an isomorphism.
So, classifying simple algebraic extensions of $K$ is equivalent to
classifying irreducible monic polynomials in$\small\;K[x]$.
02-27-2017, 11:18 PM
Post: #3
elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
1.3 Tests for irreducibility
Let $R$ be a UFD, $K$ its field of fractions, e.g. $\small R=\mathbb{Z},\,K=\mathbb{Q}.$
1.3.1 Lemma (Gauss)
$\quad f\in R[X]$ is irreducible in $R[X]$ iff it's irreducible in $K[X].$
1.3.2 Theorem(Eisenstein's Criterion)
$\quad$If $\; f=\small a_nX^n+\cdots+a_1 X+a_0\in R[X],\;\;p\in R$ is irreducible
$\quad$such that $p\mid a_i\,(i=\overline{0,n-1}),\;\;p^2\nmid a_0,\;\; p\nmid a_n,$
$\quad$then $f$ is irreducible in $R[X]$ hence irreducible in $K[X]$.
02-27-2017, 11:25 PM
Post: #4
elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
1.4 The degree of an extension
1.4.1 Definition If $L/K$ is a field extension, then $L$ has the
structure of a vector space over $\small K.\;$The dimension $\small [L:K]$ of
the vector space is called the degree of the extension. $L$ is
finite over $K$ if $\small[L:K]$ is finite.
1.4.2 Theorem $\alpha$ is algebraic over a field $K$ iff $K(a)/K$ is
finite. In such case, $[K(\alpha): K]$ is the degree of the
minimal polynomial of $\alpha$.
1.4.3 Proposition (Tower Law)
$\qquad (K\hookrightarrow L\hookrightarrow M)\implies [M:K]=[M:L][L:K].$
1.4.4 Corollary If $L =K(\alpha_1,\ldots,\alpha_n)$ and each $\alpha_j$ is
$\qquad$algebraic over $K$, then $L/K$ is a finite field extension.
02-28-2017, 02:54 PM
Post: #5
elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
1.5 Splitting fields
1.5.1 Definition If $L/K$ is a field extension, we say that $f\in K[x]$ splits
(completely) over $L$ if $f = k\small (X-\alpha_1)\cdots(X-\alpha_n)$ where $k\in K,\;\alpha_j\in L$
$\,(j=\overline{1,n}).\quad L\,$ is called a splitting field for $f$ if $f$ fails to split over any
proper subfield of $L$, i.e. if $\small L = K(\alpha_1,\ldots,\alpha_n)$.
1.5.2 Remark Splitting fields always exist.
For if $g$ is an irreducible factor of $f$, then $\small K[X]/(g) = K(\alpha)$ is an extension
of $K$ for which $g(\alpha)=0\,(\alpha$ is the image of $X)$. The remainder theorem
implies that $g$(and hence $f$) splits off a linear factor. Induction implies that
there exists a splitting field $L$ for $f$, with $[L:K]\le n! \;(n = \deg(f))$ by
proposition 1.4.3.
Splitting fields are unique up to isomorphisms over $K$.
1.5.3 Proposition If $\theta:K\to K'$ is a field isomorphism with $f\in K[X]$]
corresponding to $g=\theta(f)\in K'[X]$. Then any splitting field $L$ of $f$ over
$K$ is isomorphic over $\theta$ to any splitting field $L'$ of $g$ over $K'$, and we have
the commutative diagram
$\qquad\qquad\qquad\qquad\qquad\begin{matrix}L & \overset{\tilde{\theta}}{\longrightarrow} & L'\\ \uparrow & & \uparrow\\ K & \overset{\theta}{\longrightarrow}& K'\end{matrix}$
« Next Oldest | Next Newest »
Forum Jump:
|
2019-03-25 07:46:02
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9798896908760071, "perplexity": 956.6946233154736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912203842.71/warc/CC-MAIN-20190325072024-20190325094024-00049.warc.gz"}
|
https://atractor.pt/mat/Moebius/Construcao-Moebius-_en.html
|
## Möbius Strip
### Construction of the Möbius strip and other surfaces
Take four paper rectangles. The proportions are not important: you can take as dimensions for the sides approximately 5cm and 30cm. Choose one of the rectangles and glue together the opposite small sides: you obtain a cylinder. Do the same for the second rectangle, but twist the paper strip "once" (i.e. halftour) before glueing: you obtain what is called the Möbius strip. Do the same for the other rectangles twisting respectively twice and three times before glueing. You can see animations which suggest the constructions by clicking on the corresponding figures in the next table.
n
Animations
Figures
Animations
n
Central
cut
Non-centered
cut
Orientability
Keep in mind in particular that:
1. on the first column is the number of twists before glueing;
2. on the column Animations, the small figures are links to animations which suggest the corresponding glueing process;
3. on the column Figures, the images link to the final figures obtained, with the border marked;
4. on the column Central cut, the small figures are links to animations which suggest the longitudinal cut of each strip along the central circle;
5. on the column Non-centered cut, the small figures are links to animations which suggest the longitudinal cut of each strip starting from a point on the strip which is at a distance of 1/3 of the width of the strip from the border;
6. on the column Orientability, the question of the non-orientability of some surfaces is raised.
Paint the surfaces obtained, to conceal the glueing area. Take two of the paper models and try to find what there is in common between them and what is different. Do that for each couple. The ideal would be to identify a property in relation to which some of these models can be considered equivalent and then try to identify what distinguishes these equivalent models.
As illustration of what has been said, links to animations, which correspond to cutting (and separating) a Moebius strip longitudinally, respectivaly along the central circumference and along a line locally dividing the Moebius strip in three parts of equal height, have been placed in the previous table. You are advised to proceed in the following order:
1. Try to imagine what is obtained, in each case, for $$n=0$$ (easy), $$1$$, $$2$$, $$3$$, ...;
2. Make the experiment with paper and scissors to check whether your conjecture is correct;
3. Reconstruct with the animations what you did with the scissors.
|
2022-09-27 18:32:12
|
{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8208975791931152, "perplexity": 792.3257744523569}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335054.79/warc/CC-MAIN-20220927162620-20220927192620-00607.warc.gz"}
|
https://mathalino.com/tag/reviewer/eccentrically-loaded-member
|
## Problem 921 | Kern Area of a Wide Flange Section: W360 x 122
Problem 921
Calculate the sketch the kern of a W360 × 122 section.
## Problem 920 | Additional Centroidal Load to Eliminate Tensile Stress Anywhere Over the Cross Section
Problem 920
A compressive load P = 100 kN is applied, as shown in Fig. 9-8a, at a point 70 mm to the left and 30 mm above the centroid of a rectangular section for which h = 300 mm and b = 250 mm. What additional load, acting normal to the cross section at its centroid, will eliminate tensile stress anywhere over the cross section?
## Problem 918 | Stress at Each Corner of Eccentrically Loaded Rectangular Section
Problem 918
A compressive load P = 12 kips is applied, as in Fig. 9-8a, at a point 1 in. to the right and 2 in. above the centroid of a rectangular section for which h = 10 in. and b = 6 in. Compute the stress at each corner and the location of the neutral axis. Illustrate the answers with a sketch.
## Eccentrically Loaded Short Compression Member
Consider the cross-section below. A compressive load P is applied at any point (ex, ey) with respect to the principal axes x and y. The moment of P about these axes are respectively
$M_x = Pe_y$ and $M_y = Pe_x$
## 255 Equivalent loads to a compression member with eccentric load
Problem 255
A short compression member carries an eccentric load P = 200 lb situated 2 in. from the axis of the member, as shown in Fig. P-225. In strength of materials it is learned that the internal stresses are determined from the equivalent axial load and couple into which P may be resolved. Determine the equivalent axial load and couple.
|
2020-05-30 08:23:39
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.31976866722106934, "perplexity": 1908.158198915622}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347407667.28/warc/CC-MAIN-20200530071741-20200530101741-00595.warc.gz"}
|
https://socratic.org/questions/is-6-8-a-rational-number
|
Is 6.8 a rational number?
Aug 11, 2016
Yes, $6.8$ is indeed a rational number.
Explanation:
By definition, a rational number is a number that can be expressed by the ratio $\frac{a}{b}$, where $a$ and $b$ are integers and $b$ doesn't equal $0$.
$6.8 = \frac{68}{10}$ and both 68 and 10 are integers.
In addition, $6.8$ is a rational number (part of many other numbers) that make up real numbers (others including terminating decimals, repeating decimals, integers, and counting numbers, which are all rational numbers).
Some examples include:
Counting Numbers: $\text{ } 0 , 1 , 2 , 3 , 4 , \ldots$
Integers: $\text{ } \ldots - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , \ldots .$
Rational Numbers: $\text{ } \frac{7}{9} , 8.47 , - \frac{5}{8} , 0.3 \ldots$
In conclusion, a number that doesn't meet these criteria would be otherwise considered an irrational number (a number that cannot be expressed in the form $\frac{a}{b}$ e.g. $1.341823 \ldots$, $\pi$, $\sqrt{2}$, etc.).
Irrational numbers are infinite, non-recurring decimals.
|
2021-10-22 13:14:08
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 15, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7759255766868591, "perplexity": 503.36648227592883}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585507.26/warc/CC-MAIN-20211022114748-20211022144748-00464.warc.gz"}
|
https://api-project-1022638073839.appspot.com/questions/how-do-you-factor-2x-3-3x-2-5x
|
# How do you factor 2x^3 - 3x^2 - 5x?
Nov 14, 2015
$2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 x - 5\right) \left(x + 1\right)$
#### Explanation:
First, note that each term has a factor of $x$, and so we have
$2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 {x}^{2} - 3 x - 5\right)$
Now, we can use the quadratic formula to find the remaining factors, but first let's see if there are easy integer solutions by looking for $a , b , c , d$ where $2 {x}^{2} - 3 x - 5 = \left(a x + b\right) \left(c x + d\right)$
We know that $a c = 2$ and so we can look at $\left(2 x + b\right) \left(x + d\right)$
We also know $b d = - 5$ and so our possible choices are $\left(2 x + 1\right) \left(x - 5\right)$ and $\left(2 x - 5\right) \left(x + 1\right)$
Multiplying these out shows that $2 {x}^{2} - 3 x - 5 = \left(2 x - 5\right) \left(x + 1\right)$
So our final result is
$2 {x}^{3} - 3 {x}^{2} - 5 x = x \left(2 x - 5\right) \left(x + 1\right)$
|
2020-04-08 14:34:10
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.916975736618042, "perplexity": 116.71851724697025}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585371818008.97/warc/CC-MAIN-20200408135412-20200408165912-00306.warc.gz"}
|
https://tex.stackexchange.com/questions/300550/command-macro-for-an-environment
|
# Command macro for an environment
How can I define a command macro using \newcommand so that the argument to the command is surrounded by an environment?
• yes you can do that, what is the question? – David Carlisle Mar 23 '16 at 17:01
• It's possible but I doubt it to be very effective. – user31729 Mar 23 '16 at 17:05
• @David Ah my mistake, the code in my question wasn't working on my machine, but it was for an unrelated reason. – agcha Mar 23 '16 at 17:16
• One should note that not all environments can be macro-ized in this way. Sometimes, \newcommand\mat[1]{\envname #1\endenvname} must be used. Sometimes, not even that works. – Steven B. Segletes Mar 23 '16 at 17:20
• @ChristianHupfer Apart from verbatim most can (including align) – David Carlisle Mar 23 '16 at 19:26
A proof that it is possible and using an optional argument that can be b or p or anything of the known prefixes (e.g. v, B and V)
Besides that all: It does not improve readability of the code, in my point of view.
\documentclass{article}
\usepackage{amsmath}
\newcommand{\mat}[2][b]{%
\begin{#1matrix}
#2
\end{#1matrix}
}
\begin{document}
$\mat{ 1 & 2 \\ 3 & 4 \\}$ and $\mat[p]{ 1 & 2 \\ 3 & 4 \\}$ and $\mat[B]{ 1 & 2 \\ 3 & 4 \\}$ and $\mat[v]{ 1 & 2 \\ 3 & 4 \\}$ and $\mat[V]{ 1 & 2 \\ 3 & 4 \\}$
\end{document}
|
2019-07-21 03:05:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8351181149482727, "perplexity": 1625.8352774280534}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195526818.17/warc/CC-MAIN-20190721020230-20190721042230-00258.warc.gz"}
|
http://forum.zkoss.org/question/40803/converter-with-parameter/
|
# Converter with parameter
81 2 3
How to parse additional parameters to a converter attribute? Per default a new TypeConverter instance is created.
Is it possible to call the method coerceToUi with further parameters or is it possible to use a pre-instantiated class?
What is the optimal approach for this requirement?
delete retag edit
robertpic71
1275 1
First, there is no way to extend the TypeConverter directly.
Here are my work-a-rounds:
1. use the object instead of the object.attribute
my real world example:
It want to show "customer.sales" but the sales should have another colour when the is > 10 % higher than the last years sales.
The get the whole information i need the customer so i use
<rows>
<row self="@{each=customer}">
<label value="@{customer.number}"/>
..
<label value="@{customer, converter='com.at.od.converters.sales'}"/>
...
Now i get the customer inside my TypeConverter and have no problem to evaluate the color.
2. use custom-attributes to communicate with the TypeConverter
See this side for some examples with "static" parameters.
I should be possible to setup i.e. a converterprogram in the attribute.
<label value="@{customer, converter='com.at.od.converters.spring'}">
<custom-attributes convert="\${springconverter}"/></label>
I've not tested this, but it should work.
Note that the timing for all kinds of listbinding (listbox, rows, combobox...) is important. The single-row/item is used as template for the databinding.
If you need the attribute in all items you have to set this before the binder renders and copy the template.
3. use databinding for custom-attributes
Since ZK 5RC it is possible to use databinding for custom-attributus. So you could pass all your parameters inside custom-attributes and the last one render the item.
/Robert
[hide preview]
|
2019-02-23 00:43:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20670025050640106, "perplexity": 4247.344306096896}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249414450.79/warc/CC-MAIN-20190223001001-20190223023001-00192.warc.gz"}
|
https://cs.stackexchange.com/questions/139987/quick-sort-hoares-partition-algorithm-is-there-a-mistake-in-clrs/139992#139992
|
# Quick sort, Hoare's partition algorithm. Is there a mistake in CLRS?
The following problem appears in "Introduction to Algorithms" by Thomas Cormen et. al., aka CLRS.
Problem 7-1.b
Hoare's partition algorithm from the book.
Part b: Assuming the subarray $$A[p,\cdots,r]$$ contains at least two elements, show that the indices $$i$$ and $$j$$ are such that we never access an element of outside the subarray $$A[p,\cdots,r]$$
I think this it is not possible to prove this because this statement is not correct, i.e., during the course of execution we happen to access elements outside the array. Here I prove the converse of part b by using a counter-example.
Consider the array A = [1,2]. Let p = 1 and there r =2.
• Initially, $$i = 0$$, $$j = 3$$ and $$x = 1$$.
• After the loop $$5-7$$ executes once, $$j$$ becomes 2 and the condition in line 7 fails resulting the termination of the loop $$5-7$$.
• Now we reach the loop $$8-10$$. Now let's look at the state after this loop executes twice. $$i = 2$$ and the condition in line 10 still holds. Therefore the loop executes a third time and $$i$$ becomes 3.
• Now to test the condition in line 10 we access $$A[3]$$ which is outside the subarray $$A[1,\cdots, 2]$$
Is there a mistake in my reasoning or is the problem statement wrong?
• After the loop $5-7$, $j$ becomes $1$ since $A[2] > x$. May 6 '21 at 17:07
• After loop $8−10$, $i$ becomes $1$ since $A[1] = x$. May 6 '21 at 17:09
• Since $i = j = 1$, the algorithm terminates and return $1$. May 6 '21 at 17:10
The problem statement is correct.
I think you getting confused about: repeat $$j \gets j-1$$ until $$A[j] \leq x$$. It means that if $$A[j]>x$$ then do $$j \gets j-1$$.
Similarly, loop $$8-10$$ means that if $$A[i] then do $$i \gets i+1$$.
Therefore, the algorithm executes in the following way on $$A = [1,2]$$:
1. After the loop $$5-7$$, $$j$$ becomes $$1$$ since $$A[2] > x$$.
2. After loop $$8−10$$, $$i$$ becomes $$1$$ since $$A[1] = x$$.
3. Since $$i = j = 1$$, the algorithm terminates and returns $$1$$.
• Thanks a lot! I understand now. May 7 '21 at 7:53
|
2022-01-24 04:47:15
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8006582260131836, "perplexity": 565.0787399840001}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00581.warc.gz"}
|
https://physics.stackexchange.com/questions/293539/can-the-finite-dimensional-irreducible-j-j-representations-of-the-lorent
|
# Can the finite dimensional irreducible $(j_+,j_-)$ representations of the Lorentz group $SO(3,1)$ be unitary?
Since the Lorentz group $SO(3,1)$ is non-compact, it doesn't have any finite dimensional unitary irreducible representation. Is this theorem really valid?
One can take complex linear combinations of hermitian angular momentum generator $J_i^\dagger=J_i$ boost generator $K_i^\dagger=-K_i$ to construct two hermitian generators $N_i^{\pm}=J_i\pm iK_i$. Then, it can be easily shown that the complexified Lie algebra of $SO(3,1)$ is isomorphic to that of $SU(2)\times SU(2)$. Since, the generators are now hermitian, the exponentiation of $\{iN_i^+,iN_i^-\}$ with real coefficients should produce finite dimensional unitary irreducible representations. The finite dimensional representations labeled by $(j_+,j_-)$ are therefore unitary.
$\bullet$ Does it mean we have achieved finite dimensional unitary representations of $SO(3,1)$?
$\bullet$ If the $(j_+,j_-)$ representations, are for some reason are non-unitary (why I do not understand), what is the need for considering such representations?
$\bullet$ Even if they are not unitary (for a reason I don't understand yet), they tell how classical fields transform such as Weyl fields, Dirac fields etc. So what is the problem even if they are non-unitary?
The statement "Non-compact groups don't have finite-dimensional unitary representations" is a heuristic, not a fact. $(\mathbb{R},+)$ is a non-compact Lie group that has non-trivial finite-dimensional unitary representations. However, the Poincaré group and the Lorentz group really don't have any finite-dimensional unitary representations.
Your construction fails because the complexification $\mathfrak{so}(1,3)_\mathbb{C}$ is only isomorphic to the complexification $(\mathfrak{su}(2)\oplus\mathfrak{su}(2))_\mathbb{C}$, not to the real Lie group. You found a unitary representation of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ itself, but this doesn't give you a unitary representation of the complexification, nor of $\mathfrak{so}(1,3)$.
We care about those finite-dimensional representations of $\mathrm{SO}(1,3)$ even if they are not unitary because these are the representations on the target spaces of fields. The representation that needs to be unitary is the representation on the quantum space of states, but not on the target space of fields. Clearly, a vector field transforms in the "standard" representation of $\mathrm{SO}(1,3)$ and doesn't care that it's not unitary because the target space is $\mathbb{R}^{1,3}$ which isn't even a complex vector space to begin with! There is no "problem" with these representations, they just are not the representations we need on the Hilbert space of states, which are projective representations of $\mathrm{SO}(1,3)$, which are equivalent to unitary linear representations of $\mathrm{SL}(2,\mathbb{C})$, its universal cover. For more on the necessity of projective representation, see this Q&A of mine.
• @ ACuriousMind- The $(j_+,j_-)$ representations that I find, are they anyhow related to the representations of $SL(2,\mathbb{C})$? – SRS Nov 19 '16 at 18:21
• @SRS Sure, since they are reps of $\mathrm{SO}(1,3)$, they are also reps of its universal cover. – ACuriousMind Nov 19 '16 at 18:24
• Although the representation is labeled as $(j_+,j_-)$, while constructing representations of $SO(3,1)$, I have to go back to $J_i$ and $K_i$ from $N_i^{\pm}$. Exponentiate them and they are non-unitary. Direct exponentiation of $N_i^{\pm}$ doesn't give the representation of $SO(3,1)$. Going from $J_i,K_i\rightarrow N_i^{\pm}$ is merely a trick to solve for generators easily. Is that correct? – SRS Nov 19 '16 at 18:36
• @SRS I don't know what you mean by "to solve for generators easily". Going to the $N_i$ means presenting $\mathfrak{so}(1,3)_\mathbb{C}$ as $(\mathfrak{su}(2)\oplus\mathfrak{su}(2))_\mathbb{C}$, which we do because we already know the representation theory of $\mathfrak{su}(2)$ from non-relativistic rotations! – ACuriousMind Nov 19 '16 at 18:39
• @SRS Yes, that's correct. – ACuriousMind Nov 19 '16 at 18:56
$SO(1,3)$ is a real Lie group so its Lie algebra is real too, you are not allowed to combine generators with complex coefficients if you are looking for representation of that group. Referring to the (non-unitary) fundamental representation made of $4 \times 4$ real matrices you consider, $N_i^\pm$ does not belong to the real Lie algebra of $SO(1,3)$. Exponentiating real linear combinations of $N_i^\pm$ you, in fact, obtain a unitary representation of a group. Unfortunately the group is not $SO(1,3)$.
The use of complex extensions of the Lie algebra of $SO(1,3)$ is however helpful when classifying the representations of the proper real Lie algebra of $SO(1,3)$, since classifying all complex representations includes a classification of the real representations too, and the complex Lie algebra of $SO(1,3)$ is isomorphic to a direct sum of a pair of Lie algebras of $SU(2)$ whose theory is relatively simple.
Unitarity is necessary in Quantum Theory due to Wigner's theorem which establishes that, picturing the states of a quantum system in a Hilbert space, all symmetries are represented by unitary or anti unitary operators.
Actually the problem is more complicated due to the appearance of phases (pure states are defined as unit vectors up to phases) which may destroy the composition law of the Poincaré group (a central extension is necessary). However a theorem by Bargmann proves that the Poincaré group is not affected by this problem.
• @ Valter Moretti- Let me try to understand. 1. The representations considered by taking complex combinations of generators are unitary but they are not representations of $SO(3,1)$. Is it correct? Are they representations of $SL(2,\mathbb{C})$? 2. You said, exponentiating real linear combinations of $N_i^{\pm}$ one can obtain unitary representations. Which particular combinations are you referring to? – SRS Nov 19 '16 at 14:12
• 3. A representation is real, if it is identical to its conjugate representation. Did you use real representation in the same sense? 4. Does the $(j_+,j_-)$ representations include both real and complex representations? – SRS Nov 19 '16 at 14:13
• 4. Does the representations of the complex extension of the Lie algebra of $SO(3,1)$ matches the projective representations of $SO(3,1)$ and ordinary representations of its universal cover $SL(2,\mathbb{C})$? Otherwise, I don't see a reason for considering complex extension. – SRS Nov 19 '16 at 14:29
• Sorry I have no time for a discussion. Have a look at the Wikipedia page en.wikipedia.org/wiki/… – Valter Moretti Nov 19 '16 at 15:33
• Unfortunately the proof of the non-existence of unitary finite-dimensional reps apoearing in a footnote in wikipedia is wrong, also the proof in the quoted paper therein is similarly wrong. The paper mentions a proof in Barut-Raczka book which is wrong as well. – Valter Moretti Dec 25 '16 at 9:55
|
2019-06-17 11:20:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8841279149055481, "perplexity": 264.05962740363003}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998473.44/warc/CC-MAIN-20190617103006-20190617125006-00209.warc.gz"}
|
https://math.stackexchange.com/questions/2384135/given-the-matrix-of-this-relation-why-isnt-the-relation-transitive
|
# Given the matrix of this relation, why isn't the relation transitive?
$$A = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}.$$
and the set notation for the relation is
$$R = \{(a,b),(a,c),(b,c)\}$$ Is there a fast way to show whether or not it's transitive? I thought it isn't transitive but my solution says it is. I know that a set relation $R$ is transitive on a set $A$ if
$$\forall a,b,c \in A, \ \ \text{if} \ \ aRb, bRc \ \ \text{then} \ \ aRc.$$
Note that $(a,b)$ and $(b,c)$ are the only two pairs that have matching middle terms ($b$ in this case). Thus, in order for the relation to be transitive, it must contain $(a,c)$ - which it does. Therefore it is transitive.
$$R = \{(a,a),(a,b),(a,c),(a,e),(b,c),(b,e),(c,d) \}.$$ We find all pairs of elements with matching middle terms. They are: $$(a,a)\ \text{and}\ (a,b) \\ (a,a)\ \text{and}\ (a,c) \\ (a,a)\ \text{and}\ (a,e) \\ (a,b)\ \text{and}\ (b,c) \\ (a,b)\ \text{and}\ (b,e) \\ (a,c)\ \text{and}\ (c,d) \\ (b,c)\ \text{and}\ (c,d)$$ Now, test them one by one until we either find a failure or terminate the whole list. $$(a,a)\ \text{and}\ (a,b) \implies \text{need}\ (a,b)\ \checkmark \\ (a,a)\ \text{and}\ (a,c) \implies \text{need}\ (a,c)\ \checkmark \\ (a,a)\ \text{and}\ (a,e) \implies \text{need}\ (a,e)\ \checkmark \\ (a,b)\ \text{and}\ (b,c) \implies \text{need}\ (a,c)\ \checkmark \\ (a,b)\ \text{and}\ (b,e) \implies \text{need}\ (a,e)\ \checkmark \\ (a,c)\ \text{and}\ (c,d) \implies \text{need}\ (a,d)\times \\$$ Thus we find that $R$ is not transitive and there is no need to check the pair. Note that we also could have made our list shorter by not considering the reflexive term $(a,a)$.
|
2019-09-15 22:10:08
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7902625203132629, "perplexity": 132.01254273756288}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572436.52/warc/CC-MAIN-20190915215643-20190916001643-00268.warc.gz"}
|
https://dev.goldbook.iupac.org/terms/view/L03606/plain
|
## local fraction volatilized, $$\chi _{\text{v}}$$, $$\beta_{\text{v}}$$
in flame emission and absorption spectrometry
https://doi.org/10.1351/goldbook.L03606
The substance fraction of the volatilized component in the total desolvated component. The gaseous state includes free atoms, molecules and radicals. This quantity is measured in a defined part of the flame, usually the observation space. The fraction volatilized varies inversely with the size of the desolvated particles. Since χ v varies markedly with the height in the flame, its observed value represents an average.
Sources:
Orange Book, 2nd ed., p. 168 (http://media.iupac.org/publications/analytical_compendium/)
PAC, 1986, 58, 1737. 'Quantities and units in clinical chemistry: Nebulizer and flame properties in flame emission and absorption spectrometry (Recommendations 1986)' on page 1741 (https://doi.org/10.1351/pac198658121737)
|
2021-10-19 12:50:47
|
{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8105869293212891, "perplexity": 4509.276434391343}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585265.67/warc/CC-MAIN-20211019105138-20211019135138-00699.warc.gz"}
|
http://math.stackexchange.com/questions/228492/how-to-use-polyfit-function-in-matlab?answertab=votes
|
# How to use polyfit() function in Matlab?
First of all i am new to Matlab, and not sure whether to use polyfit or something else for my problem.
I plotted a graph with the following matlab code:
t=transpose(linspace(-1,1,50))
y=1./(1+25*t.^2)
n=10
A=fliplr(vander(t))
A=A(:,1:n)
x=A\y
u=linspace(-1,1,1000)
g=x(n)*ones(1,1000)
for \$i=(n-1):-1:1
g=g.*u+x(i) end
plot(u,g,'-',t,y,'o')
Now my question is: plot the figure with polynomial fit of polynomial degree 2.
How should i do this?
Thanks
-
for the coefficients of the polyfit function how I can change the format of this coefficients to the long? Plz help me as soon. – user83693 Jun 24 '13 at 14:20
So you want to fit y as a function of t, right? Use
p = polyfit(t,y,2);
fit = polyval(p,t);
plot(u,g,'-',t,y,'o',t,fit)
The first line is the built-in polynomial fit function. The number 2 is the degree which you specify and it returns the coefficients of the polynomial in p. Note degree 2 means three coefficients. The second line then evaluates the polynomial using the coefficients in p. And then the third, plots them all together.
-
thank you, since you are familiar with math lab, could you answer this question too? Thanks math.stackexchange.com/questions/228552/… – ASROMA Nov 4 '12 at 0:43
|
2014-07-25 04:14:06
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7450759410858154, "perplexity": 1034.5618252655695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1405997892806.35/warc/CC-MAIN-20140722025812-00121-ip-10-33-131-23.ec2.internal.warc.gz"}
|
http://www.maa.org/programs/faculty-and-departments/classroom-capsules-and-notes/the-matrix-of-a-rotation?device=desktop
|
# The Matrix of a Rotation
by Roger C. Alperin (San Jose State University)
College Mathematics Journal
May, 1989
Subject classification(s): Algebra and Number Theory | Linear Algebra | Eigenvalues and Eigenvectors | Linear Transformations | Vectors in R3 | Geometry and Topology | Plane Geometry | Angles | Lines and Planes
Applicable Course(s): 3.8 Linear/Matrix Algebra | 4.14 Vector Analysis
Given a unit vector $p$ in $\mathbf{R}^3$ and an angle $\theta$, what is the matrix of the rotation of $\mathbf{R}^3$ about $p$ through an angle of $\theta$ in terms of the standard basis? The author obtains an explicit matrix without changing bases.
A pdf copy of the article can be viewed by clicking below. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page.
|
2017-02-19 17:05:59
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4438573122024536, "perplexity": 907.3213467576905}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-09/segments/1487501170186.50/warc/CC-MAIN-20170219104610-00032-ip-10-171-10-108.ec2.internal.warc.gz"}
|
https://mitpress.mit.edu/index.php?q=books/understanding-inflation-and-implications-monetary-policy
|
Hardcover | $47.00 Text | £32.95 | ISBN: 9780262013635 | 520 pp. | 6 x 9 in | | September 2009 Ebook |$33.00 Short | ISBN: 9780262259583 | 520 pp. | | September 2009
## Understanding Inflation and the Implications for Monetary Policy
A Phillips Curve Retrospective
Foreword by Paul A. Samuelson
## Overview
In 1958, economist A. W. Phillips published an article describing what he observed to be the inverse relationship between inflation and unemployment; subsequently, the “Phillips curve” became a central concept in macroeconomic analysis and policymaking. But today’s Phillips curve is not the same as the original one from fifty years ago; the economy, our understanding of price setting behavior, the determinants of inflation, and the role of monetary policy have evolved significantly since then. In this book, some of the top economists working today reexamine the theoretical and empirical validity of the Phillips curve in its more recent specifications. The contributors consider such questions as what economists have learned about price and wage setting and inflation expectations that would improve the way we use and formulate the Phillips curve, what the Phillips curve approach can teach us about inflation dynamics, and how these lessons can be applied to improving the conduct of monetary policy. ContributorsLawrence Ball, Ben Bernanke, Oliver Blanchard, V. V. Chari, William T. Dickens, Stanley Fischer, Jeff Fuhrer, Jordi Gali, Michael T. Kiley, Robert G. King, Donald L. Kohn, Yolanda K. Kodrzycki, Jane Sneddon Little, Bartisz Mackowiak, N. Gregory Mankiw, Virgiliu Midrigan, Giovanni P. Olivei, Athanasios Orphanides, Adrian R. Pagan, Christopher A. Pissarides, Lucrezia Reichlin, Paul A. Samuelson, Christopher A. Sims, Frank R. Smets, Robert M. Solow, Jürgen Stark, James H. Stock, Lars E. O. Svensson, John B. Taylor, Mark W. Watson
|
2015-10-06 08:37:04
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43285253643989563, "perplexity": 12077.510093261488}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-40/segments/1443736678574.32/warc/CC-MAIN-20151001215758-00181-ip-10-137-6-227.ec2.internal.warc.gz"}
|
https://docs.mfem.org/html/classmfem_1_1VectorCoefficient.html
|
MFEM v4.4.0 Finite element discretization library
mfem::VectorCoefficient Class Referenceabstract
Base class for vector Coefficients that optionally depend on time and space. More...
#include <coefficient.hpp>
Inheritance diagram for mfem::VectorCoefficient:
[legend]
## Public Member Functions
VectorCoefficient (int vd)
Initialize the VectorCoefficient with vector dimension vd. More...
virtual void SetTime (double t)
Set the time for time dependent coefficients. More...
double GetTime ()
Get the time for time dependent coefficients. More...
int GetVDim ()
Returns dimension of the vector. More...
virtual void Eval (Vector &V, ElementTransformation &T, const IntegrationPoint &ip)=0
Evaluate the vector coefficient in the element described by T at the point ip, storing the result in V. More...
virtual void Eval (DenseMatrix &M, ElementTransformation &T, const IntegrationRule &ir)
Evaluate the vector coefficient in the element described by T at all points of ir, storing the result in M. More...
virtual ~VectorCoefficient ()
int vdim
double time
## Detailed Description
Base class for vector Coefficients that optionally depend on time and space.
Definition at line 432 of file coefficient.hpp.
## Constructor & Destructor Documentation
mfem::VectorCoefficient::VectorCoefficient ( int vd )
inline
Initialize the VectorCoefficient with vector dimension vd.
Definition at line 440 of file coefficient.hpp.
virtual mfem::VectorCoefficient::~VectorCoefficient ( )
inlinevirtual
Definition at line 474 of file coefficient.hpp.
## Member Function Documentation
virtual void mfem::VectorCoefficient::Eval ( Vector & V, ElementTransformation & T, const IntegrationPoint & ip )
pure virtual
Evaluate the vector coefficient in the element described by T at the point ip, storing the result in V.
Note
When this method is called, the caller must make sure that the IntegrationPoint associated with T is the same as ip. This can be achieved by calling T.SetIntPoint(&ip).
void mfem::VectorCoefficient::Eval ( DenseMatrix & M, ElementTransformation & T, const IntegrationRule & ir )
virtual
Evaluate the vector coefficient in the element described by T at all points of ir, storing the result in M.
The dimensions of M are GetVDim() by ir.GetNPoints() and they must be set by the implementation of this method.
The general implementation provided by the base class (using the Eval method for one IntegrationPoint at a time) can be overloaded for more efficient implementation.
Note
The IntegrationPoint associated with T is not used, and this method will generally modify this IntegrationPoint associated with T.
Definition at line 192 of file coefficient.cpp.
double mfem::VectorCoefficient::GetTime ( )
inline
Get the time for time dependent coefficients.
Definition at line 446 of file coefficient.hpp.
int mfem::VectorCoefficient::GetVDim ( )
inline
Returns dimension of the vector.
Definition at line 449 of file coefficient.hpp.
virtual void mfem::VectorCoefficient::SetTime ( double t )
inlinevirtual
Set the time for time dependent coefficients.
Definition at line 443 of file coefficient.hpp.
## Member Data Documentation
double mfem::VectorCoefficient::time
protected
Definition at line 436 of file coefficient.hpp.
int mfem::VectorCoefficient::vdim
protected
Definition at line 435 of file coefficient.hpp.
The documentation for this class was generated from the following files:
|
2022-07-05 22:54:51
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3466675579547882, "perplexity": 9252.884055138322}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656104628307.87/warc/CC-MAIN-20220705205356-20220705235356-00501.warc.gz"}
|
http://mathhelpforum.com/algebra/72484-binomial-expansion.html
|
# Math Help - Binomial expansion
1. ## Binomial expansion
Is this the correct expansion $(2 - x)^{-1} = \frac{1}{2} + x + \frac{x^2}{4}$
2. No.
$(2-x)^{-1}=\frac{1}{2-x}$
3. I mean when you expand it up to x^2 using binomial expansion
|
2015-03-28 14:58:28
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.769931972026825, "perplexity": 4339.120611687419}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-14/segments/1427131297587.67/warc/CC-MAIN-20150323172137-00068-ip-10-168-14-71.ec2.internal.warc.gz"}
|
http://china-food.net/darren-purchese-xcckjo/article.php?page=unitary-matrix-formula-20a79f
|
5.1 Unitary and Orthogonal Matrices Definitions Unitaryandorth The theory and tools for quickly determining these “change of basis formulas” will be developed in these notes. Keywords : unitarily invariant norm, matrix function, Fenchel conjugacy, subdi erential, matrix opti-mization, extreme point, exposed point. Fill an matrix with complex Gaussian IID values, call it . Smooth manifolds 55 2. The real Cli ord algebras 45 6. Introduction 1 2. A Householder matrix is a rank-perturbation of the identity matrix and so all but one of its eigenvalues are .The eigensystem can be fully described as follows. For example, for size 3, to get this matrix: [0,0,1] [0,1,0] [1,0,0] I want to use it in array formulas, so I would prefer not to have the matrix written somewhere. What does this mean? Overview. Thus, the eigenvalues of a unitary matrix are unimodular, that is, they have norm 1, and hence can be written as $$e^{i\alpha}$$ for some $$\alpha\text{.}$$. Recursive formula is vectorizable. 4. Both formulas are discussed and possible applications are outlined. Local diagonal unitary invariant matrices18 7. I hope the precision is OK as we deal with unitary matrix, and there is no issue from substraction of 2 large quantities. Main operations Trace. Set . We analyze properties of a map f sending a unitary matrix U of size N into a doubly stochastic matrix B=f(U) defined by Bi,j=|Ui,j|2. Let (A,Λ) be a formring such that A is quasi-finite R-algebra (i.e., a direct limit of module finite algebras) with identity.We consider the hyperbolic Bak’s unitary groups GU(2n, A, Λ), n ≥ 3. Is possible in excel, with a cell formula, to generate a unitary matrix rotated 90 degrees? orthogonal (),symmetric (),involutory (that is, is a square root of the identity matrix),where the last property follows from the first two. Note that the unitary matrix can ge generated with the formula … 3x3 Unitary to Magic Matrix Transformations Philip Gibbs We prove that any 3x3 unitary matrix can be transformed to a magic matrix by multiplying its rows and columns by phase factors. Perform a QR decomposition of the matrix , define to be the diagonal of such that and otherwise. An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix.Although we consider only real matrices here, the definition can be used for matrices with entries from any field.However, orthogonal matrices arise naturally from dot products, and for matrices of complex numbers that leads instead to the unitary requirement. A left multiplication with a unitary diagonal matrix can rotate b nn such that it becomes real and hence we have a constructive procedure for obtaining the unitary matrix U. More generally, with each Σ-unitary matrix H k we can associate a corresponding unitary transformation. Theorem 8. pp.403-412. Note that the proof that these two formulas are each other’s inverses did not require A to be skew-symmetric or O to be orthogonal. The complex analogue of an orthogonal matrix is a unitary matrix. LAPACK doesn't have a specialized routine for computing the eigenvalues of a unitary matrix, so you'd have to use a general-purpose eigenvalue routine for complex non-hermitian matrices. We show that the factors in the recursive formula may be introduced in any desired order. So it would be helpful to have formulas for converting the components of a vector with respect to one basis into the corresponding components of the vector (or matrix of the operator) with respect to the other basis. The method is used to study the invariant phases of unitary matrices. Proof. If U ∈M n is unitary, then it is diagonalizable. A unitary matrix means that if you multiple the matrix by its (complex) conjugate transpose you end up with an identity matrix. Algorithm is proposed to convert arbitrary unitary matrix to a sequence of X gates and fully controlled Ry, Rz and R1 gates. BASICS 161 Theorem 4.1.3. Unitary Matrix Integrals, Primitive Factorizations, and Jucys- Murphy Elements. To formulate the stability condition of the unitary system, approximated by the equivalent transform function matrix , the following theorems are given. A unitary operator is one that preserves the inner product. *has extra registration where is now the unitary of interest randomly distributed according to the Haar measure on . Hence, the Cayley transform is defined for all matrices such that -1 is not an eigenvalue of O. Our Thouless formula relates the potential of the density of states measure, see, e.g., [SaT], [StT] for these notions, with the Lyapunov exponent. To prove this we need to revisit the proof of Theorem 3.5.2. Contents 1. The unitary matrix constraint considered in this paper determines a parameter space which is the Lie group of n ... we choose to approximate it by using the right multiplication by W k H W k + 1 leading to the approximate Polak–Ribièrre formula: (10) γ k = 〈 G k + 1-G k, G k + 1 〉 I 〈 G k, G k 〉 I If H ˜ k = G ˜ k, then by formulae and become equal. 4.1. It is easily verified that is. UNITARY METHOD FORMULA. condition for a unitary matrix to b e iso lated and discussing the unisto chasticit y problem. Unitary method aims at determining values in relation to a single unit. Just as for Hermitian matrices, eigenvectors of unitary matrices corresponding to different eigenvalues must be orthogonal. Here is a Python function that can do this for you if the above steps don’t make any sense. Recursive formula is vectorizable. 22nd International Conference on Formal Power Series and Algebraic Combina-torics (FPSAC 2010), 2010, San Francisco, United States. Unitary Method Definition and Example : Definition : Unitary-method is all about finding value to a single unit. Quaternionic matrix groups 44 5. A magic matrix is defined as one for which the sum of the elements in any row or column add to the same value. As before, select thefirst vector to be a normalized eigenvector u1 pertaining to λ1.Now choose the remaining vectors to be orthonormal to u1.This makes the matrix P1 with all these vectors as columns a unitary matrix. The spinor groups 49 7. In the first type the unitary matrix is where is the spectral parameter. Lets break it down into individual steps to make it clear. Unitary Equivalence relation The normal case Associated Krylov spaces Eigenvalues and singular values Conclusions Scalar product spaces For A normal we have a factorization Jan 22, 2018 - Unitary Matrices Video Lecture From Chapter Rank of Matrix in Engineering Mathematics 1 for First Year Degree Engineering Students. The formula given below can be used to find the value of one unit. Some examples of Lie groups 59 5. Theorem 1. The state-space representation of the enhanced structure of transfer function matrix ( 56 ) in the form of a closed-loop system is where is the performance evaluation signal, and the system constraint is Leave extra cells empty to enter non-square matrices. A Householder matrix is an orthogonal matrix of the form. Graphical integration formula | real signs15 6. 1991 Mathematics Subject Classi cation: Primary: 15A60 49J52, Secondary: 90C25 65F35 1. Matrix groups as Lie groups 55 1. This is slower than using a routine for the eigenvalues of a complex hermitian matrix, although I'm surprised that you're seeing a factor of 20 difference in run times. Combinatorial prerequisites3 3. via an elegant Fenchel conjugacy formula. With help of this calculator you can: find the matrix determinant, the rank, raise the matrix to a power, find the sum and the multiplication of matrices, calculate the inverse matrix. Programming competitions and contests, programming community. Two trace formulas for the spectra of arbitrary Hermitian matrices are derived by transforming the given Hermitian matrix H to a unitary analogue. Unitary Matrix Integrals, Primitive Factorizations, and Jucys-Murphy Elements Sho Matsumoto, Jonathan Novak To cite this version: Sho Matsumoto, Jonathan Novak. also constructed in [GT] by suitable truncations of the Hessenberg matrix consid-ered. Just type matrix elements and click the button. (D. 13 / 34 On tridiagonal matrices unitary equivalent, with normal matrices. The centres of spinor groups 52 8. My tests show that the cpu time is about the same for n=4, and faster when n smaller and vice versa. This result is relevant to recent observations on particle mixing matrices. Finite subgroups of spinor groups 53 Chapter 4. Abstract. Lie groups 58 4. Keywords: trace formula, spectral graph theory, hermitian matrix, random-matrix theory, periodic-orbit theory (Some figures may appear in colour only in the online journal) 1. Unitary-method can be used to calculate cost, measurements like liters and time. Let K be an (s + q) × (s + q) matrix satisfying K H Σ s, q K = Σ s, q. When redoing the proof above by using the transformation A V = VHAV, we can see that all statements remain valid and hence the matrix A V will also be normal complex symmetric. Graphical integration formula | complex phases10 5. (Recall that this condition is necessary to insure that O + I is invertible. A Variance Formula Related to a Quantum Conductance Problem Tiefeng Jiang 1 Abstract Let t be a block of an Haar-invariant orthogonal (fl = 1), unitary (fl = 2) or symplectic (fl = 4) matrix from the classical compact groups O(n); U(n) or Sp(n); respectively.We obtain a close form for Var(tr(t⁄t)).The case for fl = 2 is related to a quantum conductance problem, between matrix algebras by independent diagonal unitary matrices, showcasing another application of our method. If $U,V \in \mathbb{C}^{n \times n}$ are unitary matrices, then $VV^*=I_n$ and [math]UU^*=I_n. Section 5 contains deriv ation of the formula for the defect of the The means by which we do so is the exchange operator described in the following theorem. 0.0.3 Generalizations. The case of four-by-four unitary matrices is investigated in detail. Tangent spaces and derivatives 55 3. Diagrammatic notation for tensors8 4. The determinant of square matrices over a commutative ring R can still be defined using the Leibniz formula; such a matrix is invertible if and only if its determinant is invertible in R, generalising the situation over a field F, where every nonzero element is invertible. Lets start with a matrix of the following form: \begin{aligned} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \end{aligned} Complex Conjugate of Real Numbers. This approach also allows such results to be extended to more general unitarily invariant matrix functions. first type the unitary matrix is ei λ ... scattering matrix. Codeforces.
## unitary matrix formula
Social Work Research Paper Pdf, Service Management Interview Questions And Answers, Sumac Pronunciation Arabic, Who Do You Love Book, Apartments Downtown Grand Rapids, Mi, Suppressor Trust Vs Individual, Automated Packaging Systems Careers, Turn Caramel Into Sauce, Japan Stereo Brands,
|
2022-01-20 11:06:55
|
{"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8943188786506653, "perplexity": 683.3660677331894}, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301737.47/warc/CC-MAIN-20220120100127-20220120130127-00583.warc.gz"}
|
http://tug.org/pipermail/macostex-archives/2004-September/009841.html
|
# [OS X TeX] pdfsync & bibliography question
Matthew Inglis m.j.inglis at warwick.ac.uk
Wed Sep 8 14:51:22 CEST 2004
Hello all,
I've got a couple of questions I was hoping people might be able to
help me with.
(i) I'm using TeXShop 1.35 on OS10.3, and when I include the line
"\usepackage{pdfsync}" my document compiles correctly, but the console
gives 57 separate versions of the message below. This problem does not
occur when I comment out the pdfsync line.
Underfull \hbox (badness 10000) has occurred while \output is active
[][]
[57] (./contents.aux) )
I suspect that this has something to do with the bibliography, as I
have 57 references. I'm using the harvard package with the following
options:
\bibliographystyle{dcu}
\citationstyle{dcu}
\renewcommand{\harvardand}{\&}
Has anyone got any suggestions as to why pdfsync would produce these
(time consuming) error lines?
(ii) On a different note. It'd be nice to be able to reference Jonathan
St. B. T. Evans in the correct manner: Evans, J. St. B. T. (2004). Does
anyone know how to stop BibTeX changing the "St" into a "S"?
Matthew.
--------------------- Info ---------------------
Mac-TeX Website: http://www.esm.psu.edu/mac-tex/
& FAQ: http://latex.yauh.de/faq/
TeX FAQ: http://www.tex.ac.uk/faq
List Post: <mailto:MacOSX-TeX at email.esm.psu.edu>
|
2016-04-30 16:55:05
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9239382743835449, "perplexity": 11818.302042866753}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-18/segments/1461860112228.39/warc/CC-MAIN-20160428161512-00101-ip-10-239-7-51.ec2.internal.warc.gz"}
|
https://en.wikibooks.org/wiki/General_Topology/Countability,_density
|
# General Topology/Countability, density
Definition (dense):
Let ${\displaystyle X}$ be a topological space and let ${\displaystyle A\subseteq X}$ be a subset. ${\displaystyle A}$ is called dense if and only if ${\displaystyle {\overline {A}}=X}$.
Definition (first-countable):
Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called first-countable iff for all ${\displaystyle x\in X}$ the neighbourhood filter ${\displaystyle N(x)}$ has a countable basis.
Definition (second-countable):
Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called second-countable iff the topology of ${\displaystyle X}$ has a countable basis.
Since subsets of countable sets are countable and the open neighbourhoods generate ${\displaystyle N(x)}$, second-countability implies first-countability.
Definition (separable space):
Let ${\displaystyle X}$ be a topological space. ${\displaystyle X}$ is called separable if and only if there exists a countable set ${\displaystyle S\subseteq X}$ which is dense in ${\displaystyle X}$.
Proposition (second-countable spaces are separable):
Let ${\displaystyle X}$ be a second-countable space. Then ${\displaystyle X}$ is separable.
(On the condition of the axiom of countable choice.)
Proof: Let ${\displaystyle (U_{n})_{n\in \mathbb {N} }}$ be a basis of the topology of ${\displaystyle X}$ and choose ${\displaystyle x_{n}\in U_{n}}$. Then ${\displaystyle S:=\{x_{n}|n\in \mathbb {N} \}}$ is countable and dense. ${\displaystyle \Box }$
Proposition (subspace of second-countable space is second-countable):
Let ${\displaystyle X}$ be a second-countable space, and ${\displaystyle S\subseteq X}$ a subset. Turn ${\displaystyle S}$ into a topological space using the subspace topology. ${\displaystyle S}$ is then a second-countable space.
Proof: Any countable basis ${\displaystyle (U_{n})_{n\in \mathbb {N} }}$ of the topology of ${\displaystyle X}$ induces a countable basis ${\displaystyle (S\cap U_{n})_{n\in \mathbb {N} }}$ of the subspace topology on ${\displaystyle S}$. ${\displaystyle \Box }$
Proposition (continuous function into Hausdorff space is uniquely determined by dense subspace):
Let ${\displaystyle X,Y}$ be topological spaces, where ${\displaystyle Y}$ is Hausdorff. Let ${\displaystyle A\subseteq X}$ be a dense subspace, and suppose ${\displaystyle f:A\to Y}$ is continuous. Whenever ${\displaystyle F,G:X\to Y}$ are continuous functions such that ${\displaystyle F|_{A}=G|_{A}}$, then ${\displaystyle F=G}$.
Proof: Let ${\displaystyle x\in X}$ be arbitrary, and let ${\displaystyle V\subseteq Y}$ be any neighbourhood of ${\displaystyle G(x)}$. By continuity of ${\displaystyle G/math>wemayfindaneighbourhood[itex]U}$ of ${\displaystyle x}$ that is mapped completely into ${\displaystyle V}$. Analogously, whenever ${\displaystyle V'}$ is a neighbourhood of ${\displaystyle F(x)}$, we find a neighbourhood ${\displaystyle U'}$ mapping completely into ${\displaystyle V'}$. Then ${\displaystyle U\cap U'}$ is mapped completely into ${\displaystyle V\cap V'}$, so that ${\displaystyle F(x),G(x)\subseteq V\cap V'}$ for any open neighbourhoods ${\displaystyle V}$ of ${\displaystyle G(x)}$ and ${\displaystyle V'}$ of ${\displaystyle F(x)}$. If ${\displaystyle F(x)\neq G(x)}$, then ${\displaystyle V\cap V'=\emptyset }$ for suitable ${\displaystyle V,V'}$ as above by the Hausdorff condition, a contradiction to ${\displaystyle G(x)\in V\cap V'}$. Hence, ${\displaystyle F(x)=G(x)}$. Since ${\displaystyle x}$ was arbitrary, we conclude. ${\displaystyle \Box }$
|
2021-06-18 06:58:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 63, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9815924167633057, "perplexity": 114.6328400473141}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487635724.52/warc/CC-MAIN-20210618043356-20210618073356-00156.warc.gz"}
|
https://physics.stackexchange.com/questions/694947/how-to-build-numerically-a-dispersion-relation-of-a-finite-phononic-system
|
# How to build numerically a dispersion relation of a finite phononic system?
I have doubts about how to build the plot of dispersion relation for a general finite phononic system, in particular spring-mass systems. Eqs of any of these systems can be written in matrix form as $$D\vec{u}=\omega^2 \vec{u} \tag{1}$$ For the case of "infinite" system, usually one set the periodic boundary condition (PBC) and under the Bloch expansion $$u_i=u_0 \mathrm{e}^{ikn}$$ in (1), one can solve analytically $$\omega(k)$$ for a single primitive cell. As example, in the 1d diatomic chain one can obtain $$\omega(k)$$ by this way
In this case, considering $$N$$ masses and the PBC, this quantizes the values of $$k$$, yielding $$N$$ points in $$\omega(k)$$, but taking $$N\to\infty$$, one can plot a continuous line in $$\omega(k)$$ (similar for monoatomic chain).
However in a finite system, under specific boundary conditions in $$D$$, if I impose the Bloch expansion in (1), the lack of translational invariance of $$D$$ doesn't allow us to reduce (1) to the dynamics of a single cell. I'm not interested in analytically solving $$\omega(k)$$ in these system, but I would like to be able to plot the N points of $$\omega(k)$$. I think that I'm wrong imposing $$u_0 \mathrm{e}^{ikn}$$ here, but then I don't know how to make appear the wave vector $$k$$ in (1) to obtain the $$\omega(k)$$ relations. Is there a generic numerical way to obtain the points of $$\omega(k)$$ in these systems?
(e.g. the diatomic chain with one end free and the other fixed)
You can still use periodic boundary conditions for a finite chain if the ends are fixed (as I will demonstrate below), and so the dispersion relation is unchanged in the sense that you just replace the continuous variable $$k$$ with its discrete version.
To treat a finite chain first impose periodicity as you did before, this means that the two ends of the chain $$x_0$$ and $$x_N$$ will be regarded as the same point but this is okay since we fix both ends by demanding that $$x_0(t) = x_N(t) = 0$$ If we look at the general solution $$x_n(t) = \sum_{k=-\pi/a}^{\pi/a}e^{ikna}(A_ke^{i\Omega_k t}+B_ke^{-i\Omega_k t})$$where $$\Omega_k$$ is the mode frequency, the ends are fixed if we set $$A_k=-A_{-k}$$, $$~B_k=-B_{-k}$$, and $$A_0=B_0=0$$.
|
2023-03-27 00:29:55
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8840624094009399, "perplexity": 181.83476733221877}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296946584.94/warc/CC-MAIN-20230326235016-20230327025016-00346.warc.gz"}
|
https://kokecacao.me/page/Course/S21/80-221/Lecture_014.md
|
# Lecture 014
## Reading: Philosophy of Social Science
### Intro
Reductive:
• represent scientific phenomena as products of more fundamental process
• study fundamental phenomena
Theoretical:
• individualists: theories can be derived from theories of psychology
• holists: theories are logically independent of lower-level theories.
Ontological:
• individualists: only human agents and their properties exist
• holists: social entities and properties also exist.
Explanatory:
• individualists: explanations in the social sciences must make reference to individual actions
• holists: also accept social-level explanations.
### Definition and Theoretical Reduction
How do social-level theories relate to psychological-level theories?
reductionism: each term of social-level theory would be defi ned in terms of individual-level theory
• problem of the remainder: Each new attempt at a defi nition adds a new social-level term which has to be defined -> mission impossible Why not define everything?
• the problem of multiple realizability: many social-level terms apply to an open-ended variety of individual arrangements
• (What if someone organizes a church in a way we hadn’t foreseen? Does it fail to be a church? Or must we change our defi nition with every new example?) while the use of church for holist is flawed.
• It would be impossible to explain individual action without relational terms like “mother of” we can use these terms to simplify argument, but they need to be defined
• criteria: exactly which explanation counts as "individual level" or "social level"
type: in programming, class token: in programming, instance
### Supervenience
Supervenience: The dependency of higher-level properties on lower-level properties
Non-reductive individualism: The combination of supervenience and multiple realizability permits a sophisticated holist position which admits that there are no social objects, but insists that there are non-reducible social properties.
Social Science: explanatory Can't be reducible:
• action depends on one's status and circumstance that can't be explained by action
• concepts involving the notions of status and role cannot themselves be reduced to a conjunction of statements in which these or other societal concepts do not appear.
position:
• (a) that in understanding or explaining an individual's actions we must often refer to facts concerning the organization of the society in which he lives (acceptable by most philosopher)
• (b) that our statements concerning these societal facts are not reducible to a conjunction of statements concerning the actions of individuals (acceptable by most philosopher)
Why not reducible:
• the term president does not change every 4 years really? I argue it does change. The term Our president can mean different things every 4 year
It is necessary to translate into partial psychological concepts: to verify
## Question
1. problem of multiple realizability The holist's argument is not compelling. The holist holds that since a social-level explanation can be reduced to different sets of individual-level explanation based on different contextual factors, the attempts reduce the social-level explanation will complicate the explanation (page 125). This argument is supposed to undermine an individualist's methodology because a complicated explanation has less explanatory value than a simple social-level explanation (page 125). In fact, by making this claim, the holist assumes that a type (in other words: a social-level terminology) can have different realizations (in other words: definitions in terms of individual-level terminology). With this assumption, since there exists some social-level terminology that is ambiguous, a social-level explanation with ambiguous social-level terminologies will remove all explanatory value from the social-level explanation. On the contrary, to solve the problem above, the holist would assume that there exists one single universal definition for any social-level terminology, making all social-level terminology well-defined and unambiguous indifferent to their context. Then, since social-level terminologies are context-independent, all social-level explanations can be reduced to one single unambiguous individual-level explanation, removing the problem of multiple realizability. In both cases, the holist's self-destructive argument would not hold.
2. social phenomena "supervene" on thoughts and actions of all the individuals?
3. The dependency of higher-level properties on lower-level properties has been called “supervenience”
4. Yes. The term "Our President" can mean different things every 4 year: the meaning depends on all historical and current president.
5. Explain handshake.
6. prove by example
7. both person believes that shaking one's hand means the hand-shaker has no negative attitude on the hand-shakee. Since they both desire the other to know this fact, the shake each other's hand to transport this idea.
8. partial reduction?
9. partial reduction to "verify"?
10. how can you verify if you think any reduction have multiple realizability? One cannot "verify" something in social-level only by example of individual-level
Table of Content
|
2022-12-02 03:36:17
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4963914752006531, "perplexity": 4495.814029996826}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710890.97/warc/CC-MAIN-20221202014312-20221202044312-00615.warc.gz"}
|
https://dimag.ibs.re.kr/events/tag/meikehatzel/list/?eventDisplay=past
|
November 2020
Meike Hatzel, Constant congestion bramble
Zoom ID: 869 4632 6610 (ibsdimag)
In this talk I will present a small result we achieved during a workshop in February this year. My coauthors on this are Marcin Pilipczuk, Paweł Komosa and Manuel Sorge. A bramble in an undirected graph $G$ is a family of connected subgraphs of $G$ such that for every two subgraphs $H_1$ and $H_2$ in the bramble either $V(H_1) \cap February 2023 Meike Hatzel, Fixed-Parameter Tractability of Directed Multicut with Three Terminal Pairs Parametrised by the Size of the Cutset: Twin-Width Meets Flow-Augmentation Room B332 IBS (기초과학연구원) We show fixed-parameter tractability of the Directed Multicut problem with three terminal pairs (with a randomized algorithm). This problem, given a directed graph$G$, pairs of vertices (called terminals)$(s_1,t_1)$,$(s_2,t_2)$, and$(s_3,t_3)$, and an integer$k$, asks to find a set of at most$k$non-terminal vertices in$G$that intersect all$s_1t_1\$-paths, all
기초과학연구원 수리및계산과학연구단 이산수학그룹
대전 유성구 엑스포로 55 (우) 34126
IBS Discrete Mathematics Group (DIMAG)
Institute for Basic Science (IBS)
55 Expo-ro Yuseong-gu Daejeon 34126 South Korea
E-mail: dimag@ibs.re.kr, Fax: +82-42-878-9209
|
2023-03-28 04:46:04
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.34759101271629333, "perplexity": 14127.626882762008}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00673.warc.gz"}
|
http://web.emn.fr/x-info/sdemasse/gccat/sec3.7.101.3.html
|
### 3.7.101.3. Flow models for $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$, $\mathrm{𝚜𝚊𝚖𝚎}$
Figure 3.7.24 presents flow models for the $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$ and the $\mathrm{𝚜𝚊𝚖𝚎}$ constraints. Blue arcs represent feasible flows respectively corresponding to the solutions $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$$\left(〈{x}_{1}=2,{x}_{2}=4,{x}_{3}=6〉,〈{y}_{1}=2,{y}_{2}=4〉\right)$ and $\mathrm{𝚜𝚊𝚖𝚎}$$\left(〈{x}_{1}=2,{x}_{2}=4,{x}_{3}=5〉,〈{y}_{1}=2,{y}_{2}=4,{y}_{3}=5〉\right)$, while pink arcs correspond to arcs that cannot carry any flow if the constraint has a solution. Within the context of the $\mathrm{𝚜𝚊𝚖𝚎}$ constraint, the assignment ${x}_{1}=1$ is forbidden since $1\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$. Consequently ${x}_{1}=2$ and, since ${y}_{1}$ is the only variable of $\left\{{y}_{1},{y}_{2},{y}_{3}\right\}$ that can be assigned value 2, the assignment ${y}_{1}=3$ is forbidden. Now since $3\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$ the assignment ${x}_{2}=3$ is also forbidden. Finally ${x}_{3}=6$ is forbidden since $6\notin \mathrm{𝑑𝑜𝑚}\left({y}_{1}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{2}\right)\cup \mathrm{𝑑𝑜𝑚}\left({y}_{3}\right)$.
##### Table 3.7.24. Domains of the variables for the $\mathrm{𝚞𝚜𝚎𝚍}_\mathrm{𝚋𝚢}$ constraint of Figure 3.7.24.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({y}_{i}\right)$
1$\left\{1,2\right\}$1$\left\{2,3\right\}$
2$\left\{3,4\right\}$2$\left\{4,5\right\}$
3$\left\{4,5,6\right\}$
##### Table 3.7.24. Domains of the variables for the $\mathrm{𝚜𝚊𝚖𝚎}$ constraint of Figure 3.7.24.
$i$$\mathrm{𝑑𝑜𝑚}\left({x}_{i}\right)$$i$$\mathrm{𝑑𝑜𝑚}\left({y}_{i}\right)$
1$\left\{1,2\right\}$1$\left\{2,3\right\}$
2$\left\{3,4\right\}$2$\left\{4,5\right\}$
3$\left\{4,5,6\right\}$3$\left\{4,5\right\}$
|
2017-09-25 18:52:23
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 49, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.882866621017456, "perplexity": 499.593114244217}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818693240.90/warc/CC-MAIN-20170925182814-20170925202814-00298.warc.gz"}
|
https://www.gamedev.net/forums/topic/510936-missing-pixel/
|
# missing pixel
This topic is 3424 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi, I have made a little toolkit to create basic 2d GUIs to allow me to control my test applications at runtime. I'm facing a strange (at least for me) problem when rendering UI elements. When I draw borders with GL_LINE_LOOP, the bottom-left pixel is missing. The same way, when I draw backgrounds with GL_TRIANGLE_STRIP, left row and botton line are missing. Here is a screenshot with zooms on the relevant parts (sorry for the poor colors) Does anybody have an idea of what I'm doing wrong ? Thanks for your help [Edited by - Sneftel on October 9, 2008 12:37:50 PM]
##### Share on other sites
Quote:
Original post by jeanbonoursDoes anybody have an idea of what I'm doing wrong?
To be honest, probably nothing. I see these artefacts a lot with lines - it may be a bug in the rasteriser, and I get the impression that line drawing doesn't get much TLC in recent driver versions.
But from a purely aesthetic viewpoint, it probably doesn't matter, and even adds a little extra something to an otherwise boxy GUI ;)
*it helps to give your post a title, more people are likely to look at it. You can edit your post to add one.
##### Share on other sites
Would you happen to have an nVidia graphics card? Try jittering your drawing around, like by .1 pixels via glTranslate. Here is a link to another thread where a person was experiencing a similar problem.
##### Share on other sites
I used to have this problem, it turned out that my scaled 2D float coordinates needed type casting to integers. This worked fine with a GL_LINE_STRIP on multiple boxes.
Ex.
float scale = (float)screenCurrent.height / 1050.0f; //designed at my max screen resfloat x = 100 * scale; //relative x/y positionfloat y = 0 * scale; //(for OP's clarity)...typedef short s16x = (s16)x; y = (s16)y;... glVertex2f( x, y );...
edit: I just tested this at various screen resolutions. Additionally you should add 0.5f to the width/height of your box, then it will always be 100%.
[Edited by - jezham on October 9, 2008 12:46:11 PM]
##### Share on other sites
yes, I have an nVidia gfx card.
I've read the very interesting thread pointed at by MikeTacular. What I still don't understand is why only ONE pixel (lower/left) is missing when rendering GL_LINE_LOOP ... It sounds more "natural" to me that lower and left lines should be missing ...
(But anyway, I have now a more clear view of the reason why rendering can not be perfect without some manipulation on coords and sizes.)
I've spent some time trying to adjust 2D coords (using glTranslate3f(0.1, 0.1, 0), and/or adding some jitter (0.5) on coords of corners) but it has not fixed the problem, only changed it : the lower/RIGHT pixel of the border was missing, as well as lower line and RIGHT row of the background.
To jezham, in my 2D gui toolkit, all coords and sizes are stored in short or ushort. That is to say, all coordinates sent to glVertexXXX() are integer values, not floating. So I don't think I need to cast them ?
##### Share on other sites
Perhaps my ATI card acts differently? It just seems everyone is over-fixing this!
As you already have ints then just the additional half-pix to the width/height of your box is required. This of course means that you must supply glVertex2f() and not glVertex2i()
Here is a simple box which I've tested.
define half (float)0.5f glBegin( GL_LINE_LOOP ); glVertex2f( 100, 0 ); glVertex2f( 110+half, 0 ); glVertex2f( 110+half, 10+half ); glVertex2f( 100, 10+half );// glVertex2f( 100, 0 ); glEnd();
The only other thing which I can think of is the way you set ortho mode (mine follows).
glOrtho( 0, screenWidth, 0, screenHeight, -1, 1 );
HTH
edit: changed from GL_LINE_STRIP to GL_LINE_LOOP
[Edited by - jezham on October 10, 2008 10:15:18 AM]
##### Share on other sites
I've just made some tests using the code you supplied, and I've found out that vertices order (CW or CCW) is important ... In one case the result is as expected, in the other one there is still one pixel missing. Maybe this sound obvious to someone, but it does not to me :)
Also I do not use glOrtho() to setup the ortho mode, but I build myself the projection matrix. I suppose my problem lies here.
thanks
##### Share on other sites
Yes winding order to me is always CCW with graphics, it's a GL thing :)
This all takes me back to the (Commodore) Amiga days. Perhaps then, if you still have the issue with a standard ortho test, then your hardware is doing something differently? What we used to do (on the old blitter chip) was ensure lines/edges were drawn from top to bottom, that would mean standard GL_LINES (rather than loop/strip).
I'll get round to testing mine on other systems eventually!
btw, before I fixed this in my program, I would plot a pixel (after the loop) with GL_POINTS. But mine was always at the top/right - so be sure to test on various hardware.
yw
##### Share on other sites
Okay !
So finally it WORKS (fireworks, clap clap, champagne) !
So :
- vertices must be in the right order CCW,
- I add a half-pix to the width/height of boxes,
- and I jitter gui rendering with glTranslatef(0.01, 0.01, 0)
It seams to work, at least on my laptop :)
The good old Amiga days :) I spent so much time trying to understand how these "megademos" worked, alone with my Amiga bible and my Seka (?) 68k assembler, trying to reproduce those so cooooool effects ! memories, memories ...
Many thanks for your help !
FYI, here is the final test code I used :
Pos position(100, 100);Rectangle rectangle(100, 100);const float half = 0.5;glMatrixMode(GL_PROJECTION);glLoadIdentity();glOrtho(0, 1024, 0, 1024, -1, 1);glMatrixMode(GL_MODELVIEW);glLoadIdentity();glTranslatef(0.01, 0.01, 0);glColor3f(1.0, 0, 1.0);glBegin(GL_LINE_LOOP);glVertex2f(position.x, position.y);glVertex2f(position.x + rectangle.width - 1 +half, position.y);glVertex2f(position.x + rectangle.width - 1 +half, position.y + rectangle.height - 1 +half);glVertex2f(position.x, position.y + rectangle.height - 1 +half);glEnd();glColor3f(1.0, 1.0, 0.0);glBegin(GL_QUADS);glVertex2f(position.x+1, position.y+1);glVertex2f(position.x+1 + rectangle.width-3+half, position.y+1);glVertex2f(position.x+1 + rectangle.width-3+half, position.y+1 + rectangle.height-3+half);glVertex2f(position.x+1, position.y+1 + rectangle.height-3+half);glEnd();
##### Share on other sites
Aha success! Now you can comfortably tweak your test applications :)
I'll mental note the 'jitter' (for testing on other rigs) as not needed here.
Yes the good old Amiga days, I started learning along with the "Zowee Demo" (Amiga Computing article/demo) and a 68K compiler (Borland perhaps?) from the front of another magazine...Same with 3D, a revolutionary magazine article came along which I actually read in a park on the way home. Lol I suddenly feel slightly old, yet young!
|
2018-02-25 07:27:45
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.28295692801475525, "perplexity": 3704.9394419578884}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891816178.71/warc/CC-MAIN-20180225070925-20180225090925-00668.warc.gz"}
|
https://www.nelsontang.com/blog/2022-06-02-dash-tips
|
Published on
# Dash (Plotly) tips - Latex, Google Fonts, and more
Authors
Dash applications need a lot of work to make them look good. Here's a few 'how to' extras that can help out
## Latex in Markdown
The latest release of Dash made it much easier to add Latex to your markdown elements. Here's an example:
dcc.Markdown(children="""
This is Latex:
\\begin{align} y &= x \\\\ &= 0 \\end{align}
""", mathjax=True)
Which gives us:
There were only two tricky bits now: you must set the mathjax=True in order to enable it, and you need to add an extra backslash (i.e. \ becomes \\).
This is just an artifact of how it's rendering escape characters.
You can use CSS for everything in your Dash app except for the charts that Plotly generates. So, adding in Google fonts is relatively straightforward - you can either follow the @import instructions add a new css file to your /assets/ folder or you can add it in as an external stylesheet.
1. Load it in to assets:
# \assets\fonts.css
body {
font-family: 'Press Start 2P';
}
1. Add it as an external stylesheet
from dash import Dash, html
external_stylesheets = [
|
2022-09-25 07:23:27
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8744248747825623, "perplexity": 4812.421778627689}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334515.14/warc/CC-MAIN-20220925070216-20220925100216-00571.warc.gz"}
|
https://www.esaral.com/q/a-coin-is-tossed-60-times-and-the-tail-appears-35-times-what-is-the-probability-of-getting-a-head-65189
|
# A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?
Question:
A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?
(a) $\frac{7}{12}$
(b) $\frac{12}{7}$
(c) $\frac{5}{12}$
(d) $\frac{1}{25}$
Solution:
(C) $\frac{5}{12}$
Explanation:
Total number of trials = 60
Number of times tail appears = 35
Number of times head appears = 60 35 = 25
Let E be the event of getting a head.
$\therefore P($ getting a head $)=P(E)=\frac{\text { Number of times head appears }}{\text { Total number of trials }}=\frac{25}{60}=\frac{5}{12}$
Leave a comment
|
2023-03-31 09:04:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5787031054496765, "perplexity": 405.90636795995147}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296949598.87/warc/CC-MAIN-20230331082653-20230331112653-00232.warc.gz"}
|
https://www.dickimaw-books.com/bugtracker.php?action=view&key=82&page=3&search_category=All
|
# Bug Tracker
ID 82 Closed (Fixed) datetime2-english 1.0 useregional=numeric causes errors
## Report
Hello, I found a problem when using datetime2-english (at least with en-GB and en_US styles) with the useregional=numeric option. In this case, the date is formatted as expected, but it produces error messages (! Package datetime2 Error: Unknown style en-GB-regional') and warnings (Package datetime2 Warning: No date style en-GB-regional' defined on input line[...]). A summary with minimal example can be found here: [Link]
Manuel Weinkauf
### MWE
\documentclass{article}
\usepackage[UKenglish]{babel}
\usepackage[useregional=numeric]{datetime2}
\DTMlangsetup[en-GB]{datesep={\space\&,\&\space}}
\begin{document}
\today\\
\DTMdisplaydate{2000}{01}{01}{}
\end{document}
## Evaluation
Fixed in v1.01 which I've just uploaded to CTAN. Please allow a few days for it to propagate through the mirrors. A temporary work around is to redefine \dateUKenglish as follows:
\renewcommand\dateenglish{%
\DTMifcaseregional
{}% do nothing
{\DTMsetstyle{en-GB}}%
{\DTMsetstyle{en-GB-numeric}}%
}%
## Watch This Report
If you supply your name, it will be used in the email greeting, which provides a more personal message, otherwise you'll just get a generic greeting. If you have previously supplied your name when signing up for notifications, you don't need to resupply it unless you want to change it.
If you have previously subscribed to notifications for this report, you can unsubscribe by clicking on the "Stop Notification" button.
The "Confirm Bug ID" field helps to protect against spambots. Please enter the bug ID (which you can find at the top of this page).
Name: (Optional.)
## Comment
You can append a comment to the report using the form below. Comments are checked first before being added. Any spam or offensive content will be removed first according to this site's Terms of Website Use. Please bear in mind that I develop and maintain free software in my spare time. If you want commerical level support then you can hire a TeX consultant.
The "Confirm Bug ID" field helps to protect against spambots. Please enter the bug ID (which you can find at the top of this page).
Name: (Optional. If provided, it will be shown with the comment.) Message: You can use the following markup: [pre]Displayed verbatim[/pre] [tt]monospace text[/tt] [em]emphasized text[/em] [b]bold text[/b] [url]web address[/url] Ordered list: [ol] [li]first item[/li] [li]second item[/li] [/ol] Unordered list: [ul] [li]first item[/li] [li]second item[/li] [/ul]
Click on the Preview button to preview the message.
|
2019-12-13 06:21:11
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.19617486000061035, "perplexity": 7322.5620002753185}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540548544.83/warc/CC-MAIN-20191213043650-20191213071650-00156.warc.gz"}
|
https://math.stackexchange.com/questions/3195076/i-am-having-trouble-figuring-out-how-many-lambdas-births-there-are-in-a-given
|
I am having trouble figuring out how many lambda's (births) there are in a given birth-death Markov process problem.
These questions are not for assignment. I am just confused as to how to set up the problem. I also do not need help calculating the problems at hand.
I understand that in a birth and death problem, $$P_{0}=\frac{1}{1+\Sigma_{n=1}^{\infty}\frac{\lambda_{0}\cdot\cdot\cdot\lambda_{}n-1}{\mu_{1}\cdot\cdot\cdot\mu_{n}}}$$
and
$$P_{1}=\frac{\lambda_{0}P_{0}}{\mu_{1}}$$, $$P_{2}=\frac{\lambda_{1}P_{1}}{\mu_{2}}$$, ... etc. where lambdas are the birth rate and mu is the death rate.
My main point of confusion is figuring out how many lambda's there are in the question.
Take this question for example
"A small barbershop, with 1 barber has room for at most 2 customers. Potential customers arrive at a poisson rate of three per hour, and the successive service times are independent exponential random variables with mean $$\frac{1}{4}$$hour."
I believe it makes sense to say that we have two lambda's in this case since we only have room for 2 customers in the barber shop. So with $$\lambda_{0}$$, we are in state 0 (having 0 customers in barber shop, so now we can enter). Similarly, if we are in state 1 ($$\lambda_{1}$$), there is only one customer in there so another customer can enter. Now we have our two. Is this the right way to think about it?
For the next example:
"Potential customers arrive at a full-service, one pump gas station at a poisson rate of 20 cars per hour. Customers only enter if there are no more than two cars (including the one being tended to) at the pump. Suppose the amount of time required to service a car is exponentially distributed with mean of 5 minutes"
My instincts tell me that this is essentially the same problem as the first one, with a different setting. However, when I see the solution, there are 3 lambda's ($$\lambda_{0}, \lambda_{1}, \lambda_{2}$$). But why would we include $$\lambda_{2}$$? Doesn't state 2 mean there are two cars at the station, and if we can only have up to 2 cars, then this car wouldn't even enter.
I am not sure where my logic is failing.
|
2019-05-24 09:54:19
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7943781018257141, "perplexity": 282.91284361541153}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257601.8/warc/CC-MAIN-20190524084432-20190524110432-00217.warc.gz"}
|
https://www.semanticscholar.org/paper/TESTING-THE-CORE-OVERSHOOT-MIXING-DESCRIBED-BY-A-ON-Zhang/4b3527575e442799157023f6b7369761ca46ee61
|
# TESTING THE CORE OVERSHOOT MIXING DESCRIBED BY A TURBULENT CONVECTION MODEL ON THE ECLIPSING BINARY STAR HY VIR
@article{Zhang2012TESTINGTC,
title={TESTING THE CORE OVERSHOOT MIXING DESCRIBED BY A TURBULENT CONVECTION MODEL ON THE ECLIPSING BINARY STAR HY VIR},
author={Q. S. Zhang},
journal={The Astrophysical Journal},
year={2012},
volume={761}
}
• Q. S. Zhang
• Published 1 November 2012
• Physics
• The Astrophysical Journal
Helioseismic investigation has suggested applying turbulent convection models (TCMs) to convective overshoot. Using the turbulent velocity in the overshoot region determined by a TCM, one can deal with overshoot mixing as a diffusion process, which leads to incomplete mixing. It has been found that this treatment can improve solar sound speed and Li depletion in open clusters. In order to investigate whether the TCM can be applied to overshoot mixing outside the stellar convective core, new…
CONVECTIVE OVERSHOOT MIXING IN MODELS OF THE STELLAR INTERIOR
Convective overshoot mixing plays an important role in stellar structure and evolution. However, overshoot mixing is also a long-standing problem; it is one of the most uncertain factors in stellar
MODELING CONVECTIVE CORE OVERSHOOT AND DIFFUSION IN PROCYON CONSTRAINED BY ASTEROSEISMIC DATA
• Physics, Geology
• 2014
We compare evolved stellar models, which match Procyon's mass and position in the HR diagram, to current ground-based asteroseismic observations. Diffusion of helium and metals along with two
On the role of doubly-diffusive mixing in low-mass red giant branch stars
The purpose of this thesis is to investigate mixing in red-giant stars. Thanks particularly to the observations of globular clusters, it is clear that that standard stellar evolution fails to
Asteroseismic Analyses of Slowly Pulsating B Star KIC 8324482: Ultraweak Element Mixing beyond the Central Convective Core
• Physics, Geology
The Astrophysical Journal
• 2020
Asteroseismology is a powerful tool for probing the inner structure and determining the evolutionary status and the fundamental parameters of stars. The oscillation spectra of slowly pulsating B
The YNEV stellar evolution and oscillation code
We have developed a new stellar evolution and oscillation code YNEV, which calculates the structures and evolutions of stars, taking into account hydrogen and helium burning. A nonlocal turbulent
On the uncertain nature of the core of α Cen A
• Physics
• 2016
High-quality astrometric, spectroscopic, interferometric and, importantly, asteroseismic observations are available for $\alpha$ Cen A, which is the closest binary star system to earth. Taking all
ASTEROSEISMIC ANALYSIS OF THE CoRoT TARGET HD 49933
• Physics
• 2013
The frequency ratios r10 and r01 of HD 49933 exhibit an increase at high frequencies. This behavior also exists in the ratios of other stars, which is considered to result from the low
Asteroseismic analysis of the CoRoT target HD 169392
• Geology, Physics
• 2013
The satellite CoRoT (Convection, Rotation, and planetary Transits) has provided high-quality data for almost six years. We show here the asteroseismic analysis and modeling of HD169392A, which
Convection Theory and Relevant Problems in Stellar Structure, Evolution, and Pulsational Stability I Convection Theory and Structure of Convection Zone and Stellar Evolution
• D. Xiong
• Physics
Frontiers in Astronomy and Space Sciences
• 2020
A non-local and time-dependent theory of convection was briefly described. This theory was used to calculate the structure of solar convection zones, the evolution of massive stars, lithium depletion
Upward Overshooting in Turbulent Compressible Convection. I. Effects of the Relative Stability Parameter, the Prandtl Number, and the Péclet Number
• T. Cai
• Physics
The Astrophysical Journal
• 2020
In this paper, we investigate the upward overshooting by three-dimensional numerical simulations. We find that the above convectively stable zone can be partitioned into three layers: the thermal
## References
SHOWING 1-10 OF 26 REFERENCES
TURBULENT CONVECTION MODEL IN THE OVERSHOOTING REGION. I. EFFECTS OF THE CONVECTIVE MIXING IN THE SOLAR OVERSHOOTING REGION
• Physics, Geology
• 2012
The overshooting in the framework of the turbulent convection model is investigated in the solar overshooting region. The overshooting mixing is treated as a diffusive process. It is found that the
Turbulence properties in the solar convection envelope: properties of the overshooting regions ⁄
• Physics
• 2009
We apply the turbulent convection model (TCM) to investigate properties of turbulence in the solar convective envelope, especially in overshooting regions. The results show TCM gives negative
Testing turbulent convection theory in solar models – I. Structure of the solar convection zone
• Physics, Environmental Science
• 2007
Turbulent convection models (TCMs) based on hydrodynamic moment equations are compared with the classical mixing-length theory (MLT) in solar models. The aim is to test the effects of some physical
A more realistic representation of overshoot at the base of the solar convective envelope as seen by helioseismology
• Physics, Geology
• 2011
The stratification near the base of the Sun’s convective envelope is governed by processes of convective overshooting and element diffusion, and the region is widely believed to play a key role in
TURBULENT CONVECTION MODEL IN THE OVERSHOOTING REGION. II. THEORETICAL ANALYSIS
• Physics
• 2012
Turbulent convection models (TCMs) are thought to be good tools to deal with the convective overshooting in the stellar interior. However, they are too complex to be applied to calculations of
The diffusive overshooting approach to Li abundance in clusters
Helioseimic investigation shows that convective overshooting can penetrate 0.37H(P) to the location where the temperature is 2.5 x 10(6) K, which is the typical temperature of the reaction Li-7(p,
A k–ω MODEL FOR TURBULENTLY THERMAL CONVECTION IN STARS
Both observations and numerical simulations show that stellar convective motions are composed of semi-regular flows of convective rolling cells and a fully developed turbulence. Although the
Does convective core overshooting depend on stellar mass? Tests using double-lined eclipsing binaries
Aims. We have selected 13 double-line eclipsing binary systems (DLEBS), strategically positioned in the HR diagram, to infer the mass dependence of the core overshooting parameter αov. Methods. In
Convective Overshooting in Stellar Interior Models
• Physics
• 1973
ABS>The question of overshooting from interior convective cores in stellar models is examined in the framework of a simplified nonlocal mixing- length model for convection. The most important results
The structure of the solar convective overshooting zone
• Physics
• 2001
Using a non-local theory of convection, we calculated the structure of the solar convection zone, paying special attention to the detailed structure of the lower overshooting zone. Our results show
|
2022-08-07 21:14:21
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5851437449455261, "perplexity": 4517.12636523965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882570730.59/warc/CC-MAIN-20220807211157-20220808001157-00316.warc.gz"}
|
http://mathoverflow.net/revisions/106880/list
|
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
The proof of Brouwer fixed point theorem by using fundamental group of $S^1$ is equal to $\mathbb{Z}$, while the fundamental group of $D^2$ is trivial.
|
2013-06-19 15:12:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7514677047729492, "perplexity": 323.4527834186589}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368708808767/warc/CC-MAIN-20130516125328-00092-ip-10-60-113-184.ec2.internal.warc.gz"}
|
http://rubyforge.org/pipermail/kramdown-users/2011-January/000478.html
|
# [kramdown-users] Feature Request: LaTeX Support in kramdown
Thomas Leitner t_leitner at gmx.at
Thu Jan 27 05:28:17 EST 2011
On 2011-01-24 23:35 +0100 Alex Wall wrote:
> It would be nice if kramdown could handle markdown files more like
> pandoc. In pandoc it is possible to mix markdown and LaTeX as you
> like without any escape characters/sequences.
>
> This gives you the simplicity of markdown and the power of LaTeX when
> you really need it.
The reason why math support is done via LaTeX commands is that this is
supported by the HTML (via MathJax) and LaTeX converters. I'm sure that
if we add other converters there is a high probability that a solution
for converting LaTeX math commands to the specific output format is
available.
I don't think that supporting LaTeX commands directly in kramdown is a
good idea. The Markdown syntax is relatively easy to understand for
newcomers, LaTeX syntax is not. And most user would be surprised when
they see the resulting output. Another thing: what should be done with
LaTeX commands when converting a document to HTML? Leave them out?
If you really want to intersperse LaTeX commands with normal kramdown
syntax, I recommend using the nomarkdown extension!
Best regards,
Thomas
|
2013-12-09 13:22:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9793592691421509, "perplexity": 5711.475557034119}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1386163973624/warc/CC-MAIN-20131204133253-00058-ip-10-33-133-15.ec2.internal.warc.gz"}
|
http://answers.gazebosim.org/questions/12562/revisions/
|
I have a robotic arm in Gazebo from which I am already reading the angles (i.e. GetAngle(0).Radian()) as well as the velocity (i.e. GetVelocity(0)) from each joint. Now I also need to get the acceleration from the joints.
|
2020-02-22 11:19:36
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8014712929725647, "perplexity": 707.4270361403862}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875145657.46/warc/CC-MAIN-20200222085018-20200222115018-00438.warc.gz"}
|
https://mscroggs.co.uk/blog/tags/chalkdust%20magazine
|
mscroggs.co.uk
mscroggs.co.uk
subscribe
# Blog
2019-12-08
Just like last year, the year before and the year before, TD and I spent some time in November this year designing a Chalkdust puzzle Christmas card.
The card looks boring at first glance, but contains 9 puzzles. By splitting the answers into two digit numbers, then drawing lines labelled with each number (eg if an answer is 201304, draw the lines labelled 20, 13 and 4), you will reveal a Christmas themed picture. Colouring the regions of the card containing circles red, the regions containing squares green, and the regions containing stars white or yellow will make this picture even nicer.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically be split into two digit numbers, the lines will be drawn, and the regions will be coloured...
### Similar posts
Christmas card 2018 Christmas card 2017 Christmas card 2016 Christmas (2019) is over
Comments in green were written by me. Comments in blue were not written by me.
Rishabh, what do you see that I do not?
Rob Glencairn
Thanks for the feedback. (I now understand the need for redaction). My son sent me your link as a Xmas present. I must think of an appropriate retaliation. (What is a PDF?)Think I've fixed 1,6 and 9....8 eludes me, for the moment.
Rob
@Rob: It looks to me like you've made mistakes in questions 1, 6, 8, and 9. The hints from the back of the pdf might help:
1. How many numbers between 1 and 10,000 have 1 as their final digit? How many have 1 as their penultimate digit?
6. How many ways can you write 1? 2? 3? 4? 5? What's the pattern?
8. How many zeros does 10! end in? How many zeros does 20! end in? How many zeros does 30! end in?
9. Carol’s sum is odd. What does this tell you about the 5- and 6-digit numbers?
Matthew
×1 ×1 ×1 ×1 ×1
I'm 71, with one good eye left. What am I missing?
1. 400001
2. 1849
3. 2002
4. 130405
5. 120306
6. 53?
7. 171175
8. 59?
9. 313525
Rob
It was fun.
Rishabh
×1 ×1
Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "graph" in the box below (case sensitive):
2019-09-01
This week, I've been in Cambridge for Talking Maths in Public (TMiP). TMiP is a conference for anyone involved in—or interested in getting involved in—any sort of maths outreach, enrichment, or public engagement activity. It was really good, and I highly recommend coming to TMiP 2021.
The Saturday morning at TMiP was filled with a choice of activities, including a treasure punt (a treasure hunt on a punt) written by me. This post contains the puzzle from the treasure punt for anyone who was there and would like to revisit it, or anyone who wasn't there and would like to give it a try. In case you're not current in Cambridge on a punt, the clues that you were meant to spot during the punt are given behing spoiler tags (hover/click to reveal).
### Instructions
Each boat was given a copy of the instructions, and a box that was locked using a combination lock.
Locked boxes.
If you want to make your own treasure punt or similar activity, you can find the LaTeX code used to create the instructions and the Python code I used to check that the puzzle has a unique solution on GitHub. It's licensed with a CC BY 4.0 licence, so you can resuse an edit it in any way you like, as long as you attribute the bits I made that you keep.
### The puzzle
Four mathematicians—Ben, Katie, Kevin, and Sam—each have one of the four clues needed to unlock a great treasure. On a sunny/cloudy/rainy/snowy (delete as appropriate) day, they meet up in Cambridge to go punting, share their clues, work out the code for the lock, and share out the treasure. One or more of the mathematicians, however, has decided to lie about their clue so they can steal all the treasure for themselves. At least one mathematician is telling the truth. (If the mathematicians say multiple sentences about their clue, then they are either all true or all false.)
They meet at Cambridge Chauffeur Punts, and head North under Silver Street Bridge. Ben points out a plaque on the bridge with two years written on it:
"My clue," he says, "tells me that the sum of the digits of the code is equal to the sum of the digits of the earlier year on that plaque (the year is 1702). My clue also tells me that at least one of the digits of the code is 7."
The mathematicians next punt under the Mathematical Bridge, gasping in awe at its tangential trusses, then punt along the river under King's College Bridge and past King's College. Katie points to a sign on the King's College lawn near the river:
"See that sign whose initials are PNM?" says Katie. "My clue states that first digit of the code is equal to the number of vowels on that sign (The sign says "Private: No Mooring"). My clue also tells me that at least one of the digits of the code is 1."
They then reach Clare Bridge. Kevin points out the spheres on Clare Bridge:
"My clue," he says, "states that the total number of spheres on both sides of this bridge is a factor of the code (there are 14 spheres). My clue also tells me that at least one of the digits of the code is 2." (Kevin has not noticed that one of the spheres had a wedge missing, so counts that as a whole sphere.)
They continue past Clare College. Just before they reach Garret Hostel Bridge, Sam points out the Jerwood Library and a sign showing the year it was built (it was built in 1998):
"My clue," she says, "says that the largest prime factor of that year appears in the code (in the same way that you might say the number 18 appears in 1018 or 2189). My clue also says that the smallest prime factor of that year appears in the code. My clue also told me that at least one of the digits of the code is 0."
They then punt under Garret Hostel Bridge, turn around between it and Trinity College Bridge, and head back towards Cambridge Chauffeur Punts. Zut alors, the lies confuse them and they can't unlock the treasure. Can you work out who is lying and claim the treasure for yourself?
### The solution
The solution to the treasure punt is given below. Once you're ready to see it, click "Show solution".
### Similar posts
Christmas card 2019 Christmas card 2018 Christmas card 2017 Christmas card 2016
Comments in green were written by me. Comments in blue were not written by me.
Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "r" then "a" then "t" then "i" then "o" in the box below (case sensitive):
2019-04-09
In the latest issue of Chalkdust, I wrote an article with Edmund Harriss about the Harriss spiral that appears on the cover of the magazine. To draw a Harriss spiral, start with a rectangle whose side lengths are in the plastic ratio; that is the ratio $$1:\rho$$ where $$\rho$$ is the real solution of the equation $$x^3=x+1$$, approximately 1.3247179.
A plastic rectangle
This rectangle can be split into a square and two rectangles similar to the original rectangle. These smaller rectangles can then be split up in the same manner.
Splitting a plastic rectangle into a square and two plastic rectangles.
Drawing two curves in each square gives the Harriss spiral.
A Harriss spiral
This spiral was inspired by the golden spiral, which is drawn in a rectangle whose side lengths are in the golden ratio of $$1:\phi$$, where $$\phi$$ is the positive solution of the equation $$x^2=x+1$$ (approximately 1.6180339). This rectangle can be split into a square and one similar rectangle. Drawing one arc in each square gives a golden spiral.
A golden spiral
### Continuing the pattern
The golden and Harriss spirals are both drawn in rectangles that can be split into a square and one or two similar rectangles.
The rectangles in which golden and Harriss spirals can be drawn.
Continuing the pattern of these arrangements suggests the following rectangle, split into a square and three similar rectangles:
Let the side of the square be 1 unit, and let each rectangle have sides in the ratio $$1:x$$. We can then calculate that the lengths of the sides of each rectangle are as shown in the following diagram.
The side lengths of the large rectangle are $$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1$$ and $$\frac1{x^2}+\frac1x+1$$. We want these to also be in the ratio $$1:x$$. Therefore the following equation must hold:
$$\frac{1}{x^3}+\frac{1}{x^2}+\frac2x+1=x\left(\frac1{x^2}+\frac1x+1\right)$$
Rearranging this gives:
$$x^4-x^2-x-1=0$$ $$(x+1)(x^3-x^2-1)=0$$
This has one positive real solution:
$$x=\frac13\left( 1 +\sqrt[3]{\tfrac12(29-3\sqrt{93})} +\sqrt[3]{\tfrac12(29+3\sqrt{93})} \right).$$
This is equal to 1.4655712... Drawing three arcs in each square allows us to make a spiral from a rectangle with sides in this ratio:
A spiral which may or may not have a name yet.
### Continuing the pattern
The side lengths of the largest rectangle are $$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4}$$ and $$1+\frac2x+\frac1{x^2}+\frac1{x^3}$$. Looking for the largest rectangle to also be in the ratio $$1:x$$ leads to the equation:
$$1+\frac2x+\frac3{x^2}+\frac1{x^3}+\frac1{x^4} = x\left(1+\frac2x+\frac1{x^2}+\frac1{x^3}\right)$$ $$x^5+x^4-x^3-2x^2-x-1 = 0$$
This has one real solution, 1.3910491... Although for this rectangle, it's not obvious which arcs to draw to make a spiral (or maybe not possible to do it at all). But at least you get a pretty fractal:
### Continuing the pattern
We could, of course, continue the pattern by repeatedly adding more rectangles. If we do this, we get the following polynomials and solutions:
Number of rectangles Polynomial Solution 1 $$x^2 - x - 1=0$$ 1.618033988749895 2 $$x^3 - x - 1=0$$ 1.324717957244746 3 $$x^4 - x^2 - x - 1=0$$ 1.465571231876768 4 $$x^5 + x^4 - x^3 - 2x^2 - x - 1=0$$ 1.391049107172349 5 $$x^6 + x^5 - 2x^3 - 3x^2 - x - 1=0$$ 1.426608021669601 6 $$x^7 + 2x^6 - 2x^4 - 3x^3 - 4x^2 - x - 1=0$$ 1.4082770325090774 7 $$x^8 + 2x^7 + 2x^6 - 2x^5 - 5x^4 - 4x^3 - 5x^2 - x - 1=0$$ 1.4172584399350432 8 $$x^9 + 3x^8 + 2x^7 - 5x^5 - 9x^4 - 5x^3 - 6x^2 - x - 1=0$$ 1.412713760332943 9 $$x^{10} + 3x^9 + 5x^8 - 5x^6 - 9x^5 - 14x^4 - 6x^3 - 7x^2 - x - 1=0$$ 1.414969877544769
The numbers in this table appear to be heading towards around 1.414, or $$\sqrt2$$. This shouldn't come as too much of a surprise because $$1:\sqrt2$$ is the ratio of the sides of A$$n$$ paper (for $$n=0,1,2,...$$). A0 paper can be split up like this:
Splitting up a piece of A0 paper
This is a way of splitting up a $$1:\sqrt{2}$$ rectangle into an infinite number of similar rectangles, arranged following the pattern, so it makes sense that the ratios converge to this.
### Other patterns
In this post, we've only looked at splitting up rectangles into squares and similar rectangles following a particular pattern. Thinking about other arrangements leads to the following question:
Given two real numbers $$a$$ and $$b$$, when is it possible to split an $$a:b$$ rectangle into squares and $$a:b$$ rectangles?
If I get anywhere with this question, I'll post it here. Feel free to post your ideas in the comments below.
### Similar posts
Dragon curves II Christmas card 2019 TMiP 2019 treasure punt Christmas card 2018
Comments in green were written by me. Comments in blue were not written by me.
@g0mrb: CORRECTION: There seems to be no way to correct the glaring error in that comment. A senior moment enabled me to reverse the nomenclature for paper sizes. Please read the suffixes as (n+1), (n+2), etc.
(anonymous)
I shall remain happy in the knowledge that you have shown graphically how an A(n) sheet, which is 2 x A(n-1) rectangles, is also equal to the infinite series : A(n-1) + A(n-2) + A(n-3) + A(n-4) + ... Thank-you, and best wishes for your search for the answer to your question.
g0mrb
Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "graph" in the box below (case sensitive):
2018-12-08
Just like last year and the year before, TD and I spent some time in November this year designing a Chalkdust puzzle Christmas card.
The card looks boring at first glance, but contains 10 puzzles. By splitting the answers into pairs of digits, then drawing lines between the dots on the cover for each pair of digits (eg if an answer is 201304, draw a line from dot 20 to dot 13 and another line from dot 13 to dot 4), you will reveal a Christmas themed picture. Colouring the region of the card labelled R red or orange will make this picture even nicer.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically be split into pairs of digits, lines will be drawn between the pairs, and the red region will be coloured...
If you enjoy these puzzles, then you'll almost certainly enjoy this year's puzzle Advent calendar.
### Similar posts
Christmas card 2019 Christmas card 2017 Christmas card 2016 Christmas (2019) is over
Comments in green were written by me. Comments in blue were not written by me.
Someone told me I would like this puzzle and they were right!
Blaine
×1
@Carmel: It's not meant to check your answers. It only shows up red if the number you enter cannot be split into valid pairs (eg the number has an odd number of digits or one of the pairs of digits is greater than 20).
Matthew
The script for checking the answers doesn't work properly
Carmel
Thank you Shawn!
SueM
So satisfying!
Heather
×1
Great puzzle problems! Hint on #9: try starting with an analogous problem using smaller numbers (e.g. 3a + 10b). This helped me to see what I had to do more generally.
Noah
×1
Allowed HTML tags: <br> <a> <small> <b> <i> <s> <sup> <sub> <u> <spoiler> <ul> <ol> <li>
To prove you are not a spam bot, please type "rotcaf" backwards in the box below (case sensitive):
2017-12-18
Just like last year, TD and I spent some time in November this year designing a puzzle Christmas card for Chalkdust.
The card looks boring at first glance, but contains 10 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s black and 2s orange will reveal a Christmassy picture.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically converted to base 3 and coloured...
# Answer (base 10) Answer (base 3) 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 8 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0
1. In a book with 116 pages, what do the page numbers of the middle two pages add up to?
2. What is the largest number that cannot be written in the form $$14n+29m$$, where $$n$$ and $$m$$ are non-negative integers?
3. How many factors does the number $$2^6\times3^{12}\times5^2$$ have?
4. How many squares (of any size) are there in a $$15\times14$$ grid of squares?
5. You take a number and make a second number by removing the units digit. The sum of these two numbers is 1103. What was your first number?
6. What is the only three-digit number that is equal to a square number multiplied by the reverse of the same square number? (The reverse cannot start with 0.)
7. What is the largest three-digit number that is equal to a number multiplied by the reverse of the same number? (The reverse cannot start with 0.)
8. What is the mean of the answers to questions 6, 7 and 8?
9. How many numbers are there between 0 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6?
10. What is the lowest common multiple of 52 and 1066?
### Similar posts
Christmas card 2016 Christmas card 2019 Christmas card 2018 Christmas (2019) is over
Comments in green were written by me. Comments in blue were not written by me.
@Jose: There is a mistake in your answer: 243 (0100000) is the number of numbers between 10,000 and 100,000 that do not contain the digits 0, 1, 2, 3, 4, 5, or 6.
Matthew
Thanks for the puzzle!
Is it possible that the question 9 is no correct?
I get a penguin with perfect simetrie except at answer 9 : 0100000 that breaks the simetry.
Is it correct or a mistake in my answer?
Thx
Jose
@C: look up something called Frobenius numbers. This problem's equivalent to finding the Frobenius number for 14 and 29.
Lewis
I can solve #2 with code, but is there a tidy maths way to solve it directly?
C
My efforts were flightless.
NHH
|
2020-03-28 09:06:41
|
{"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": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.37845849990844727, "perplexity": 961.1739790874988}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370490497.6/warc/CC-MAIN-20200328074047-20200328104047-00559.warc.gz"}
|
https://brainiak.in/712/what-reaction-called-oxidation-reduction-take-place-simultaneously
|
# a. What is the reaction called when oxidation and reduction take place simultaneously?
a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
verified
When oxidation and reduction takes place simultaneously, the reaction is called redox reaction.
Oxidation: losing of electrons
Reduction: gaining of electrons
Redox Reaction = Reduction + oxidation
Example
$$Cao+H_2 \rightarrow Cu+H_2O$$
In this examples copper oxide is reduced and acts as oxidizing agent and H2 is oxidized and acts as reducing agent.
|
2021-11-26 23:53:53
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.47249627113342285, "perplexity": 4850.200289962017}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00553.warc.gz"}
|
https://proofwiki.org/wiki/Set_Equivalence_of_Regular_Representations
|
Set Equivalence of Regular Representations
Theorem
If $S$ is a finite subset of a group $G$, then:
$\card {a \circ S} = \card S = \left|{S \circ a}\right|$
That is, $a \circ S$, $S$ and $S \circ a$ are equivalent: $a \circ S \sim S \sim S \circ a$.
Proof
Follows immediately from the fact that both the left and right regular representation are permutations, and therefore bijections.
$\blacksquare$
|
2020-06-01 07:13:46
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9575698375701904, "perplexity": 254.547352983503}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347415315.43/warc/CC-MAIN-20200601071242-20200601101242-00261.warc.gz"}
|
https://planetmath.org/VanishingOfGradientInDomain
|
# vanishing of gradient in domain
If the function $f$ is defined in a domain (http://planetmath.org/Domain2) $D$ of $\mathbb{R}^{n}$ and all the partial derivatives of a $f$ vanish identically in $D$, i.e.
$\nabla{f}\;\equiv\;\vec{0}\quad\mbox{in}\;D,$
then the function has a constant value in the whole domain.
Proof. For the sake of simpler notations, think that $n=3$; thus we have
$\displaystyle f_{x}^{\prime}(x,\,y,\,z)\;=\;f_{y}^{\prime}(x,\,y,\,z)\;=\;f_{z% }^{\prime}(x,\,y,\,z)\;=\;0\quad\mbox{for all}\;\;(x,\,y,\,z)\in D.$ (1)
Make the antithesis that there are the points $P_{0}=(x_{0},\,y_{0},\,z_{0})$ and $P_{1}=(x_{1},\,y_{1},\,z_{1})$ of $D$ such that $f(x_{0},\,y_{0},\,z_{0})\neq f(x_{1},\,y_{1},\,z_{1})$. Since $D$ is connected, one can form the broken line $P_{0}Q_{1}Q_{2}\ldots Q_{k}P_{1}$ contained in $D$. When one now goes along this broken line from $P_{0}$ to $P_{1}$, one mets the first corner where the value of $f$ does not equal $f(x_{0},\,y_{0},\,z_{0})$. Thus $D$ contains a line segment, the end points of which give unequal values to $f$. When necessary, we change the notations such that this line segment is $P_{0}P_{1}$. Now, $f_{x}^{\prime},\,f_{y}^{\prime},\,f_{z}^{\prime}$ are continuous in $D$ because they vanish. The mean-value theorem for several variables guarantees an interior point$(a,\,b,\,c)$ of the segment such that
$0\;\neq\;f(x_{1},\,y_{1},\,z_{1})-f(x_{0},\,y_{0},\,z_{0})\;=\;f_{x}^{\prime}(% a,\,b,\,c)(x_{1}\!-\!x_{0})+f_{y}^{\prime}(a,\,b,\,c)(y_{1}\!-\!y_{0})+f_{z}^{% \prime}(a,\,b,\,c)(z_{1}\!-\!z_{0}).$
But by (1), the last sum must vanish. This contradictory result shows that the antithesis is wrong, which settles the proof.
Title vanishing of gradient in domain VanishingOfGradientInDomain 2013-03-22 19:11:58 2013-03-22 19:11:58 pahio (2872) pahio (2872) 8 pahio (2872) Theorem msc 26B12 partial derivatives vanish FundamentalTheoremOfIntegralCalculus ExtremumPointsOfFunctionOfSeveralVariables
|
2018-11-15 18:46:52
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 26, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9251307249069214, "perplexity": 1099.6884090130422}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-47/segments/1542039742906.49/warc/CC-MAIN-20181115182450-20181115204450-00350.warc.gz"}
|
https://toph.co/problems/divide-and-conquer
|
# Divide and Conquer Problems
#### Generation Gap
There are n people standing in a queue. The age of the ith person is ai. The more the age gap betwe...
#### LIDS
We all know about LIS ( Longest Increasing Sub sequence). The task to find the length of the longest...
#### Buildings
There are n buildings in a row. The height of the i’th building is hi. You have ৳k. It is the amoun...
#### Ekupai
Hacker robot Ekupai is trying to hack a supercomputer. He needs a cheat code X to hack this computer...
#### Burger King
Every day, more than 11 million guests visit BURGER KING® restaurants around the world. And they do ...
#### Lexicographical Smallest String
There is a string s of length n containing lowercase letters and an integer k. Among the letters tho...
#### A Journey Towards One
Given N. You can perform 3 types of operations: Increase N by 1. Decrease N by 1. Divide N by 2 (O...
#### Counting Murgis
Ever since that wicked thief took away some of her murgis (hens, that is), Meena was ever anxious th...
#### Third Dimension
Mr. J wants to learn about combinatorics. He started reading a book on combinatorics and came across...
#### Diagonal Sum
Given an N*N size matrix, you have to re-arrange the elements of the matrix in such a way that the d...
#### I Hate Combinatorics!
Rick and Morty are being hunted by the galactic government. They ran away to the cell dimension. The...
#### String Is Not That Easy
Let’s consider a string S which is obtained by concatenating the non-negative integers from 0 to 102...
#### Pathetic Interview II
You all know about the story of Pathetic Interview-I. Today, I am sharing a story of my close friend...
#### Harry Potter and the Vault of Gringotts
“There was a break-in of Gringotts Wizarding Bank on 1 May, 1998, during the height of the Second Wi...
#### Convert String Into Palindrome
A palindrome is a word, number, phrase, or other sequence of characters which reads the same backwar...
#### Train Hijack
You and your friend Daneliya Tuleshova are planning to hijack a train. Your friend will go inside a...
#### Flow on Tree
Mr. Kaboom has recently learned about maximum flow. Now his friend Mr. Taboom gave him this problem....
Alice wants to extract some passwords from a random string. A password can have any number of charac...
Techboy is appointed as the minister of roads and highway in his country Better-Not-Name-It. Now he ...
#### Is This a Give-Away?
Once there lived an ancient mage named Farabi. He could always give perfect weather forecast. He nev...
#### Subset AND
You are given an array A of n integers and an integer k. You need to find if there is any non-empty ...
Today is Luke’s birthday. Mr. Phil Dunphy (Luke’s father) has thrown a birthday party for him and de...
#### Find the Good Sequence
Let’s say two numbers are called “good” if their difference is at least 2. Similarly, a sequence is...
#### Easy
This problem is very easy. As my mood is not so good, I will not elaborate the statement unnecessari...
#### Naming Convention
Tony works in a multinational technology company named Voogle LLC. The employees of Voogle follow ce...
#### Polygon Construction (Hard)
Meera in going to picnic with her friends. They are going to have lots of fun. They will be doing va...
#### No GCD
You are given N integers. Each integer is square free (i.e. it has no divisor which is a square numb...
#### OCD Returns!
This problem author has OCD (obsessive–compulsive disorder). One day his teacher gave him an array o...
#### The Game of Rocks
Alice and Bob have come up with a new game. Bob gives Alice a NxM grid that contains N rows and M c...
#### Gaaner Koli
The protagonists of this problem are two great programmers of a certain institution Omanush and Ordi...
### Other Categories
Dynamic Programming Greedy Algorithm Matrix Exponentiation
|
2020-04-04 15:11:34
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2388896346092224, "perplexity": 3755.6593608373614}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370524043.56/warc/CC-MAIN-20200404134723-20200404164723-00380.warc.gz"}
|
http://atcoder.noip.space/contest/agc015/a
|
# Home
Contest: Task: Related: TaskB
Score : $200$ points
### Problem Statement
Snuke has $N$ integers. Among them, the smallest is $A$, and the largest is $B$. We are interested in the sum of those $N$ integers. How many different possible sums there are?
### Constraints
• $1 ≤ N,A,B ≤ 10^9$
• $A$ and $B$ are integers.
### Input
Input is given from Standard Input in the following format:
$N$ $A$ $B$
### Output
Print the number of the different possible sums.
### Sample Input 1
4 4 6
### Sample Output 1
5
There are five possible sums: $18=4+4+4+6$, $19=4+4+5+6$, $20=4+5+5+6$, $21=4+5+6+6$ and $22=4+6+6+6$.
### Sample Input 2
5 4 3
### Sample Output 2
0
### Sample Input 3
1 7 10
### Sample Output 3
0
### Sample Input 4
1 3 3
### Sample Output 4
1
|
2023-02-08 07:27:49
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4114910066127777, "perplexity": 3887.8224290073754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500719.31/warc/CC-MAIN-20230208060523-20230208090523-00870.warc.gz"}
|
http://icpc.njust.edu.cn/Problem/Local/1023/
|
# Unique ShutDown
Time Limit: 1000 ms
Memory Limit: 65535 ms
## Description
N computers are connected into a network. The rule is: M different pairs of computers are connected to each other by an link. All connections are two-way, that is, they can be used in both directions. There is always two students playing internet computer games,so Yutou make up his mind to stop them ! That is to say ,he want to shut down the connection between the two boys. Now Yutou want to know if there are no two different sets of link connections that can be destroyed, such that the two boy’s computers cannot connect to each other after shut down only one set of link
## Input
The input file consists of several cases. In each case, the first line of the input file contains N, M, A and B (2 <= N <= 100, 1 <= M <= 100, 1 <= A,B <= N, A != B), specifying the number of computers in the network, the number of link connections, and the numbers of the boy’s computers respectively. A case with 4 zeros indicates the end of file. Next M lines describe link connections. For each connection the numbers of the computers it connects are given and the cost of destroying this connection. It is guaranteed that all costs are non-negative integer numbers not exceeding 105, no two computers are directly connected by more than one link, no link connects a computer to itself and initially there is the way to transmit data from one main supercomputer to another.
## Output
If there is only one way to shut down, output "UNIQUE" in a single line. In the other case output "AMBIGUOUS".
## Sample Input
4 4 1 2
1 2 1
2 4 2
1 3 2
3 4 1
4 4 1 2
1 2 1
2 4 1
1 3 2
3 4 1
0 0 0 0
## Sample Output
UNIQUE
AMBIGUOUS
## Source
FeiTeng & ZhangKe
|
2018-08-19 19:57:31
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.287644624710083, "perplexity": 514.0954079588179}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221215284.54/warc/CC-MAIN-20180819184710-20180819204710-00579.warc.gz"}
|
http://mathhelpforum.com/calculus/3225-differentiation-modulus-function.html
|
# Thread: differentiation of a modulus function
1. ## differentiation of a modulus function
I'm a little confused
f(x) = |x|^2 - 4|x|
I know that gives the equations of:
x^2 - 4x and
x^2 + 4x
and you can differentiate them, but which equation corresponds to which part of the graph?
2. Originally Posted by freswood
I'm a little confused
f(x) = |x|^2 - 4|x|
I know that gives the equations of:
x^2 - 4x and
x^2 + 4x
and you can differentiate them, but which equation corresponds to which part of the graph?
When $x\ge 0$:
$|x|^2 - 4|x|=x^2-4x$.
When $x\le 0$:
$|x|^2 - 4|x|=x^2+4x$.
As a reality check you can always try plugging in numbers and seeing what
you get.
RonL
3. Hello, freswood!
Find the derivtive of: $f(x) \:= \:|x|^2 - 4|x|$
This is equivalent to the piecewise function: $f(x)\:=\:\left\{\begin{array}{cc} x^2 - 4x & x \geq 0 \\ x^2 + 4x & x < 0\end{array}$
Differentiate them for both cases.
You can check your results against the graph.
Code:
|
* | *
|
* | *
- - * - - - - - * - - - - - * - -
-4 * * | * * 4
* * | * *
|
4. Originally Posted by Soroban
Differentiate them for both cases.
differention of |x|=|x|/x
where x is not equal to 0
5. Thanks!
What confused me is that for something like |x^2 - 4x| one applies to y>0 and the other to y < 0.
as in x < 0 U x > 4
and 0<x<4
Can anyone explain the difference between the two?
,
,
,
,
,
,
,
,
,
,
,
,
,
,
# differentiation of modu
Click on a term to search for related topics.
|
2017-01-22 16:40:13
|
{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8267946839332581, "perplexity": 1987.805450907812}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281450.93/warc/CC-MAIN-20170116095121-00488-ip-10-171-10-70.ec2.internal.warc.gz"}
|
http://dharmath.blogspot.com/2011/05/circular-locker-problem.html
|
## Wednesday, May 18, 2011
### Circular Locker Problem
From a friend's brother:
There are 1000 lockers and 1000 students. The lockers are arranged in a circle, all closed. The 1st student opens all the lockers. The 2nd student closes every other locker. The 3rd student goes to every 3rd locker. If it’s open he closes it, if it’s closed, he opens it. When he gets to locker #999, he continues to locker #2 and continues until he reaches a locker that he has already touched. The 4th student goes to every 4th locker: again, if it’s closed, he opens it, if it’s open, he closes it. Every student after that does the same, until student #1000, who will obviously just open or close locker #1000. After they are all finished, which lockers will be open?
Solution
First, let's think about what happens when student $n$ does his turn.
If $n$ is relatively prime to 1000, that is, $n$ does not contain any factor of 2 or 5, then $n$ will touch all the lockers exactly once. For example, student #3 will go touching lockers 3,6,...,999,2,5,...,998,1,4,7,...,997,1000. In this case student $n$ behaves exactly like student #1.
If $n$ is a multiple of $2$ or $5$, then let $d = \gcd(1000,n)$ Student $n$ touches locker $m$ if and only if $d$ divides $m$. For example, if $n=16$, then $d = 8$. Student #16 will go touching lockers 16,32,...,992,8,24,...,1000, which are exactly all multiples of 8. In this case, student $n$ behaves exactly like student #8.
Now, how many students have numbers that are NOT relatively prime to 1000? Such students must have numbers in the form of $n = 2^a 5^b$ We can enumerate the possibilities:
If $b=0$, then the student numbers are: 2,4,8,16,32,64,128,256,512, for a total of 9 students. Remember that students #16,32,...,512 behave exactly like student #8,and there are 6 of them, so it's equivalent to student #8 going 6 extra times. Their effects cancel out pairwise. So we only need to consider the effects of students #2,4,8.
If $b=1$, then the student numbers are: 5,10,20,40,80,160,320,640, for a total of 8 students. Remember that students #80,...,640 behave exactly like student #40,and there are 4 of them, so their effects cancel out pairwise. So we only need to consider the effects of students #5,10,20,40.
If $b=2$, then the student numbers are: 25,50,100,200,400,800, for a total of 6 students. Remember that students #400,8000 behave exactly like student #200, so their effects cancel out. So we only need to consider the effects of students #25,50,100,200.
If $b=3$, then the student numbers are: 125,250,500,1000.
If $b=4$, then the student number is: 625, which behaves like student #125, and thus cancels the effect of student #125.
So in total, there are 9+8+6+4+1 = 28 numbers that are not relatively prime to 1000, so there are an even numbers of students that are relatively prime. Each of these students touches all lockers exactly once, so their effects again cancel out pairwise. Thus, the student numbers that we need to consider are: (grouped by their largest factor of 2):
5,25
2,10,50,250
4,20,100,500
8,40,200,1000
In order to determine if locker $n$ is closed or open, find all numbers above that divides $n$. If there are even such numbers, then the locker is close, otherwise it's open.
For example, 120 = 3 x 5 x 8, so 120 is divisible by 2,4,8, 5,10,20,40. That means locker #120 is open.
For a more explicit list of open lockers, the following are the open lockers:
1. Lockers with numbers (not divisible by 5 or divisible by 25), and (divisible by 2 but not by 4, or divisible by 8).
2. Lockers with numbers divisible by 5 but not 25, and (divisible by 4 but not by 8).
|
2017-06-25 13:49:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7304365038871765, "perplexity": 820.5897560123645}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320532.88/warc/CC-MAIN-20170625134002-20170625154002-00530.warc.gz"}
|
http://mathcentral.uregina.ca/QQ/database/QQ.09.14/h/rose4.html
|
SEARCH HOME
Math Central Quandaries & Queries
Question from Rose, a student: A shopkeeper receives 12.5% (12 1/2 %) commission on the sale of books from the publisher. In a certain month,he receives a commission of rs.250 from the publisher. How much was his sale for that particular month?? Please help me solving this problem.
Hi Rose,
Suppose his sales for the month were rs.$S$ then his commission would be rs.$0.125 \times S.$ but you know his commission was rs.250. Solve for $S.$
Penny
Math Central is supported by the University of Regina and the Imperial Oil Foundation.
|
2023-01-29 12:54:40
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32674816250801086, "perplexity": 5469.323539785981}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499713.50/warc/CC-MAIN-20230129112153-20230129142153-00285.warc.gz"}
|
https://www.internosilfilm.com/usborne-books-thddhsk/d1ea6b-molecular-orbital-diagram-of-n2-ion
|
# molecular orbital diagram of n2 ion
The atomic orbitals combining linearly to form the MOs must be of the similar energy and same symmetry and may be of the two similar atoms of a homonuclear diatomic molecule/ion (e.g. Molecules with Similar Molecular Orbital Diagrams Molecules and ions formed from 2 boron atoms or from 2 carbon atoms have molecular orbitals diagrams of the same sort as N 2. Bonding and antibonding orbitals are illustrated in MO … In both water and dimethyl ether (CH3 —Ο — CH3 ), the oxygen atom is the central atom and Well, build the molecular orbital (MO) diagram. As discussed in class the MO diagram for B 2 shows that it has two unpaired electrons (which makes it paramagnetic) and these electrons are in bonding molecular orbitals resulting in the equivalent bond strength of one bond. Organic Chemistry Hybridization and Atomic and Molecular Orbitals Molecular Orbitals and Hybridizations. Bond Order= 3. • Atomic orbitals must have the proper symmetry and energy to interact and form molecular orbitals. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The double bond in C 2 consist of both Pi bonds because the four electrons are present in the two pi molecular orbitals. number of electrons in the sigma2p molecular orbital is. Join thousands of students and gain free access to 46 hours of Chemistry videos that follow the topics your textbook covers. Draw a molecular orbital diagram for Ar2+. The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure 9). The bond order you get from molecular orbital theory is more specific and accurate than seeing a single, double, or triple bond in a … Construct the Molecular orbital Diagram for N2 and then Identify the Bond order. Two superpositions of these two orbitals can be formed, one by summing the orbitals and the other by taking their difference. You can follow their steps in the video explanation above. By registering, I agree to the Terms of Service and Privacy Policy. Our expert Chemistry tutor, Dasha took 3 minutes and 39 seconds to solve this problem. 0 0. Our tutors have indicated that to solve this problem you will need to apply the Bond Order concept. 10) N 2. molecular orbital diagram for N2. H2, H2+, N2, O2, O22- etc.) orbitals the 3s and 3p rather than 2s and 2p. In the molecular orbital diagram for the molecular ion, $\ce{N_2{^{+}}}$, the number of electrons in the $\sigma_{2p}$ molecular orbital is : In Wolff‐Kishner reduction, the carbonyl group of aldehydes and ketones is converted into. Hello! Write the molecular orbital diagram of N2+ and calculate their bond order why nitrogen have different structure of molecular orbital theory An atomic orbital is monocentric while a molecular orbital is polycentric. Leila. Question: The Following Is The Molecular Orbital Diagram Showing The Energy Ordering For The Molecular Orbitals. Each hydrogen atom contributes one electron, and thus, "H"_2^(-) has three electrons while "H"_2^(+) has one. RE: Draw an MO diagram for the cyanide ion CN-1? Classification of Elements and Periodicity in Properties. Molecular orbital (MO) diagram for N2 and N2^- $-$\mathrm{p}$interaction moving from$\ce{Li2}$to$\ce{F2}$. CO, NO, CN-, … How long does this problem take to solve? JEE Main 2018: In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the σ2p molecular orbital is: (A) 0 (B) 1 (C) (b) 2 *2 2 *2 2 … The decomposition of$\ce{N2O4}$, in equilibrium mixture of$\ce{NO2(g)}$and$\ce{N2O4(g)}$, can be increased by : Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is : (Atomic wt. 99% (88 ratings) Problem Details. We illustrate how to use these points by constructing a molecular orbital energy-level diagram for F 2.We use the diagram in part (a) in Figure $$\PageIndex{1}$$; the n = 1 orbitals (σ 1 s and σ 1 s *) are located well below those of the n = 2 level and are not shown. 4- Theoretically it would not be possible to form a molecule from two hydrides because the anti-bonding and bonding orbitals would cancel each other out. What is the bond order of C2-?Express the bond order numerically.The blank molecular orbital diagram shown here applies to the valence of diatomic lit... Construct the molecular orbital diagram for He2 + and then identify the bond order. The atomic radiusis: Find out the solubility of$Ni(OH)_2$in 0.1 M NaOH. Atomic Orbitals Molecular Orbitals Atomic Orbitals 22 2p 2p Energy 2s 25 B2, C2, N2 Part A … 2. Structure anions of acids,$ HNO_{3}, H_{3}PO_{4} $and$ H_{2}SO_{4} $are, respectively. of Cl=35.5 u). Indicate if it is diamagnetic or paramagnetic. The radicals allyl: and pentadienyl: For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Explain What is the relationship between bond … Figure 8.37 This shows the MO diagrams for each homonuclear diatomic molecule in the second period. The first photo is straight from a 2006 edition Pearson general chemistry textbook, and it shows you what the molecular orbital (MO) diagram for O2 is. Bonding and Antibonding Molecular Orbitals. is it diamagnetic or paramagnetic. All molecular orbitals except the highest would be occupied by electron pairs, and the highest orbital ( u*) would be singly occupied, giving a bond order of 0.5. Molecular orbital theory gives a much more detailed and specific understanding of the type of bonding occurring through two atoms in a covalent bond. Write the molecular orbital diagram of N2+ and calculate their bond order why nitrogen have different structure of molecular orbital theory An atomic orbital is monocentric while a molecular … ... of a delocalized bonding pi molecular orbital system extending above the plane of the,sigma system of the carbonate ion and an antibonding pi molecular orbital system … If you forgot your password, you can reset it. Coordination compound - Coordination compound - Ligand field and molecular orbital theories: Since 1950 it has been apparent that a more complete theory, which incorporates contributions from both ionic and covalent bonding, is necessary to give an adequate account of the properties of coordination compounds. Password must contain at least one uppercase letter, a number and a special character. • Photoelectron spectroscopy provides useful information on the energies of atomic orbitals. What scientific concept do you need to know in order to solve this problem? IF5, the Iodine atom has 7 valence electrons in molecular orbitals it will form 5 bonds with 5 Cl atoms using 5 electrons from its molecular orbital, two electrons will form one lone pair on Iodine atom, which gives the square pyramidal geometry. What is the hybridization and geometry of the compound$ XeOF_4 $? If you need more Bond Order practice, you can also practice Bond Order practice problems. what is the bond order of this species. They weren't drawn that way on this diagram, but they should be. This ion has been observed in the gas phase. Here$ B $is bond pair and$ L $is lone pair. Because the antibonding ortibal is filled, it destabilizes the structure, making the "molecule" H 2 2-very non-stable. In hydrogen azide (above) the bond orders of bonds (I) and (II) are : The most polar compound among the following is : Identify the pair in which the geometry of the species is T-shape and squarepyramidal, respectively : The incorrect geometry is represented by : What will be the geometry of the compound$ MB_4L_2 $? Problem: Draw the molecular orbital diagram for N2- ion, and calculate the bond order. showing all of the electrons . Which of the following conversions involves change in both shape and hybridisation ? As discussed in class it is not a bond. The p-molecular orbitals are extended over the whole molecule. Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. For butadiene, the p manifold contains four electrons, leading to an electronic configuration of p 1 2 p 2 2. What professor is this problem relevant for? or may be of two different atoms of a heterodiatomic molecule/ion (e.g. Use This Diagram To Answer The Question. Cn Molecular Orbital Diagram. Its molecular orbitals are constructed from the valence-shell orbitals of each hydrogen atom, which are the 1s orbitals of the atoms. Vapour pressure of liquid ‘S’ at 500 K is approximately equal to :$\ce{(R=2 \, cal \, K^{-1} \, mol^{-1})}$. Get a better grade with hundreds of hours of expert tutoring videos for your textbook.$\begingroup$Yes, in the molecular orbitals of N2, the highest energy occupied orbital is sigma, while the two filled pi orbitals are below it. write molecular electron configuration for … Each horizontal line … (a) 2 2 2 *2 *2 *2. Draw the molecular orbital diagram for N2- ion, and calculate the bond order. 1 Answer Truong-Son N. Nov 2, 2015 If we build the MO diagram for #"N"_2#, it looks like this: First though, notice that the #p# orbitals are supposed to be degenerate. There are many descriptions online of why this is so and more generally how to determine the orbital energy rankings in diatomics$\endgroup$– … FREE Expert Solution. So, the bond order is zero. Our tutors rated the difficulty ofDraw the molecular orbital diagram for N2- ion, and calculat...as medium difficulty. Q. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule. • Next we’ll see that symmetry will help us treat larger molecules in Indicate if it is diamagnetic or paramagnetic. In order to test whether there may be$\ce{s}$-$\ce{p}$mixing and also to check whether there may be actually a bound state, I optimised this molecule (at PBE0-D3/def2-TZVPD) using NWChem 6.6 and analysed it using natural bonding orbitals (NBO 5.9).I found following minimum geometry: Figure 1: Minimum geometry found for$\ce{N2^{-2}}$(distance in angstrom) The atomic number of nitrogen is 7. 99% (88 ratings) FREE Expert Solution. Which of the following is the correct electron configuration for C2? 5 years ago. If each molecule of the hydrocarbon has one chlorine atom only; chlorine atoms present in 1 g of chlorohydrocarbon are : The gas phase reaction$\ce{2NO2(g) -> N2O4(g)}$is an exothermic reaction. Summary MO Theory • LCAO-MO Theory is a simple method for predicting the approximate electronic structure of molecules. Because the bond in Ar2 + would be weaker than in Cl 2, the Ar–Ar distance would be According to molecular orbital theory, which of the following will not be a viable molecule ? I have attached 2 photos to help answer this question. Based on our data, we think this problem is relevant for Professor Bindell's class at UCF. When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is : For which of the following processes,$\Delta S$is negative ? In the molecular orbital diagram for the molecular ion, N2+, the number of electrons in the σ2p molecular orbital is : All of the following share the same crystal structure except : The de-Broglie’s wavelength of electron present in first Bohr orbit of ‘H’ atom is :$\Delta_f G^{\circ}$at 500 K for substance ‘S’ in liquid state and gaseous state are$\ce{+ 100.7 \, kcal \, mol^{-1}}$and$\ce{+103 \, kcal \, mol^{-1}}$, respectively. For the energy diagram and pictorial view of the p-molecular orbitals - please see below: Allyl radical. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Which of the following set of molecules will have zero dipole moment ? Chemical bonding - Chemical bonding - Molecular orbitals of H2 and He2: The procedure can be introduced by considering the H2 molecule. N b = 8, Na= 2. Clutch Prep is not sponsored or endorsed by any college or university. A molecular orbital diagram, or MO diagram, is a qualitative descriptive tool explaining chemical bonding in molecules in terms of molecular orbital theory in general and the linear combination of atomic orbitals (LCAO) method in particular. Q. Construct the molecular orbital diagram for He2 + and then identify the bond order. The electronic configuration of N2 is KK (σ(2s)) 2 (σ ∗ (2s)) 2 (π(2p x)) 2 (π(2p y)) 2 (σ(2p z)) 2. Source(s): https://shrink.im/a0Ch6. I actually just covered this question in my gen chem class this week. Each hydrogen atom contributes one 1s atomic orbital, and thus, the orbitals overlap according to MO theory to form one sigma_(1s) and one sigma_(1s)^"*" MO by conservation of orbitals. Given that the ionic product of$Ni(OH)_2$is$2 \times 10^{-15}$. Which one of the following is not a property of physical adsorption ? An unknown chlorohydrocarbon has 3.55% of chlorine. This example was covered in class to show the rare exception that this single bond is a bond. 37. Compound X in the sigma2p molecular orbital ( MO ) diagram electronic configuration of 1... Us treat larger molecules in Hello of bonding occurring through two atoms a. Ratings ) FREE Expert Solution detailed and specific understanding of the following will not a. Electronic structure of molecules each homonuclear diatomic molecule in the second period ratings ) FREE Expert Solution us treat molecules. The correct electron configuration for C2 FREE Expert Solution - please see below: Allyl radical and of. Ll see that symmetry will help us treat larger molecules in Hello MO theory • theory... Be a viable molecule solve this problem you will need to apply the bond order are constructed from valence-shell! Provides useful information on the energies of atomic and molecular orbitals are extended over the molecule. More bond order change in both shape and hybridisation 2 photos to help this... Have the proper symmetry and energy to interact and form molecular orbitals, to! Identify the bond order the rare exception that this single bond is a simple molecular orbital diagram of n2 ion for predicting the electronic... Information on the energies of atomic and molecular orbitals of the atoms the bond order structure of molecules think. ( 88 ratings ) FREE Expert Solution: Draw the molecular orbital diagram ( Figure 9 ) more... Sequence of reactions: identify a molecule which does not exist because the antibonding ortibal is filled it... The energy Ordering for the cyanide ion CN-1 class to show the rare exception that this single bond is simple. Other by taking their difference view of the following is the correct electron configuration for C2 and special... Theory, which are the 1s orbitals of H2 and He2: the will! Shown in a molecular orbital diagram ( Figure 9 ) the energy Ordering for the cyanide ion?! A property of physical adsorption ( MO ) diagram and 2p the following is hybridization! Viable molecule it is not a bond, build the molecular orbital diagram for N2-,. College or university FREE access to 46 hours of Chemistry videos that follow the topics textbook! On our data, we think this problem is relevant for Professor Bindell 's at! Have attached 2 photos to help answer this question diagrams for each homonuclear molecule... Theory is a simple method for predicting the approximate electronic structure of molecules a property of physical adsorption not.. Order to solve this problem you will need to know in order to solve this problem for C2 this,... Four electrons, leading to an electronic configuration of p 1 2 2. Photos to help answer this question in my gen chem class this week and. As discussed in class it is not a property of physical adsorption 2 p 2 2.... • atomic orbitals must have the proper symmetry and energy to interact and form molecular orbitals are typically in! Us treat larger molecules in Hello identify the bond order diagrams for each homonuclear molecule! Involves change in both shape and hybridisation for each homonuclear diatomic molecule in the second period provides useful on! Of the p-molecular orbitals - please see below: Allyl radical has been observed in the sigma2p orbital... Molecules in Hello textbook covers our tutors rated the difficulty ofDraw the molecular orbital diagram for N2 and identify... Attached 2 photos to help answer this question orbitals - please see below: Allyl radical filled, destabilizes! Have indicated that to solve this problem is relevant for Professor Bindell 's at! The p-molecular orbitals are extended over the whole molecule a much more detailed and specific understanding of the atoms ortibal. 3S and 3p rather than 2s and 2p diagrams for each homonuclear diatomic molecule in the following involves! Simple method for predicting the approximate electronic structure of molecules see below: Allyl radical sigma2p... Well, build the molecular orbital ( MO ) diagram for C2 are typically in. M NaOH type of bonding occurring through two atoms in a molecular orbital diagram N2! 2 p 2 2 but they should be bond order to apply the order. Simple method for predicting the approximate electronic structure of molecules • Photoelectron spectroscopy provides useful information on the of. The ionic product of$ Ni ( OH ) _2 $is bond pair$! Least one uppercase letter, a number and a special character one uppercase letter a... Two orbitals can be introduced by considering the H2 molecule to the Terms of Service Privacy! A heterodiatomic molecule/ion ( e.g • Photoelectron spectroscopy provides useful information on the energies of atomic and orbitals! 2 * 2 * 2 * 2 * 2 a molecular orbital diagram for N2- ion, calculate! Our Expert Chemistry tutor, Dasha took 3 minutes and 39 seconds to solve this is. Levels of atomic and molecular orbitals a covalent bond they were n't drawn that way on diagram... Need to apply the bond order N2 and then identify the bond order,... Here $B$ is $2 \times 10^ { -15 }$ following the... That the ionic product of $Ni ( OH ) _2$ in 0.1 M NaOH orbitals the and... Ni ( OH ) _2 $is lone pair may be of two different atoms of a molecule/ion... Please see below: Allyl radical is not sponsored or endorsed by any college university. Considering the H2 molecule on the energies of atomic orbitals must have the proper symmetry and energy to interact form. Covered this question based on our data, we think this problem is relevant for Professor Bindell class. Radiusis: Find out the solubility of$ Ni ( OH ) _2 $in 0.1 M NaOH taking difference. Service and Privacy Policy students and gain FREE access to 46 hours of Chemistry videos that follow the your. Lcao-Mo theory is a bond will not be a viable molecule constructed from the valence-shell orbitals of each atom... Shown in a covalent bond$ B $is$ 2 \times 10^ { }. The difficulty ofDraw the molecular orbitals are extended over the whole molecule 0.1 M NaOH practice. The procedure can be introduced by considering the H2 molecule diatomic molecule in the following of... On our data, we think this molecular orbital diagram of n2 ion ( 88 ratings ) FREE Expert Solution in my gen class. For predicting the approximate electronic structure of molecules this week should be summary theory. Special character p-molecular orbitals are typically shown in a covalent bond to molecular orbital diagram for the molecular orbital MO! That symmetry will help us treat larger molecules in Hello cyanide ion CN-1... medium! That this single bond is a simple method for predicting the approximate electronic structure molecules! The molecular orbital diagram for He2 + and then identify the bond.! This shows the MO diagrams for each homonuclear diatomic molecule in the following is the hybridization and geometry of p-molecular... Bindell 's class at UCF write molecular electron configuration for C2 diatomic molecule in video... For He2 + and then identify the bond order the 1s orbitals of each hydrogen atom, which the! In both shape and hybridisation for Professor Bindell 's class at UCF, and calculate the bond order practice.! It destabilizes the structure, making the molecule '' H 2 2-very non-stable if need. Of physical adsorption and then identify the bond order diagram ( Figure 9 ) we think this problem is for. Be a viable molecule 39 seconds to solve this problem you will need to in! By considering the H2 molecule we ’ ll see that symmetry will help us treat larger molecules Hello... Not be a viable molecule well, build the molecular orbital diagram Showing the energy diagram and pictorial view the. And 3p rather than 2s and 2p, making the molecule '' H 2 2-very non-stable your covers! Terms of Service and Privacy Policy have indicated that to solve this problem relevant... Bonding - molecular orbitals of each hydrogen atom, which of the p-molecular orbitals - please see below Allyl... Orbital theory, which of the following conversions involves change in both shape and hybridisation endorsed by college! Through two atoms in a covalent bond to molecular orbital diagram for N2 and then identify bond. Summary MO theory • LCAO-MO theory is a bond of $Ni ( OH )$. 39 seconds to solve this problem not sponsored or endorsed by any college or university different atoms of a molecule/ion. Tutor, Dasha took 3 minutes and 39 seconds to solve this problem Terms Service! Need more bond order a bond $in 0.1 M NaOH destabilizes the structure making! In Hello ionic product of$ Ni ( OH ) _2 \$ in 0.1 M NaOH that.: the procedure can be introduced by considering the H2 molecule calculate bond! Or university ) diagram the gas phase attached 2 photos to help answer this question should be gives a more... Energy to interact and form molecular orbitals Bindell 's class at UCF electronic structure of molecules energies atomic! Ion, and calculate the bond order a property of physical adsorption over the whole molecule been! Join thousands of students and gain FREE access to 46 hours of Chemistry videos that follow topics. Molecule which does not exist He2 + and then identify the bond order practice, you can also bond. College or university theory, which of the following is the correct electron for. What scientific concept do you need to know in order to solve this problem you will to... Much more detailed and specific understanding of the p-molecular orbitals are constructed from the valence-shell orbitals of each atom. Have indicated that to solve this problem you will need to know in order to solve this problem will... The H2 molecule single bond is a simple method for predicting the approximate electronic structure of molecules He2 the. Leading to an electronic configuration of p 1 2 p 2 2 2.! Homonuclear diatomic molecule in the sigma2p molecular orbital diagram for N2- ion, and calculate the bond order practice....
|
2021-06-22 02:03:19
|
{"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": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5711374878883362, "perplexity": 1852.7312624874153}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488504969.64/warc/CC-MAIN-20210622002655-20210622032655-00603.warc.gz"}
|
http://yueliu.wang/2018/12/30/Newton-X-on-Hyak/
|
Newton-X is a general-purpose program package for excited-state molecular dynamics, which is used to simulate absorption spectrum with GAUSSIAN09 in our group.
There is a tutorial written by Andy Dang about why we need Newton-X, how to run it on Hyak and how to analyse its data. One can find another tutorial on Newton-X website, which is also the reference of the following content. A fast setup could be achieved by my python script. Scripts attached here can run both in python2 and in python3.
Newton-X Setup
Procedures
1. geometry and normal mode input
In the working directory prepare two files: one is the optimized geometry in gaussian input format and the other is the normal mode calculation output file (gaussian frequency log file, usually calcuated with b3lyp method):
opt.gjf freq.log
To work in the newton-x environment on Hyak, run module load contrib/newtonX or module load contrib/newton-x. Their difference is that newton-x contains other package like gaussin09.
4. newton-x working directory
Creat a new directory TDDFT_SPEC in the working directory, copy/move geom to it, copy/move freq.log to it with a new name freq.out:
5. energy and transition moment input
Move to the directory TDDFT_SPEC and create a new subdirectory called JOB_AD. Move into JOB_AD and prepare two files named, basis and gaussian$.$com.
basis contains the basis set information, like 6-31+g(d,p). gaussian$.$com is same with the very first optimized geometry file opt.gjf but with different link command lines and route card, like (%rwf and %nosave could be deleted):
Note that the subdirectory must be named with JOB_AD and the name of these two files must be basis and gaussian$.$com since Newton-X will search for them.
Back to the directory TDDFT_SPEC, use command $NX/nxinp and answer several quesitons by instructions to genetrate the newton-x input file initqp_input. Answers to the questions are 1 (Generate initial condition); 2 (Winger); numer of atoms; 300 (number of initial conditions); geom; 4 (gaussian output); freq.out; 0.975 (modified frequency); 310 (temperature); n; 1 (check energy); 1 (ground state); number of states; 1 ; 100 (de, width of restriction); 6.5; 0 (seed value); 1; 7 (exit), respectively. Here, the large “de” implies that this restriction will not be used. It can be imposed later on. 7. splitting jobs This step is to split the job among several computers (nodes), could be skipped and go directly to run the newton-x by $NX/initcond.pl > initcond.log.
To split the job, run NX/split_initcond.pl in the directory TDDFT_SPEC. Two questions will be asked, the first one is the number of directories to split the job and the second one is if the job run in a batch system. The answer to the second question is “n”. This program produces one file named split_initcond.log and a directory called INITIAL_CONDITIONS. If the answer to the first quesiton is 10, 10 subdirectories called I1,I2,…,I10 are inside INITIAL_CONDITIONS, each one containing a complete set of input files but with 30 (300 $\div$ 10) initial conditions and iseed=-1 not 0.
8. submit newton-x job
In every directory containing a complete set of input files (geom, freq.out, initqp_input and JOB_AD), create a sbatch file to submit the job to Hyak node.
Scripts
python script newtonx.py can do exactly what step 2 to step 8 do.
• Usage
• python newtonx.py geometry-gif-file freq-log
• Descriptions
• TDDFT_SPEC folder is created, which contains geom, freq.log, initqp_input, split_initcond.log, newtonx$.$sh, JOB_AD, INITIAL_CONDITIONS
• I1 I2 … in INITIAL_CONDITIONS, each subfolder containing a complete set input – geom, freq.out, initqp_input (iseed=1234,2468,…), JOB_AD and a sbatch file – nx_submit.sh
• nx_submit.sh is the sbatch file same with what is listed in step 8, but with partition=ckpt, account=stf-ckpt.
• newtonx$.$sh contains a list of bash command like: cd absolute-path-of-I1; sbatch -p ckpt -A stf-ckpt --time=20:00:00 nx_submit.sh
• after satisfied with everything, run bash newtonx.sh to submit all jobs to hyak nodes; the final partition, account and time are decided by the setting in newtonx$.$sh even if newtonx$.$sh is different from nx_submit.sh
• Note
• why ckpt?
• increasing the number of splitting jobs speeds up the task greatly
• ckpt queue is a good choice to run short jobs that finish within 4 hours
• why iseed=1234,2468,…,1234*n?
• different iseed values guarantee no repeated jobs
• iseed=-1 may generate a super large number not suit for ckpt queue
Newton-X Result
Procedures
After all sub-tasks finish, go to the directory INITIAL_CONDITIONS and run $NX/merge_initcond.pl. This program will ask the number of jobs to be merged and it will create a new directory called I_merged with merged results. All important data are in final_optput.1.N file, which contains transition information from state 1 to state N. spectrum simulation Move to this directory and proceed with the spectrum simulation by command $NX/nxinp. The answers to the questions it will ask are 5 (spectra); 1 (spectra); 1 (initial state); 2-N (array of final states); F (Absorption); 0 (no restriction); -1(read osc strength from file); local; 1 (random seed); lorentz; 0.1 (delta); 310 (temperature); 1 (refraction index); 0.005 (distance between consecutive points in the spectrum); 3 (kappa: the range of the spectrum is defined between Emin-kappa*delta and Emax+kappa*delta); 7 (exit), among which delta contronls the width of the curve.
The simulated cross section using a Lorentzian line shape with phenomenological broadening $\delta=0.1eV$ is written to cross-section.dat, containing four columns of data – DE/ev, lambda/nm, sigma/A^2 and +/-error/A^2.
Script
All the above steps can be achieved by script nxplot.py. (This script is to be updated, especially for plot function)
• Usage
• python nxplot.py
• run inside the directory INITIAL_CONDITIONS
• Descriptions
• check if jobs completed
• merge splitting jobs
• spectrum simulation
• extract the lamda and sigma columns if lamda within 0-1200nm from cross-section.dat and written to cross-section.tsv
• plot cross-section.tsv if in python3 environment
|
2021-04-10 21:04:19
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2220619171857834, "perplexity": 10228.714372944183}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038059348.9/warc/CC-MAIN-20210410210053-20210411000053-00485.warc.gz"}
|
https://web2.0calc.com/questions/how-do-i-solve-for-x-in-this-xy-5-x-7-i-cant-figure-it-out-and-i-really-need-help
|
+0
# how do I solve for x in this... xy+5=x+7? I cant figure it out and I really need help..
0
725
7
+49
how do I solve for x in this... xy+5=x+7? I cant figure it out and I really need help.
Oct 12, 2014
#7
+99356
+5
Hi Braksess,
I really like your dogged determination to understand this!
I think you are having problems with this
$$xy-x=x(y-1)$$
so that is the bit I am going to look at.
Look at the rectangle below.
The area of the whole rectangle is $$x\times y \quad units^2$$
I have cut a little rectangle off the end. The area of the cut off bit is $$1\times x=x \quad units^2$$
so the area that is left when you take away the litle bit at the end is
$$\\x\times y - x\\ or\; just\\ xy-x$$
The area that is left after you take away the bit at the end is displayed in yellow below.
The length of the top is y-1 (because 1 unit was cut off)
The side is still x units
The area is x(y-1)
SO $$xy-x=x(y-1)$$
LETS look at it the other way around
you probably know that 3(x-1)=3*x-3*1 = 3x-3
It works with letters too
$$x(y-1)=x\times y - x\times 1 = xy-x$$
NOW lets look at how to factorise it going in the other direction.
$$\\xy-x=x\times y - x\times 1\\ x is a common factor so we can factor it out and we will be left with y-1 in the bracket.\\ =x(y-1)$$
.
Oct 14, 2014
#1
+17746
+5
To solve for x, get all the x-terms to one side and everything else to the other side:
xy + 5 = x + 7
Subtract 5 from both sides:
xy = x + 2
Subtract x from both sides:
xy - x = 2
Factor out the x:
x(y - 1) = 2
Divide both sides by y - 1:
x = 2 / (y - 1)
Oct 12, 2014
#2
+49
+5
how did you keep the x after you subtracted it from both sides?
Oct 12, 2014
#3
+98196
+5
I think you're getting a little confused here, Braksess....here's what geno did
xy = x + 2 ( subtract x from both sides )
-x -x
------------
xy - x = 2 (note that he didn't "divide away" x......I think this is what you were thinking......!)
Can you take it from here??
Oct 13, 2014
#4
+49
0
y=2? i know im wrong but im still really confused, xy/-x=x-2/x im trying to find x. maybe its x-x=-y-2...so....0=y-2
Oct 13, 2014
#5
+98196
+5
Let's try this again.............
Note that we have......
xy + 5 = x + 7 now....just subtract 5 from both sides....
-5 = -5
--------------------
xy = x + 2
Now...I want to get rid of the x on the right side.....so I can just subtract it from both sides...so we have
xy = x + 2
-x -x
------------------
xy - x = 2 Notice how the x "disappeared from the right and ended up on the left as a "negative??"
Now...I just want to "factor the x on the left out of both terms....so we have....
x(y - 1) = 2 do you see that???
So this is like having
x * (y - 1) = 2 and since I want the "x" by itself on the left....we can divide both sides by (y - 1)
So
x* (y - 1)/(y-1) = 2/(y-1) and the (y -1)s on the left "cancel" leaving us with......
x = 2/(y - 1) !!!!
Does that help ???
(We can't get any "numerical" values for x or y, because we don't know the value of either !!!! )
Oct 13, 2014
#6
+49
+5
i still dont get how an x turned into a 1? i get it all up till there
Oct 13, 2014
#7
+99356
+5
Hi Braksess,
I really like your dogged determination to understand this!
I think you are having problems with this
$$xy-x=x(y-1)$$
so that is the bit I am going to look at.
Look at the rectangle below.
The area of the whole rectangle is $$x\times y \quad units^2$$
I have cut a little rectangle off the end. The area of the cut off bit is $$1\times x=x \quad units^2$$
so the area that is left when you take away the litle bit at the end is
$$\\x\times y - x\\ or\; just\\ xy-x$$
The area that is left after you take away the bit at the end is displayed in yellow below.
The length of the top is y-1 (because 1 unit was cut off)
The side is still x units
The area is x(y-1)
SO $$xy-x=x(y-1)$$
LETS look at it the other way around
you probably know that 3(x-1)=3*x-3*1 = 3x-3
It works with letters too
$$x(y-1)=x\times y - x\times 1 = xy-x$$
NOW lets look at how to factorise it going in the other direction.
$$\\xy-x=x\times y - x\times 1\\ x is a common factor so we can factor it out and we will be left with y-1 in the bracket.\\ =x(y-1)$$
Melody Oct 14, 2014
|
2019-03-26 01:15:31
|
{"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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7621810436248779, "perplexity": 3198.383334334453}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00140.warc.gz"}
|
https://rethomics.github.io/damr.html
|
# DAM2 data, in practice
## Aims
In this practical chapter, we will use a real experiment to learn how to:
• Set the zeitgeber reference (ZT0)
• Assess graphically the quality of the data
• Use good practices to exclude individuals from our experiments
## Prerequisites
library(devtools)
install_github("rethomics/behavr")
install_github("rethomics/damr")
install_github("rethomics/ggetho")
## Background
Drosophila Activity Monitors (DAMs) are a wildely used tool to monitor activity of fruit flies over several days. I am assuming that, if you are reading this tutorial, you are already familiar with the system, but I will make a couple of points clear before we start something more hands-on:
• This tutorial is about single beam DAM2 but will adapt very well to multibeam DAM5.
• We work with the raw data (the data from each monitor is in one single file, and all the monitor files are in the same folder)
## Getting the data
For this tutorial, you need to download some DAM2 data that we have made available. This is just a zip archive containing four files. Download and extract the files from the zip into a folder of your choice. Store the path in a variable. For instance, adapt something like:
DATA_DIR <- "C:/Where/My/Zip/Has/been/extracted
Check that all four files live there:
list.files(DATA_DIR, pattern= "*.txt|*.csv")
## [1] "metadata.csv" "Monitor11.txt" "Monitor14.txt" "Monitor64.txt"
For this exercise, we will work with the data and metadata in the same place. However, in practice, I recommend to:
• Have all raw data from your acquisition platform in the same place (possibly shared with others or a network drive)
• Have one folder per “experiment”. That is a folder that contains one metadata file, your R scripts, your figures regarding a set of consistent experiment.
For now, we can just set our working directory to DATA_DIR:
setwd(DATA_DIR)
## From experiment design to metadata
### Our toy experiment
In this example data, we were interested in comparing the behaviour of populations of fruit flies, according to their sex and genotype. We designed the experiment as shown is the figure above. In summary, we have:
• three genotypes (A, B and C)
• two sexes (male and female)
• two replicates (2017-07-01 -> 2017-07-04 and 2017-07-11 -> 2017-07-14)
• Altogether, 192 individuals
It is crucial that you have read metadata chapter to understand this part. Our goal is to encode our whole experiment in a single file in which:
• each row is an individual
• each column is a metavariable
Luckily for you, I have already put this file together for you as metadata.csv! Lets have a look at it (you can use R, excel or whatever you want). If you are using R, type this commands:
library(damr)
metadata
## file start_datetime stop_datetime region_id sex
## 1: Monitor11.txt 2017-07-01 08:00:00 2017-07-04 1 M
## 2: Monitor11.txt 2017-07-01 08:00:00 2017-07-04 2 M
## 3: Monitor11.txt 2017-07-01 08:00:00 2017-07-04 3 M
## 4: Monitor11.txt 2017-07-01 08:00:00 2017-07-04 4 M
## 5: Monitor11.txt 2017-07-01 08:00:00 2017-07-04 5 M
## ---
## 188: Monitor64.txt 2017-07-11 08:00:00 2017-07-14 28 F
## 189: Monitor64.txt 2017-07-11 08:00:00 2017-07-14 29 F
## 190: Monitor64.txt 2017-07-11 08:00:00 2017-07-14 30 F
## 191: Monitor64.txt 2017-07-11 08:00:00 2017-07-14 31 F
## 192: Monitor64.txt 2017-07-11 08:00:00 2017-07-14 32 F
## genotype replicate
## 1: A 1
## 2: A 1
## 3: A 1
## 4: A 1
## 5: A 1
## ---
## 188: C 2
## 189: C 2
## 190: C 2
## 191: C 2
## 192: C 2
Each of the 192 animals (rows) is defined by a set of mandatory columns (metavariables):
• file – the data file (monitor) that it has been recorded in
• start_datetime – the date and time (YYYY-MM-DD HH:MM:SS) of the start of the experiment. Time will be considered ZT0, see note
• stop_datetime – the last time point of the experiment (time is optional)
• region_id – the channel ([1, 32])
For our experiment, we also defined custom columns:
• sex – M and F for male and female, respectively
• genotype – A, B or C (I just made up the names for the sake of simplicity)
• replicate – so we can analyse how replicates differ from one another
Note that this format is very flexible and explicit. For instance, if we decided to do a third replicate, we would just need to add new rows. We could also add any condition we want as a new column (e.g. treatment, temperature, matting status and so on)
Linking is the one necessary step before loading the data. It allocates a unique identifier to each animal.
metadata <- link_dam_metadata(metadata, result_dir = DATA_DIR)
metadata
As result_dir, we just use the directory where the data lives, which you decided when you extracted your data (DATA_DIR).
Importantly, you do not need to cut the relevant parts of your DAM files (this is an error-prone step that should be avoided). In other words, no need to use the DAMFileScan utility or manipulate in any way the original data.
You can keep all the data in one file per monitor. rethomics will use start and stop datetime to find the appropriate part directly from your metadata.
In order to work with the data the last step is to load it into a behavr structure. To do that simply use load_dam function (as shown below). This function will store all data in dt (or any other given name)
dt <- load_dam(metadata)
summary(dt)
## behavr table with:
## 192 individuals
## 8 metavariables
## 2 variables
## 7.37472e+05 measurements
## 1 key (id)
That is it, all our data is loaded in dt.
## Note on datetime
### ZT0
In the circadian and sleep field, we need to align our data to a reference time of the day. Typically, when the light (would) turn on (ZT0). In damr, the time part of the start_datetime is used as a circadian reference. For instance, if you specify, in your metadata file 2017-01-01 09:00:00, you imply that ZT0 is at 09:00:00. The time is looked-up in the DAM file, so it will be at on same time zone settings as the computer that recorded the data.
### Start and stop time
When fetching some data, date and time are always inclusive.
When only the date is specified:
• start time will be at 00:00:00
• stop time will be at 23:59:59
For instance, start_date = 2017-01-01 and stop_date = 2017-01-01 retrieves all the data from the first of January 2017.
## Quality control
### Detecting anomalies
Immediatly after loading your data, it is a good idea to visualise it, in order to detect anomalies or at least to be sure that everything looks ok. We can use ggetho for that, for example the following code will create an activity tile plot, useful to detect dead animals.
library(ggetho)
## Loading required package: ggplot2
# I only show fisrt replicate
ggetho(dt[xmv(replicate) == 1 ], aes(z=activity)) +
stat_tile_etho() +
stat_ld_annotations()
Here, instead of ploting everything, I show how you can subset data according to metadata in order to display only replicate one (dt[xmv(replicate) == 1]). In practice, you could also plot everything. You can do a lot more with ggetho (see the visualisation chapter)
What does this tile plot tell us? Each row is an animal (and is labelled with its corresponding id). Each column is a 30min window. The colour intensity indicates the activity.
There are two things that we can immediatly notice:
• For most animals, the activity is rhythmic and synchronised with the light phase transisitions.
• Some animals are dead or missing. For instance take a look at channel 26 in Monitor64.txt.
In other chapters, we will learn how to group individuals, visualise and compute statistics.
### How to exclude animals?
We suggest to exclude animals a priori (e.g. because they died) by recording them as dead in the metadata. This way data is not modified or omited and can easily be recovered if needed. For instance, you can add a column status in your metadata file and put a default value such as "OK". If an animal is to be removed, you can replace "OK" by a reason (e.g. "dead", "escaped", …). Then, you can load your data without those animals load_dam_data(metadata[status == "OK"], ...). This practice has the advantage of making it very transparent, why some individuals where excluded. Also, as stated before, it can easily be reversed.
Finaly, we may want to apply a function on the data as it is loaded, in order to preprocess it, saving time. This pre-processing will annotate the data, i.e create new information (new columns) based on the original data. As an example, we can perform a sleep (bouts of immobility of 5 min or more), from our sleepr package (which you will have installed).
library(sleepr)
dt
##
##
## id file_info region_id
## <fctr> <list> <int>
## 1: 2017-07-01 08:00:00|Monitor11.txt|01 <list> 1
## 2: 2017-07-01 08:00:00|Monitor11.txt|02 <list> 2
## 3: 2017-07-01 08:00:00|Monitor11.txt|03 <list> 3
## 4: 2017-07-01 08:00:00|Monitor11.txt|04 <list> 4
## 5: 2017-07-01 08:00:00|Monitor11.txt|05 <list> 5
## ---
## 188: 2017-07-11 08:00:00|Monitor64.txt|28 <list> 28
## 189: 2017-07-11 08:00:00|Monitor64.txt|29 <list> 29
## 190: 2017-07-11 08:00:00|Monitor64.txt|30 <list> 30
## 191: 2017-07-11 08:00:00|Monitor64.txt|31 <list> 31
## 192: 2017-07-11 08:00:00|Monitor64.txt|32 <list> 32
## experiment_id start_datetime stop_datetime
## <char> <POSc> <char>
## 1: 2017-07-01 08:00:00|Monitor11.txt 2017-07-01 08:00:00 2017-07-04
## 2: 2017-07-01 08:00:00|Monitor11.txt 2017-07-01 08:00:00 2017-07-04
## 3: 2017-07-01 08:00:00|Monitor11.txt 2017-07-01 08:00:00 2017-07-04
## 4: 2017-07-01 08:00:00|Monitor11.txt 2017-07-01 08:00:00 2017-07-04
## 5: 2017-07-01 08:00:00|Monitor11.txt 2017-07-01 08:00:00 2017-07-04
## ---
## 188: 2017-07-11 08:00:00|Monitor64.txt 2017-07-11 08:00:00 2017-07-14
## 189: 2017-07-11 08:00:00|Monitor64.txt 2017-07-11 08:00:00 2017-07-14
## 190: 2017-07-11 08:00:00|Monitor64.txt 2017-07-11 08:00:00 2017-07-14
## 191: 2017-07-11 08:00:00|Monitor64.txt 2017-07-11 08:00:00 2017-07-14
## 192: 2017-07-11 08:00:00|Monitor64.txt 2017-07-11 08:00:00 2017-07-14
## sex genotype replicate
## <char> <char> <int>
## 1: M A 1
## 2: M A 1
## 3: M A 1
## 4: M A 1
## 5: M A 1
## ---
## 188: F C 2
## 189: F C 2
## 190: F C 2
## 191: F C 2
## 192: F C 2
##
## ====== DATA ======
##
## id t activity moving asleep
## <fctr> <num> <int> <lgcl> <lgcl>
## 1: 2017-07-01 08:00:00|Monitor11.txt|01 0 0 FALSE TRUE
## 2: 2017-07-01 08:00:00|Monitor11.txt|01 60 0 FALSE TRUE
## 3: 2017-07-01 08:00:00|Monitor11.txt|01 120 0 FALSE TRUE
## 4: 2017-07-01 08:00:00|Monitor11.txt|01 180 0 FALSE TRUE
## 5: 2017-07-01 08:00:00|Monitor11.txt|01 240 0 FALSE TRUE
## ---
## 737468: 2017-07-11 08:00:00|Monitor64.txt|32 230160 0 FALSE TRUE
## 737469: 2017-07-11 08:00:00|Monitor64.txt|32 230220 0 FALSE TRUE
## 737470: 2017-07-11 08:00:00|Monitor64.txt|32 230280 0 FALSE TRUE
## 737471: 2017-07-11 08:00:00|Monitor64.txt|32 230340 0 FALSE TRUE
## 737472: 2017-07-11 08:00:00|Monitor64.txt|32 230400 0 FALSE TRUE
As you can see, we now have additional columns in the data.
|
2019-09-19 16:45:09
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4407462775707245, "perplexity": 7929.7713168383325}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514573561.45/warc/CC-MAIN-20190919163337-20190919185337-00287.warc.gz"}
|
https://www.hackmath.net/en/math-problem/53813
|
# Evaluate fractions
The difference of 5 1/2 and 2/3 is added to the product of 5/6 and 1/2
x = 5.25
### Step-by-step explanation:
Did you find an error or inaccuracy? Feel free to write us. Thank you!
Tips to related online calculators
Need help to calculate sum, simplify or multiply fractions? Try our fraction calculator.
## Related math problems and questions:
• Fractions and mixed numerals
(a) Convert the following mixed numbers to improper fractions. i. 3 5/8 ii. 7 7/6 (b) Convert the following improper fraction to mixed number. i. 13/4 ii. 78/5 (c) Simplify these fractions to their lowest terms. i. 36/42 ii. 27/45 2. evaluate following ex
• Evaluate expression
Evaluate expression using BODMAS rule: 1 1/4+1 1/5÷3/5-5/8
3 3/4 + 2 3/5 + 5 1/2 Show your solution.
• Evaluate mixed expressions
Which of the following is equal to 4 and 2 over 3 divided by 3 and 1 over 2? A. 4 and 2 over 3 times 3 and 2 over 1 B. 14 over 3 times 2 over 7 C. 14 over 3 times 7 over 2 D. 42 over 3 times 2 over 31
• Evaluate 17
Evaluate 2x+6y when x=- 4/5 and y=1/3. Write your answer as a fraction or mixed number in simplest form.
• Simplify expression with mixed
When (7¹/3 + 2¾)÷(2 + 2¼ × 1¹/3) is simplified the result is?
• Product increased
When the product of 2/3 and 6/10 is increased by 2/5, the result is?
• The sum 12
The sum of 3 mixed numbers is 20 13/15. two of the numbers are 6 1/3 and 7 5/6. what is the third number?
• Complicated sum minus product
What must be subtracted from the sum of 3/8 and 5/16 to get difference equal to the product of 5/8 and 3/16?
• Quotient and product
If the quotient of [8/5 divided by 8/10] is added to the product of [8/14 x 7/12 x 3/8], what is the sum?
• Difference mixed fractions
What is the difference between 4 2/3 and 3 1/6?
• Simplify 7
Simplify. 7-1/3÷ 3-2/3 of 2+ 4-1/2÷ 2-1/4+ 1/2 solution and by step by step
• Sum of 18
Sum of two fractions is 4 3/7. If one of the fractions is 2 1/5 find the other one .
• The product 2
The product of two functions is 10. If one of them is 2 1/3, find the other one.
• Mixed numbers
Rewrite mixed numbers, so the fractions have the same denominator: 5 1/5 - 2 2/3
• Divide 11
Divide the product of 4 and 5/8 by 1 1/2. Write your answer as a mixed number.
• Quotient 3
If the quotient of 8/13 and 2 is subtracted from the product of 1 3/4 and 8/21, what is the difference?
|
2021-09-19 10:16:27
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.801480770111084, "perplexity": 2394.032260360054}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780056856.4/warc/CC-MAIN-20210919095911-20210919125911-00253.warc.gz"}
|
https://www.jiskha.com/archives/2014/09/17
|
# Questions Asked onSeptember 17, 2014
1. ## Ed. Tech
A blank is a way for students to keep track of information during research. A. Credible website B. Search log C. Boolean operator D. Online database Is the answer D?
2. ## chemistry
1a) why is it advisable to prepare a slightly more conc. soln. of kMnO4 when preparing its standard solution? b) what else do u have to do? (2). what will happen if NH4Cl is used to standardized the KMnO4. (3a). A student noticed traces of brown deposit
3. ## social studies check
1. Which of the following is NOT a reason that Oglethorpe and the trustees wanted to start the colony of Georgia? A. to help the poor who could not pay their bills B. to help the englich economy C. to form large plantations D. to provide defense I think B?
Maritza is baking cookies to bring to school that requires 3 eggs and 2 cups of sugar. To have enough for everyone she bought 12 eggs will 6 cups of sugar be enough? I tried multiplying 3 x 4 = 12 so then multiplied 2 x 4 which is 8 so there would not be
5. ## Math
In the number 436,621, which places contain digits where one digit is ten times as great as the other ?
6. ## math
Sarah has 10 stuffed animals. Explain two different ways she can group the stuffed animals so each group has the same number and no ... number and no stuffed animals ...
7. ## Algebra
Choose them compound inequality that represents all real numbers that are less then -3 or greater than or equal to 5.
8. ## Social Studies
2. Which of the following would describe a hardship of early Georgia life? A. Disease from the alligators B. Cold and rainy weather C. clearing tree roots from the farm land D. bad relations with the creek indians I honestly don't know the answer at all
9. ## Math
Students surveyed boys and girls separately to determine which sport was enjoyed the most. After completing the boy survey, it was determined that for every 3 boys who enjoyed soccer, 5 boys enjoyed basketball. The girl survey had a ratio of the number of
10. ## statistics
The data represents the daily rainfall (in inches) for one month. Construct a frequency distribution beginning with a lower class limit of 0.00 and use a class width of 0.20. The frequency of the first class limit (0.00 – 0.19) is: 0.44 0 0 0.24 0 0.65 0
11. ## Social Studies
Which of the following was a contribution of Tomochichi to the success of the Georgia colony? A. He kept peace between british and the creek Indians B. He had a cure for disease caused by the mosquitoes C. He kept most of the land that Oglethorpe wanted
12. ## Social Studies
Which of the following would describe John and Mary Musgrove? A. They owned a trading post close to the early Georgia settlement. B. The knew both the English and Indian Languages. C. They acted as interpreters for Oglethorpe and Tomochichi D. all of the
13. ## fractions
When Kaitlin divided a fraction by 1/2, the result was a mixed number. Was the original fraction less than or greater than 1/2? Explain reasoning. THANK YOU!
what is the effective interest rate of a simple discount note for $2700 at a bank discount rate of 12.25%, for 36 months? asked by Anonymous 15. ## math Luke is making pancakes. The recipe calls for 0.5 quart of milk and 2.5 cups of flour. He has 3/8 quart of milk and 18/8 cups of flour. Luke makes the recipe with the milk and flour that he has. Explain his error. asked by Eleyana 16. ## Algebra 1 A parking garage has floors above and below ground level. For a scavenger hunt, galas friends are given a list of objects they need to find on the third floor and fourth level below ground, the first and fourth above ground, and ground level. Question A . asked by Ms. Stephie 17. ## Algebra Marco is making mosaic garden stones using red, yellow, and blue tiles. He has 45 red tiles, 90 blue tiles, and 75 yellow tiles. Each stone must have the same number of each color tile. What is the greatest number of stones Marco can make? A. How many of asked by Ms. Stephie 18. ## physics A motorist is driving at 20m/s{\rm m/s} when she sees that a traffic light 200m{\rm m} ahead has just turned red. She knows that this light stays red for 15s{\rm s} , and she wants to reach the light just as it turns green again. It takes her 1.0s{\rm s} asked by ehiuof 19. ## curriculum Which of the following is an example of a project-approach activity for a preschool class? A. At 9:00 in the morning every day, half the children in the class do sorting and counting activities while the other half work on an art project. After 20 minutes, asked by Daniela 20. ## social studies check In which period of Native American history did permanent settlements develop for the first time? A. Paelo-Indian B. Archaic C. Woodland D. Mississipian Is the answer B? asked by chris 21. ## Language Arts What is a boolean operator? A. A credible website described as having current information about topic. B. A tool used to organize a search log when conducting online search. C. Search parameters set to identify specific information during internet searches asked by Gwen 22. ## physics You drop a rock from rest out of a window on the top floor of a building, 50.0 m above the ground. When the rock has fallen 3.80 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the asked by chloe 23. ## Physics A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and her takeoff point is 1.10 m above the pool. (a) How long are her feet in the air? s (b) What is her highest point above the asked by Joshua 24. ## physics You drop a rock from rest out of a window on the top floor of a building, 50.0 m above the ground. When the rock has fallen 3.80 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the asked by Anonymous 25. ## Create math equation Anthony leaves kingstons at 2:00 pm and drives to queensville, 160mi distant, at 45 mph. At 2:15 pm Helen leaves queensville and drives to kingston at 40mph. At what time do they pass each other on the road? asked by Mo 26. ## science am confused. don't know what formula to use for this problem. A car traveling at a speed of 30.0m/s encounters an emergency and comes to a complete stop. How much time will it take for the car to stop if it decelerates at -4.0m/s? asked by taylor 27. ## Fractions The length of a ribbon is 3/4 meter. Tina needs pieces measuring 1/3 meter for an art project. A)What is the greatest number of pieces measuring 1/3 meter that can be cut from the ribbon? B)How much ribbon will be left after Tina cuts the ribbon? Explain asked by Kymmy 28. ## World History What was the significance of the Code of Hammurabi? The laws were developed from an ancient Sumerian code of laws. The laws were displayed throughout the empire for all to see. The laws guaranteed many freedoms for ordinary citizens. The laws guaranteed asked by Sara 29. ## Math Aiden know the lenght of only one side of his garden. He says he will be able to find the area knowing only one side. Explain how this can be true. asked by Bryce 30. ## math kate davidson signed a simple discount note for$9000. the discount rate is 5%, and the term of the note is 18 months. what is the annual percentage yield (apy)?
Human reaction times are worsened by alcohol. How much farther would a drunk driver's car travel before he hits the brakes than a sober driver's car? Assume both cars are initially traveling at 47.0 mi/h, the sober driver takes .33 s and the drunk driver
32. ## REACTION KINECTICS
The autocatalytic reaction of A to form Q is one that accelerates with conversion. An example of this is shown below: A+Q~Q+Q However, the rate decreases at high conversion due to the depletion of reactant A. The liquid feed to the reactor contains 1 mol
33. ## REACTION KINECTICS
Propylene can be produced from the dehydrogenation of propane over a catalyst. At atmospheric pressure, what is the fraction of propane converted to propylene at 400, 500, and 600°C if equilibrium is reached at each temperature? Assume ideal behavior.
34. ## math
The distance remaining for a half marathon race over several minutes is shown in table. Use the information to determine the constant rate change in the minutes per mile. Table: Time (min) Distance Remaining (mi) 40 8 56 6 72 4 88 2
35. ## maths
by how much does the 3 in 34685 exceed the 3 in 23487
36. ## Social Studies
1. Which of the following was important about the city of Savannah in the settlement of early Georgia? A. It was laid out in a beautiful and orderly manner B. It was hight bluff, which was good for defense C. It was on the water D. All of the above I think
8. in which sentence is the word gesture used correctly ( 1 point) A.Smiling, she extended her hand in a gesture of friendship. B.Tired from a long day of work, I put my gesture down on the table wearily.**** C.He called for a gesture next Tuesday, and all
38. ## algebra
325 meters in 28 seconds? can you show me how to do this.
39. ## Urgent Spanish Check
A) Read the following sentence and choose which word is not correct in the context of the sentence. 1) La examen es difícil. La? 2) El árbol es verdes. el? B) The following sentence is written in singular form. Read the sentence, then choose the correct
40. ## ratio
What is the ratio in simplest form for 30 ounces:3 pints
41. ## physics
Which way and by what angle does the accelerometer in the figure deflect under the following conditions?(Figure 1). The cart is moving toward the left with a constant speed of 5.00m/s .
42. ## Physicsl Science
A heat pump releases 103.0 kcal as it removes 54.0 kcal at the evaporator coils. how much work does this heat pump ideally accomplish? Answer should be in KJ
59. ## Currculum
Which of the following is the most suitable theme for preschool age children? A. Predicting the weather B. Gravitational theory C. The hazards of electricity D. Plants in the classroom my answer is a.
60. ## physics
When sand is poured on a rotating disc what happen to its angular velocity ? explain
61. ## psychology
Carrie is a fifth grader at Gorrie Elementary school—a school near an affluent neighborhood in a suburb of a major city. Carrie, who lives with her divorced mother—a high school dropout with a steady job as a minimum wage housekeeper—seems to have
62. ## physics
You're sailboarding at 6.1m/s when a wind gust hits, lasting 6.3s accelerating your board at 0.48m/s2 at a 35∘ to your original direction
63. ## Maths
The school cafeteria serves tacos every sixth day and cheeseburgers every eight day. If tacos and cheeseburgers are served on the same day, how many days will it be before they are both on the menu again.
64. ## Math
Etienne owns a small recycling company that picks up empty glass bottles from restaurants. At the first restaurant, he picks up 50 bottles. At each restaurant after this, he picks up 4 more bottles than he picked up at the restaurant before. Assume that
65. ## REACTION KINECTICS
An alternative route to the production of propylene from propane would be through oxydehydrogenation At atmospheric pressure, what is the fraction of propane converted to propylene at 400, 500, and 600°C if equilibrium is reached at each temperature?
66. ## Math help.
1. 1 • 12 numerical expression variable expression; a is the variable variable expression; there is no variable variable expression; l is the variable 2. f ÷ 7 variable expression; there is no variable numerical expression variable expression; g is the
1.After doing considerable shopping, Eric has just decided what brand and type of athletic shoes to buy and where he's going to buy them. In what stage of the consumer buying decision process is Eric? Problem recognition Information search Evaluation of
68. ## physics
A uniform ladder, 3 m long and weighing 130 N, placed against a smooth wall, its lower end being 1.80 m from the wall. A 700N man stands on the upper most rung of the ladder located 0.45 m from the upper end of the ladder. a.) What are the vertical and
69. ## Math
Find three pairs of numbers for which the LCM (least common multiple) is the product of the two numbers; The pairs and LCMs are: 8 and 12: 24; 3 and 15: 15; 7 and 11: 77; 9 and 10: 90; 24 and 36: 72; 20 and 25: 100; 14 and 42: 42; 30 and 12: 60
70. ## chemistry
The solubility of acetanilide is 18.5 g in 100 mL of methanol at 0 °C, and 59.2 g in 100 mL of methanol at 60 °C. What is the maximum percent recovery that can be achieved for the recrystallization of acetanilide from methanol?
When managers disregard ethical concerns, the likely result is:
102. ## ss-maps *repost* (sorry!)
What is the difference between a road map and a bird's-eye view map? A. A road map shows fewer road names than a bird's eye view map. B. A road map looks more photo-realistic than a bird's-eye view map. C. A road map shows more physical features such as
103. ## math
stamjps are sold in bookiets of 100,50, and 10.find as many different ways to buy 200 stamps as you can. record your work in the chart. I don't understan what to do here. help
104. ## geography
which best describes why there is so little agriculture in the Appalachian Plateau region
105. ## Analytic Geometry
Find the vertices, foci, eccentricity and length of the latus rectum of the ellipse whose equation is x^2 + 9Y^2 = 9
106. ## Physics
1. A 0.50 kg ball is moving with a velocity of 8.0 m/s along a horizontal floor. It hits another ball with a mass of 0.80 kg and moving at 10.0 m/s in the opposite direction. If the first ball bounces back with a velocity of 12 m/s, with what velocity and
107. ## physics
A glider of length 12.7 cm moves on an air track with constant acceleration. A time interval of 0.563 s elapses between the moment when its front end passes a fixed point circled A along the track and the moment when its back end passes this point. Next, a
205. ## Physics
A light plane is flying at 25 degrees east of due magnetic north, while ascending at 15 degrees above the horizon; the plane is flying at 40 m/s. Find the planes celocity true to the delta x, delta y, and delta z?
207. ## Math
A bike rental charges a $4.00 rental fee plus$2.25per hour. Selena paid \$12.00 to rent a bike. How many hours did Selena rent the bike for? I can't figure out how to round the answer 4+2.25x=12 -4 -4 2.25x= 8 2.25 2.25 X= 3.555555555555555
208. ## math
The rectangular and polar coordinates of a point are (x, y) and (r, θ), where x = 4 and θ = 38◦ . Determine the value of r.
209. ## math
Michelle is now 50 miles ahead of John. Michelle is traveling at a constant rate. John is traveling in the same direction, at a rate 10 miles per hour faster than Michelle. In how many hours will John catch up to Micelle?
210. ## Mechanism and Machines
What is a practical example of a parallel crank 4 bar linkage?
211. ## Science
What are 3 facts that show ticks have a parasitic relationship with deer? 1..Ticks live on the skin of anomals
GCM OF 12 and 20. Please explain
213. ## Physics
vector A has a magnitude of 45.0 cm and makes an angle of 125 degrees with positive a –axis. What are the x- and y- components of this vector?
x^2+y^2+x-6y+9=0 Graph the circle using the center (h,k) and radius r. find the intercepts
215. ## Algebra help
Find an equation of the line L, where L is perpendicular to y=3x and passes through the point (1,3) Please help I'm not sure what formula to use or how to work out this problem
216. ## Math
An explicit formula for the general term for the arithmetic sequence 7/4,1, 1/4, -1/2?
217. ## Math
The sum of the geometric series 9 + 45 + 225 + 3 515 625 is ?
218. ## math
perform the division 3-8r2+r4 / r2-2
219. ## math
2/3 of the students watched tv. Of those students 3/4 watched a reality show. Of the students that watched the show 1/4 of them recorded the show. What fraction of the students watched and recorded a reality tv show? not sure what to do, thinking we need
221. ## math
your team has two x·sq-tiles, three x-tiles, and one unit tile on his desk and another person has one x·sq- tile, five x-tiles, and eight unit tiles on her desk. You decide to put all of the tiles together on one desk. What is the name for this new group
222. ## math
Write an equation relating the number of questions I get right on a test to the total score I will get if the test has 20 questions each with the same point value and the test is worth 100 points total. I know each question is worth 5 points so I would
223. ## philosopy
Milton Friedman observes that if the same argument that advocates of corporate social responsibility say should be brought to corporate shareholders is instead brought before union members, the logical bankruptcy of the argument would be clearly evident
224. ## english
combine these four sentences using complex compound sentences. • Debbie felt nervous and out-of-place at the party. • She sat by herself at the end of the sofa. • She didn’t want people to feel sorry for her. • She kept a strained smile on her
225. ## Help
Im trying to study but I need to know what (adverb phrases) are? 1.Please tell me what they are? 2.Please tell me how to identify them in a scentence? Please don't give me a link with a huge text of extra stuff...I just want to know how to identify them
226. ## English
1. Amanda is wearing big earrings. She is wearing high heels, too. 2. Amanda is wearing not only big earrings but also high heels. 3. Amanda is wearing high heels as well as big earrings. 4. Amanda is wearing both big earrings and high heels. (Are they all
227. ## English
1. Taeknwondo is an excellent exercise for both the mind and the body. 2. Taeknwondo is an excellent exercise for both the mind and body. 3. Taeknwondo is an excellent exercise for not only the mind but also the body. 4. Taeknwondo is an excellent exercise
228. ## Physical Science
You are driving. You have enough gas in your tank to go 50 miles. The next town is 72 kilometers away. Can you make it? Show work.
229. ## English
After weeks of convincing Mother finalyl agreed to watch the football game with the family on Sunday afternoon. "Ihope this doesn't INTERFERE with Sunday dinner," Mother warned "No," we explained. "We have the PORTABLE TV set up in the kitched so you won't
230. ## algebra 2
Find an equation of the linear function f having a slope of -2 if f(1)=3 My teacher never taught us this, please help me
231. ## Math
7x^2=8x
After his speech, the politician greeted his supporters. A.his supporters B.greeted his C.the politician D.After his speech
233. ## math help ASAP!!
simplify fully! 3x-2/x+4 = 18-4x/x^2-16 - 3x+2/4-x show all of your steps pleasee !!
234. ## career/jobs
Are there jobs that hire 14 year olds in New York?
235. ## Direct objects please check ASAP
1.D 2.C and A 3.canoe 4.D 5.B and E 2.Which words in the sentence are the direct objects? The woman grows tomatoes and cucumbers in her garden. Choose all answers that are correct. A.tomatoes B.woman C.cucumbers D.garden E.grows 3.Which word in the
236. ## Japan social empowerment
was the US successful in rebuilding Japan
237. ## Algebra 1
find the product (2x+1)(2x-1)
238. ## math
The rectangular and polar coordinates of a point are (x, y) and (r, θ), where x = 4 and θ = 38◦ . Determine the value of r.
239. ## art and creative development
children who are individualistic and dare to be noncomformists may be exhibiting personality traits that indicate: A. Convergent Thinking B. Creativity C. Rigidity D. Reality-Bound thinking
240. ## math
I have a 15ft board I need to cut into 2 1/8 inch pieces how many can I cut
241. ## Algebra1
-3|2x+6|=-12 How to solve for x, I relley need help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
242. ## Math
The sum of the first two terms of an arithmetic series is 15 and the sum of the next two terms is 43. What are the first four terms of the series?
243. ## Urgent Spanish Help
How are the I and E in Spanish similar, also how are they different to the I and E in English?
244. ## english
Combine these four sentences using complex compound sentences. • Mr. Blatt always mixes up the students’ names. • People get the feeling he doesn’t care. • Being called by another name is embarrassing. • It makes the students feel unimportant
245. ## Algebra 2
The line 4x + ky = -1 has a slope of 2/3. Find k. a. 3/2 b. 6 c. -3 d. -6
246. ## math
a plane flew for 4hrs, 30 miles per hour and an adverge speed of 515 mph how far did it fly?
247. ## Social Studies Help Please Mrs. Sue
What was the one product that the Salazburgers had some success with in Georgia that other settlers did not? A. Squash B. Rice C. corn D. Silk Is it A? I'm not sure
248. ## music
Sixteenth note x = whole note 1 8 # 2 16 3 1 - 4 4 4 Please Help Dr. Bob
249. ## Music
I know this is probably a dumb question but I have to write a report on a music artist and I like rock music so I want to know what artist should I choose? Marilyn Manson or Rob zombie? And can someone tell me some stuff about them?
250. ## Math
GCF of 12
step by step solving plsss im very law in math so plss to solve step by step thanks.On August 1,1990, Mr.Talamos borrows 9,500 pesos and agrees to pay the compounded amont on the day he pays the debt. If interest is at the rate of 5 1/2% compounded
252. ## Science
how might the extinction of a deer or a rabbit impact the ecosystem in which it lives?
|
2019-10-20 01:59:13
|
{"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.32897844910621643, "perplexity": 3042.810621120058}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986700560.62/warc/CC-MAIN-20191020001515-20191020025015-00391.warc.gz"}
|
https://www.rdocumentation.org/packages/base/versions/3.0.3/topics/file.info
|
file.info
0th
Percentile
Extract File Information
Utility function to extract information about files on the user's file systems.
Keywords
file
Usage
file.info(...)
Arguments
...
character vectors containing file paths. Tilde-expansion is done: see path.expand.
Details
What constitutes a ‘file’ is OS-dependent but includes directories. (However, directory names must not include a trailing backslash or slash on Windows.) See also the section in the help for file.exists on case-insensitive file systems.
The file ‘mode’ follows POSIX conventions, giving three octal digits summarizing the permissions for the file owner, the owner's group and for anyone respectively. Each digit is the logical or of read (4), write (2) and execute/search (1) permissions.
unix On most systems symbolic links are followed, so information is given about the file to which the link points rather than about the link. windows File modes are probably only useful on NTFS file systems, and it seems all three digits refer to the file's owner. The execute/search bits are set for directories, and for files based on their extensions (e.g., ‘.exe’, ‘.com’, ‘.cmd’ and ‘.bat’ files). file.access will give a more reliable view of read/write access availability to the R process.
UTF-8-encoded file names not valid in the current locale can be used.
Value
A data frame with row names the file names and columns
size
double: File size in bytes.
isdir
logical: Is the file a directory?
mode
integer of class "octmode". The file permissions, printed in octal, for example 644.
mtime, ctime, atime
integer of class "POSIXct": file modification, ‘last status change’ and last access times.
unix uidinteger: the user ID of the file's owner. gidinteger: the group ID of the file's group. unamecharacter: uid interpreted as a user name. grnamecharacter: gid interpreted as a group name. Unknown user and group names will be NA. windows execharacter: what sort of executable is this? Possible values are "no", "msdos", "win16", "win32", "win64" and "unknown". Note that a file (e.g. a script file) can be executable according to the mode bits but not executable in this sense.Entries for non-existent or non-readable files will be NA. unix The uid, gid, uname and grname columns may not be supplied on a non-POSIX Unix-alike system, and will not be on Windows.What is meant by the three file times depends on the OS and file system. On Windows native file systems ctime is the file creation time (something which is not recorded on most Unix-alike file systems). What is meant by ‘file access’ and hence the ‘last access time’ is system-dependent.The times are reported to an accuracy of seconds, and perhaps more on some systems. However, many file systems only record times in seconds, and some (e.g. modification time on FAT systems) are recorded in increments of 2 or more seconds.
Sys.readlink to find out about symbolic links, files, file.access, list.files, and DateTimeClasses for the date formats.
Sys.chmod to change permissions.
• file.info
Examples
library(base) ncol(finf <- file.info(dir())) # at least six ## Not run: finf # the whole list ## Those that are more than 100 days old : finf[difftime(Sys.time(), finf[,"mtime"], units = "days") > 100 , 1:4] file.info("no-such-file-exists")
Documentation reproduced from package base, version 3.0.3, License: Part of R 3.0.3
Community examples
Looks like there are no examples yet.
|
2020-02-26 16:58:27
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.27304807305336, "perplexity": 8598.74205184597}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146414.42/warc/CC-MAIN-20200226150200-20200226180200-00408.warc.gz"}
|
http://apostate.com/meaning-beauty-whimsy/markdown.html
|
Pubished:
Posted in: meaning-beauty-whimsy
Tagged as: Calepin / testing / markdown
## Markdown Test
Markdown Test
Calepin is based on Pelican, which uses the Python Markdown package with the codehilite, extra, and meta extensions.
#### Attribute Lists
{: #someid .someclass somekey=’some values’ }
Use a hash # to set the id of an element, and a dot . to add to an element’s list of classes.
##### Attributes: Block Level
To define attributes for a block level element, the attribute list should be defined on the last line of the block by itself.
This is a paragraph.
{: #an_id .a_class }
Results in
<p class="a_class" id="an_id">This is a paragraph.</p>
i.e.
This is a paragraph. {: #an_id .a_class }
The exception is headers, which are only allowed on one line.
A setext style header {: #setext}
=================================
### A hash style header ### {: #hash }
Results in
<h1 id="setext">A setext style header</h1>
<h3 id="hash">A hash style header</h3>
# A setext style header {: #setext}
### A hash style header ### {: #hash }
##### Attributes: Inline Level
To assign attributes inline, give the attribute list immediately after the inline element with no whitespace.
[link](http://example.com){: class="foo bar" title="Some title!" }
The above results in the following output:
<p><a class="foo bar" href="http://example.com" title="Some title!">link</a></p>
link{: class=“foo bar” title="Some title! }
#### Code highlight
##### SheBang (with path)
If the first line of the codeblock contains a shebang, the language is derived from that and line numbers are used. If the first line contains a shebang, but the shebang line does not contain a path (a single / or even a space), then that line is removed from the code block before processing. Line numbers are used.
#!/usr/bin/python
# Code goes here ...
# Code goes here …
Versus
#!python
# Code goes here ...
# Code goes here …
##### Colons
If the first line begins with three or more colons, the text following the colons identifies the language. The first line is removed from the code block before processing and line numbers are not used.
:::python
# Code goes here ...
:::python
# Code goes here …
Paragraph with normal text.
Emphasized text, also known as italic.
Strongly emphasized text, also known as bold.
Unnumbered list:
• Paul
• Whiteman
Numbered list:
1. Eddie
2. Cantor
Nested lists:
• Bix
• Spiritus
• Et
• Lasse
• Dahlquist
1. Hilmer
2. Borgeling
• Artibus
• Beiderbecke
## MarkdownExtra tests
### Headers with IDs (note: not name)
For example
Header 1 {#header1}
========
Results in
<h1 id="header1">Header 1</h1>
### Simple tables
This
First Header | Second Header
------------- | -------------
Content Cell | Content Cell
Content Cell | Content Cell
Produces
<table>
<tr>
</tr>
<tbody>
<tr>
<td>Content Cell</td>
<td>Content Cell</td>
</tr>
<tr>
<td>Content Cell</td>
<td>Content Cell</td>
</tr>
</tbody>
</table>
Content Cell Content Cell
Content Cell Content Cell
#### Alignment test
Alignment does not work
Item Value
Computer $1600 Phone$12
Pipe $1 #### Span-level formatting Function name Description help() Display the help window. destroy() Destroy your computer! #### Definition lists Definition lists work, but are not styled The code Apple : Pomaceous fruit of plants of the genus Malus in the family Rosaceae. Orange : The fruit of an evergreen tree of the genus Citrus. Results in <dl> <dt>Apple</dt> <dd>Pomaceous fruit of plants of the genus Malus in the family Rosaceae.</dd> <dt>Orange</dt> <dd>The fruit of an evergreen tree of the genus Citrus.</dd> </dl> Apple Pomaceous fruit of plants of the genus Malus in the family Rosaceae. Orange The fruit of an evergreen tree of the genus Citrus. #### Footnotes Footnotes work as expected: That's some text with a footnote.[^1] [^1]: And that's the footnote. Results in <p>That's some text with a footnote.<sup id="fnref:1"><a href="#fn:1" rel="footnote">1</a></sup></p> and later… <li id="fn:1"> <p>And that's the footnote. <a href="#fnref:1" rev="footnote" title="Jump back to footnote 1 in the text">↩</a></p> </li> That’s some text with a footnote.[1] #### Abbreviations Abbreviations also work. For example: *[HTML]: Hyper Text Markup Language *[W3C]: World Wide Web Consortium The HTML specification is maintained by the W3C. Results in: <p>The <abbr title="Hyper Text Markup Language">HTML</abbr> specification is maintained by the <abbr title="World Wide Web Consortium">W3C</abbr>.</p> Note that abbreviations are not styled. HTML: Hyper Text Markup Language [W3C]: World Wide Web Consortium The HTML specification is maintained by the W3C. ### Tests for common HTML elements #### Paragraphs & Inline Markup This paragraph contains elements. Up first is the anchor tag, which can occur as an a + href, or an <a> a + name, followed by an ABBR with a title value, followed by an ACRONYM with a title value, followed by a b (for bold) element, followed by a big element, followed by a Citation (cite) element, followed by a Computer code text (code) element, followed by a Deleted (del) element, followed by a Definition term (dfn) element, followed by Emphasized text (em), followed by an italics (i) element. This next paragraph also contains elements. Up first is the <ins> element, followed by keyboard text (kbd), followed by a <mark> element, followed by a Q element and a double nested Q (a quotation inside a quotation), followed by sample text (samp), followed by a small element, followed by a span element, followed by a strike element, followed by strong text (strong), followed by text in monospace font (tt), followed by an underlined element (u), followed by a Variable element (var, compared to code), all within a containing P. In order to test subscripts and superscripts inside running text, we need constructs like x1 and H2O and Mlle and 1st, and then some mathematical notations: ex, sin2 x, and nested exponents too: ex2 and f(x)g(x)a+b+c (where 2 and a+b+c should appear as exponents of exponents). ## A mechanism to upload new versions of your app First we have to clone our repository in our server: :::bash cd /var/www/ git clone your_repository_url.git your_app Then we will create a script in our server /etc/init.d/your_app that will allow us to stop/start/restart our app: :::bash #!/bin/bash DIR=/var/www/your_app PATH=/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin NODE_PATH=/usr/local/lib/node_modules NODE=/usr/local/bin/node test -x$NODE || exit 0
function start_app {
NODE_ENV=production nohup "$NODE" "$DIR/app.js" 1>>"$DIR/logs/your_app.log" 2>&1 & echo$! > "$DIR/pids/your_app.pid" } function stop_app { kill cat$DIR/pids/your_app.pid
}
case $1 in start) start_app ;; stop) stop_app ;; restart) stop_app start_app ;; *) echo "usage: your_app {start|stop}" ;; esac exit 0 The last part is to write a simple bash script for our development enviroment that will allows us to connect to our machine with ssh and pull changes. This is what I personally use: :::bash #!/bin/bash SERVER=( 'your_hostname.your_domain.com' ) DEPLOY_PATH=/var/www/your_app [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */_repository_url.git USER=deploy ENVIRONMENT=${1:-"production"}
REF=${2:-"master"} trap 'test -n "$SUCCESS" || echo " error: aborted"' EXIT
echo "* Deploying $ENVIRONMENT/$REF"
ssh $USER@$SERVER "cd $DEPLOY_PATH && \ git reset --hard && \ git checkout$REF && \
git pull && \
npm install && \
/etc/init.d/your_app stop"
SUCCESS=true
To deploy we just need to give execution permissions to that file (chmod +x deploy) and we are ready to go!
:::bash
git commit -am 'some changes'
git push
./deploy
## JSON
:::javascript
{
"author": "Merlin Mann",
"default_date_format": "%B %e, %Y",
"with_future_dates": true,
"with_pagination": true,
"output_source": true
}
## Python
:::python
def wrapwrite(text): sys.stdout.write(text.encode('utf8'))
## CSS
::CSS
* {
font-family: 'Baskerville', 'Garamond', 'Palatino', 'Palatino Linotype', 'Hoefler Text', 'Times New Roman', serif !important;
}
hr {
border: 10px solid red !important;
}
p {
}
body {
background-image: url("images/back40.gif");
}
h3 {
font-size: 60px !important;
}
## HTML
:::HTML
<h1>
yes, this
</h1>
<h2 id="newideas">
new ideas
</h2>
<ul>
<li>
omnioutliner for credit cards to change
<ul>
<li>
can also work with 1password and tags
</li>
</ul>
</li>
</ul>
## Ruby
:::ruby
require 'rubygems'
require 'chronic' # (gem install chronic)
if ARGV.length > 0
input = ARGV.join(" ").strip
else
print "Log entry: "
input = gets.strip
end
::ruby
require 'rubygems'
require 'chronic' # (gem install chronic)
if ARGV.length > 0
input = ARGV.join(" ").strip
else
print "Log entry: "
input = gets.strip
end
:ruby
require 'rubygems'
require 'chronic' # (gem install chronic)
if ARGV.length > 0
input = ARGV.join(" ").strip
else
print "Log entry: "
input = gets.strip
end
## Perl
:::perl
#!/usr/bin/python
import json
import datetime
import cgi
import os
import sys
input_filename = "/Users/%s/Library/Application Support/Google/Chrome/Default/Bookmarks" % user
# modify if necessary
output_filename = "/Users/%s/Documents/chrome-bookmarks.html" % user
:::python
>>> def hello():
print "Hello"
You are so good looking. Lorem ipsum dolor sit amet elit, sed diam nonummy nibh euismod tincidunt ut laoreet dolore magna aliquam erat volutpat. Ut wisi enim ad tation ullamcorper ut aliquip ex ea commodo consequat. Pretty fractal, right?
:::CSS
* {
font-family: 'Baskerville', 'Garamond', 'Palatino', 'Palatino Linotype', 'Hoefler Text', 'Times New Roman', serif !important;
}
hr {
border: 10px solid red !important;
}
p {
}
body {
background-image: url("images/back40.gif");
}
h3 {
font-size: 60px !important;
}
# h1
## h2
### Lessons Learnt
1. Don’t push the initial state to store it into the History, just use javascript for that job (or any other storage). One alternative will be use the replaceState call, but when the user clicks back/forward the javascript inside the page is executed, so, it’s simple to just load the default data from javascript.
:::javascript
var state = History.getState(), data = state.data;
if (data === null || $.isEmptyObject(data)) { data = { // Just load your default content }; } 2. Keep the content simple. This allow me to scale the use of the History API really easily in the hole web team. ### Functions Functions are the base of functional languages. In Haskell they receive 0 or more arguments and always return a value. Haskell syntax for functions is really simple: :::haskell numCardsInAPokerDeck = 53 numCards deck = length deck isAPokerDeck deck = (numCards deck) (==) numCardsInAPokerDeck biggerDeck d1 d2 = if (numCards d1) > (numCards d2) then d1 else d2 Function application is left associative. ### Purity Haskell functions are pure. This means that they cannot alter the state of the program by producing side-effects. Any pure function is also idempotent so it must return always the same result for a given set of arguments. How does Haskell achieve purity? - There are no variables: Functions can be used to define contant values but those can’t be modified. - Immutable data types: All the builtin data types defined in Haskell are immutable. Instead of altering the state of an instance, copies representing the new one are returned. I’m hoping that MathJax works. Let’s test it: $$A=\{a,b,c\}$$ $$f(x)=x^{3n+1}$$ Why green?! Displaystyle? $\int_a^b f(x)\ dx=F(b)-F(a)$ Here is the script that I’m using: :::Java <script type="text/javascript" src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML"> MathJax.Hub.Config({ extensions: ["tex2jax.js","TeX/AMSmath.js","TeX/AMSsymbols.js"], jax: ["input/TeX","output/HTML-CSS"], tex2jax: {inlineMath: [["$","$"]]} }); </script> Markdown Note: This document is itself written using Markdown; you can see the source for it by adding ‘.text’ to the URL. ## Overview ### Philosophy Markdown is intended to be as easy-to-read and easy-to-write as is feasible. Readability, however, is emphasized above all else. A Markdown-formatted document should be publishable as-is, as plain text, without looking like it’s been marked up with tags or formatting instructions. While Markdown’s syntax has been influenced by several existing text-to-HTML filters – including Setext, atx, Textile, reStructuredText, Grutatext, and EtText – the single biggest source of inspiration for Markdown’s syntax is the format of plain text email. To this end, Markdown’s syntax is comprised entirely of punctuation characters, which punctuation characters have been carefully chosen so as to look like what they mean. E.g., asterisks around a word actually look like *emphasis*. Markdown lists look like, well, lists. Even blockquotes look like quoted passages of text, assuming you’ve ever used email. ### Inline HTML Markdown’s syntax is intended for one purpose: to be used as a format for writing for the web. Markdown is not a replacement for HTML, or even close to it. Its syntax is very small, corresponding only to a very small subset of HTML tags. The idea is not to create a syntax that makes it easier to insert HTML tags. In my opinion, HTML tags are already easy to insert. The idea for Markdown is to make it easy to read, write, and edit prose. HTML is a publishing format; Markdown is a writing format. Thus, Markdown’s formatting syntax only addresses issues that can be conveyed in plain text. For any markup that is not covered by Markdown’s syntax, you simply use HTML itself. There’s no need to preface it or delimit it to indicate that you’re switching from Markdown to HTML; you just use the tags. The only restrictions are that block-level HTML elements – e.g. <div>, <table>, <pre>, <p>, etc. – must be separated from surrounding content by blank lines, and the start and end tags of the block should not be indented with tabs or spaces. Markdown is smart enough not to add extra (unwanted) <p> tags around HTML block-level tags. For example, to add an HTML table to a Markdown article: This is a regular paragraph. <table> <tr> <td>Foo</td> </tr> </table> This is another regular paragraph. Note that Markdown formatting syntax is not processed within block-level HTML tags. E.g., you can’t use Markdown-style *emphasis* inside an HTML block. Span-level HTML tags – e.g. <span>, <cite>, or <del> – can be used anywhere in a Markdown paragraph, list item, or header. If you want, you can even use HTML tags instead of Markdown formatting; e.g. if you’d prefer to use HTML <a> or <img> tags instead of Markdown’s link or image syntax, go right ahead. Unlike block-level HTML tags, Markdown syntax is processed within span-level tags. ### Automatic Escaping for Special Characters In HTML, there are two characters that demand special treatment: < and &. Left angle brackets are used to start tags; ampersands are used to denote HTML entities. If you want to use them as literal characters, you must escape them as entities, e.g. <, and &. Ampersands in particular are bedeviling for web writers. If you want to write about ‘AT&T’, you need to write ‘AT&T’. You even need to escape ampersands within URLs. Thus, if you want to link to: http://images.google.com/images?num=30&q=larry+bird you need to encode the URL as: http://images.google.com/images?num=30&q=larry+bird in your anchor tag href attribute. Needless to say, this is easy to forget, and is probably the single most common source of HTML validation errors in otherwise well-marked-up web sites. Markdown allows you to use these characters naturally, taking care of all the necessary escaping for you. If you use an ampersand as part of an HTML entity, it remains unchanged; otherwise it will be translated into &. So, if you want to include a copyright symbol in your article, you can write: © and Markdown will leave it alone. But if you write: AT&T Markdown will translate it to: AT&T Similarly, because Markdown supports inline HTML, if you use angle brackets as delimiters for HTML tags, Markdown will treat them as such. But if you write: 4 < 5 Markdown will translate it to: 4 < 5 However, inside Markdown code spans and blocks, angle brackets and ampersands are always encoded automatically. This makes it easy to use Markdown to write about HTML code. (As opposed to raw HTML, which is a terrible format for writing about HTML syntax, because every single < and & in your example code needs to be escaped.) ## Block Elements ### Paragraphs and Line Breaks A paragraph is simply one or more consecutive lines of text, separated by one or more blank lines. (A blank line is any line that looks like a blank line – a line containing nothing but spaces or tabs is considered blank.) Normal paragraphs should not be indented with spaces or tabs. The implication of the “one or more consecutive lines of text” rule is that Markdown supports “hard-wrapped” text paragraphs. This differs significantly from most other text-to-HTML formatters (including Movable Type’s “Convert Line Breaks” option) which translate every line break character in a paragraph into a <br /> tag. When you do want to insert a <br /> break tag using Markdown, you end a line with two or more spaces, then type return. Yes, this takes a tad more effort to create a <br />, but a simplistic “every line break is a <br />” rule wouldn’t work for Markdown. Markdown’s email-style blockquoting and multi-paragraph list items work best – and look better – when you format them with hard breaks. Markdown supports two styles of headers, Setext and atx. Setext-style headers are “underlined” using equal signs (for first-level headers) and dashes (for second-level headers). For example: This is an H1 ============= This is an H2 ------------- Any number of underlining =’s or -’s will work. Atx-style headers use 1–6 hash characters at the start of the line, corresponding to header levels 1–6. For example: # This is an H1 ## This is an H2 ###### This is an H6 # This is an H1 ## This is an H2 ###### This is an H6 Optionally, you may “close” atx-style headers. This is purely cosmetic – you can use this if you think it looks better. The closing hashes don’t even need to match the number of hashes used to open the header. (The number of opening hashes determines the header level.) : # This is an H1 # ## This is an H2 ## ### This is an H3 ###### ### Blockquotes Markdown uses email-style > characters for blockquoting. If you’re familiar with quoting passages of text in an email message, then you know how to create a blockquote in Markdown. It looks best if you hard wrap the text and put a > before every line: > This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, > consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. > Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. > > Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse > id sem consectetuer libero luctus adipiscing. This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. Markdown allows you to be lazy and only put the > before the first line of a hard-wrapped paragraph: > This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. > Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. Blockquotes can be nested (i.e. a blockquote-in-a-blockquote) by adding additional levels of >: > This is the first level of quoting. > > > This is nested blockquote. > > Back to the first level. This is the first level of quoting. This is nested blockquote. Back to the first level. Blockquotes can contain other Markdown elements, including headers, lists, and code blocks: > ## This is a header. > > 1. This is the first list item. > 2. This is the second list item. > > Here's some example code: > > return shell_exec("echo$input | $markdown_script"); ## This is a header. 1. This is the first list item. 2. This is the second list item. Here’s some example code: return shell_exec("echo$input | $markdown_script"); Any decent text editor should make email-style quoting easy. For example, with BBEdit, you can make a selection and choose Increase Quote Level from the Text menu. ### Lists Markdown supports ordered (numbered) and unordered (bulleted) lists. Unordered lists use asterisks, pluses, and hyphens – interchangably – as list markers: * Red * Green * Blue is equivalent to: + Red + Green + Blue and: - Red - Green - Blue • Red • Green • Blue Ordered lists use numbers followed by periods: 1. Bird 2. McHale 3. Parish It’s important to note that the actual numbers you use to mark the list have no effect on the HTML output Markdown produces. The HTML Markdown produces from the above list is: <ol> <li>Bird</li> <li>McHale</li> <li>Parish</li> </ol> If you instead wrote the list in Markdown like this: 1. Bird 1. McHale 1. Parish 1. Bird 2. McHale 3. Parish or even: 3. Bird 1. McHale 8. Parish you’d get the exact same HTML output. The point is, if you want to, you can use ordinal numbers in your ordered Markdown lists, so that the numbers in your source match the numbers in your published HTML. But if you want to be lazy, you don’t have to. If you do use lazy list numbering, however, you should still start the list with the number 1. At some point in the future, Markdown may support starting ordered lists at an arbitrary number. List markers typically start at the left margin, but may be indented by up to three spaces. List markers must be followed by one or more spaces or a tab. To make lists look nice, you can wrap items with hanging indents: * Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. * Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. • Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. • Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. But if you want to be lazy, you don’t have to: * Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. * Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. • Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. • Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. If list items are separated by blank lines, Markdown will wrap the items in <p> tags in the HTML output. For example, this input: * Bird * Magic will turn into: <ul> <li>Bird</li> <li>Magic</li> </ul> • Bird • Magic But this: * Bird * Magic will turn into: <ul> <li><p>Bird</p></li> <li><p>Magic</p></li> </ul> • Bird • Magic List items may consist of multiple paragraphs. Each subsequent paragraph in a list item must be indented by either 4 spaces or one tab: 1. This is a list item with two paragraphs. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. Donec sit amet nisl. Aliquam semper ipsum sit amet velit. 2. Suspendisse id sem consectetuer libero luctus adipiscing. It looks nice if you indent every line of the subsequent paragraphs, but here again, Markdown will allow you to be lazy: * This is a list item with two paragraphs. This is the second paragraph in the list item. You're only required to indent the first line. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. * Another item in the same list. To put a blockquote within a list item, the blockquote’s > delimiters need to be indented: * A list item with a blockquote: > This is a blockquote > inside a list item. To put a code block within a list item, the code block needs to be indented twice – 8 spaces or two tabs: * A list item with a code block: <code goes here> It’s worth noting that it’s possible to trigger an ordered list by accident, by writing something like this: 1986. What a great season. In other words, a number-period-space sequence at the beginning of a line. To avoid this, you can backslash-escape the period: 1986\. What a great season. ### Code Blocks Pre-formatted code blocks are used for writing about programming or markup source code. Rather than forming normal paragraphs, the lines of a code block are interpreted literally. Markdown wraps a code block in both <pre> and <code> tags. To produce a code block in Markdown, simply indent every line of the block by at least 4 spaces or 1 tab. For example, given this input: This is a normal paragraph: This is a code block. Markdown will generate: <p>This is a normal paragraph:</p> <pre><code>This is a code block. </code></pre> This is a normal paragraph: This is a code block. One level of indentation – 4 spaces or 1 tab – is removed from each line of the code block. For example, this: Here is an example of AppleScript: tell application "Foo" beep end tell will turn into: <p>Here is an example of AppleScript:</p> <pre><code>tell application "Foo" beep end tell </code></pre> A code block continues until it reaches a line that is not indented (or the end of the article). Within a code block, ampersands (&) and angle brackets (< and >) are automatically converted into HTML entities. This makes it very easy to include example HTML source code using Markdown – just paste it and indent it, and Markdown will handle the hassle of encoding the ampersands and angle brackets. For example, this: <div class="footer"> © 2004 Foo Corporation </div> will turn into: <pre><code><div class="footer"> &copy; 2004 Foo Corporation </div> </code></pre> Regular Markdown syntax is not processed within code blocks. E.g., asterisks are just literal asterisks within a code block. This means it’s also easy to use Markdown to write about Markdown’s own syntax. ### Horizontal Rules You can produce a horizontal rule tag (<hr />) by placing three or more hyphens, asterisks, or underscores on a line by themselves. If you wish, you may use spaces between the hyphens or asterisks. Each of the following lines will produce a horizontal rule: * * * *** ***** - - - --------------------------------------- ## Span Elements Markdown supports two style of links: inline and reference. In both styles, the link text is delimited by [square brackets]. To create an inline link, use a set of regular parentheses immediately after the link text’s closing square bracket. Inside the parentheses, put the URL where you want the link to point, along with an optional title for the link, surrounded in quotes. For example: This is [an example](http://example.com/ "Title") inline link. [This link](http://example.net/) has no title attribute. Will produce: <p>This is <a href="http://example.com/" title="Title"> an example</a> inline link.</p> <p><a href="http://example.net/">This link</a> has no title attribute.</p> This is an example inline link. This link has no title attribute. If you’re referring to a local resource on the same server, you can use relative paths: See my [About](/about/) page for details. Reference-style links use a second set of square brackets, inside which you place a label of your choosing to identify the link: This is [an example][id] reference-style link. You can optionally use a space to separate the sets of brackets: This is [an example] [id] reference-style link. Then, anywhere in the document, you define your link label like this, on a line by itself: [id]: http://example.com/ "Optional Title Here" That is: • Square brackets containing the link identifier (optionally indented from the left margin using up to three spaces); • followed by a colon; • followed by one or more spaces (or tabs); • followed by the URL for the link; • optionally followed by a title attribute for the link, enclosed in double or single quotes, or enclosed in parentheses. The following three link definitions are equivalent: [foo]: http://example.com/ "Optional Title Here" [foo]: http://example.com/ 'Optional Title Here' [foo]: http://example.com/ (Optional Title Here) Note: There is a known bug in Markdown.pl 1.0.1 which prevents single quotes from being used to delimit link titles. The link URL may, optionally, be surrounded by angle brackets: [id]: <http://example.com/> "Optional Title Here" You can put the title attribute on the next line and use extra spaces or tabs for padding, which tends to look better with longer URLs: [id]: http://example.com/longish/path/to/resource/here "Optional Title Here" Link definitions are only used for creating links during Markdown processing, and are stripped from your document in the HTML output. Link definition names may consist of letters, numbers, spaces, and punctuation – but they are not case sensitive. E.g. these two links: [link text][a] [link text][A] are equivalent. The implicit link name shortcut allows you to omit the name of the link, in which case the link text itself is used as the name. Just use an empty set of square brackets – e.g., to link the word “Google” to the google.com web site, you could simply write: [Google][] And then define the link: [Google]: http://google.com/ Because link names may contain spaces, this shortcut even works for multiple words in the link text: Visit [Daring Fireball][] for more information. And then define the link: [Daring Fireball]: http://daringfireball.net/ Link definitions can be placed anywhere in your Markdown document. I tend to put them immediately after each paragraph in which they’re used, but if you want, you can put them all at the end of your document, sort of like footnotes. Here’s an example of reference links in action: I get 10 times more traffic from [Google] [1] than from [Yahoo] [2] or [MSN] [3]. [1]: http://google.com/ "Google" [2]: http://search.yahoo.com/ "Yahoo Search" [3]: http://search.msn.com/ "MSN Search" I get 10 times more traffic from Google than from Yahoo or MSN. Using the implicit link name shortcut, you could instead write: I get 10 times more traffic from [Google][] than from [Yahoo][] or [MSN][]. [google]: http://google.com/ "Google" [yahoo]: http://search.yahoo.com/ "Yahoo Search" [msn]: http://search.msn.com/ "MSN Search" Both of the above examples will produce the following HTML output: <p>I get 10 times more traffic from <a href="http://google.com/" title="Google">Google</a> than from <a href="http://search.yahoo.com/" title="Yahoo Search">Yahoo</a> or <a href="http://search.msn.com/" title="MSN Search">MSN</a>.</p> For comparison, here is the same paragraph written using Markdown’s inline link style: I get 10 times more traffic from [Google](http://google.com/ "Google") than from [Yahoo](http://search.yahoo.com/ "Yahoo Search") or [MSN](http://search.msn.com/ "MSN Search"). I get 10 times more traffic from Google than from Yahoo or MSN. The point of reference-style links is not that they’re easier to write. The point is that with reference-style links, your document source is vastly more readable. Compare the above examples: using reference-style links, the paragraph itself is only 81 characters long; with inline-style links, it’s 176 characters; and as raw HTML, it’s 234 characters. In the raw HTML, there’s more markup than there is text. With Markdown’s reference-style links, a source document much more closely resembles the final output, as rendered in a browser. By allowing you to move the markup-related metadata out of the paragraph, you can add links without interrupting the narrative flow of your prose. ### Emphasis Markdown treats asterisks (*) and underscores (_) as indicators of emphasis. Text wrapped with one * or _ will be wrapped with an HTML <em> tag; double *’s or _’s will be wrapped with an HTML <strong> tag. E.g., this input: *single asterisks* _single underscores_ **double asterisks** __double underscores__ will produce: <em>single asterisks</em> <em>single underscores</em> <strong>double asterisks</strong> <strong>double underscores</strong> single asterisks single underscores double asterisks double underscores You can use whichever style you prefer; the lone restriction is that the same character must be used to open and close an emphasis span. Emphasis can be used in the middle of a word: un*frigging*believable But if you surround an * or _ with spaces, it’ll be treated as a literal asterisk or underscore. To produce a literal asterisk or underscore at a position where it would otherwise be used as an emphasis delimiter, you can backslash escape it: \*this text is surrounded by literal asterisks\* ### Code To indicate a span of code, wrap it with backtick quotes (). Unlike a pre-formatted code block, a code span indicates code within a normal paragraph. For example: Use the printf() function. will produce: <p>Use the <code>printf()</code> function.</p> Use the printf() function. To include a literal backtick character within a code span, you can use multiple backticks as the opening and closing delimiters: There is a literal backtick () here. which will produce this: <p><code>There is a literal backtick () here.</code></p> There is a literal backtick () here. The backtick delimiters surrounding a code span may include spaces – one after the opening, one before the closing. This allows you to place literal backtick characters at the beginning or end of a code span: A single backtick in a code span: A backtick-delimited string in a code span: foo will produce: <p>A single backtick in a code span: <code></code></p> <p>A backtick-delimited string in a code span: <code>foo</code></p> A single backtick in a code span: A backtick-delimited string in a code span: foo With a code span, ampersands and angle brackets are encoded as HTML entities automatically, which makes it easy to include example HTML tags. Markdown will turn this: Please don't use any <blink> tags. into: <p>Please don't use any <code><blink></code> tags.</p> You can write this: — is the decimal-encoded equivalent of —. to produce: <p><code>&#8212;</code> is the decimal-encoded equivalent of <code>&mdash;</code>.</p> ### Images Admittedly, it’s fairly difficult to devise a “natural” syntax for placing images into a plain text document format. Markdown uses an image syntax that is intended to resemble the syntax for links, allowing for two styles: inline and reference. Inline image syntax looks like this:   That is: • An exclamation mark: !; • followed by a set of square brackets, containing the alt attribute text for the image; • followed by a set of parentheses, containing the URL or path to the image, and an optional title attribute enclosed in double or single quotes. Reference-style image syntax looks like this: ![Alt text][id] Where “id” is the name of a defined image reference. Image references are defined using syntax identical to link references: [id]: url/to/image "Optional title attribute" As of this writing, Markdown has no syntax for specifying the dimensions of an image; if this is important to you, you can simply use regular HTML <img> tags. ## Miscellaneous Markdown supports a shortcut style for creating “automatic” links for URLs and email addresses: simply surround the URL or email address with angle brackets. What this means is that if you want to show the actual text of a URL or email address, and also have it be a clickable link, you can do this: <http://example.com/> Markdown will turn this into: <a href="http://example.com/">http://example.com/</a> http://example.com/ Automatic links for email addresses work similarly, except that Markdown will also perform a bit of randomized decimal and hex entity-encoding to help obscure your address from address-harvesting spambots. For example, Markdown will turn this: <[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */> into something like this: <a href="mailto:addre ss@example.co m">address@exa mple.com</a> address@example.com which will render in a browser as a clickable link to “[email protected]”. (This sort of entity-encoding trick will indeed fool many, if not most, address-harvesting bots, but it definitely won’t fool all of them. It’s better than nothing, but an address published in this way will probably eventually start receiving spam.) ### Backslash Escapes Markdown allows you to use backslash escapes to generate literal characters which would otherwise have special meaning in Markdown’s formatting syntax. For example, if you wanted to surround a word with literal asterisks (instead of an HTML <em> tag), you can use backslashes before the asterisks, like this: \*literal asterisks\* *literal asterisks* Markdown provides backslash escapes for the following characters: \ backslash backtick * asterisk _ underscore {} curly braces [] square brackets () parentheses # hash mark + plus sign - minus sign (hyphen) . dot ! exclamation mark \ backslash backtick * asterisk _ underscore {} curly braces [] square brackets () parentheses # hash mark • plus sign • minus sign (hyphen) . dot ! exclamation mark # MultiMarkdown Prototype table Grouping First Header Second Header Third Header Content Long Cell Content Cell Cell New section More Data And more And more Here is some text containing a footnote.[2] This is a statement that should be attributed to its source[p. 23, 3]. This is a statement that should be attributed to its source[3] And following is the description of the reference to be used in the bibliography. ### Def lists Apple Pomaceous fruit of plants of the genus Malus in the family Rosaceae. An american computer company. Orange The fruit of an evergreen tree of the genus Citrus. ### Overview An oft-requested feature was the ability to have Markdown automatically handle withindocument links as easily as it handled external links. To this aim, I added the ability to interpret [Some Text][] as a cross-link, if a header named âSome Textâ exists. As an example, [Metadata][] will take you to the section describing metadata (section 4.1). Alternatively, you can include an optional label of your choosing to help disambiguate cases where multiple headers have the same title: This is a formatted ![image][image] and a [link][] with attributes. [image]: http://path.to/image “Image title” width=40px height=400px [link]: http://path.to/link.html “Some Link” class=external style=“border: solid black 1px;” ### Images ### Backslash Escapes Markdown allows you to escape special characters that would otherwise be interpreted as Markdown code or HTML with a backslash. For example, if you want to escape Paragraph with normal text. Emphasized text, also known as italic. Strongly emphasized text, also known as bold. ## MarkdownExtra tests ## Calepin Style #### Paragraphs & Inline Markup This paragraph contains elements. Up first is the anchor tag, which can occur as an <a> a + href, or an <a> a + name, followed by an ABBR with a title value, followed by an ACRONYM with a title value, followed by a b (for bold) element, followed by a big element, followed by a Citation (cite) element, followed by a Computer code text (code) element, followed by a Deleted (del) element, followed by a Definition term (dfn) element, followed by Emphasized text (em), followed by an italics (i) element. This next paragraph also contains elements. Up first is the <ins> element, followed by keyboard text (kbd), followed by a <mark> element, followed by a Q element and a double nested Q (a quotation inside a quotation), followed by sample text (samp), followed by a small element, followed by a span element, followed by a strike element, followed by strong text (strong), followed by text in monospace font (tt), followed by an underlined element (u), followed by a Variable element (var, compared to code), all within a containing P. In order to test subscripts and superscripts inside running text, we need constructs like x1 and H2O and Mlle and 1st, and then some mathematical notations: ex, sin2 x, and nested exponents too: ex2 and f(x)g(x)a+b+c (where 2 and a+b+c should appear as exponents of exponents). # h1 ## h2 ### h3 #### h4 ##### h5 ###### h6 # Other tricks ### Using MathJax Inline math should come between $ dollars like $$\sum_{k=1}^n k = \frac{n(n+1)}{2}$$ and $$\frac{1+\sqrt{5}}{2} = [1; 1, 1, 1, \ldots]$$. Perhaps $$A=\{a,b,c\}$$ or $$f(x)=x^{3n+1}$$ are more your style. $$\forall x \in Y$$. $\int_a^b f(x)\ dx=F(b)-F(a)$
Display math goes between \$ and \$ slash-square-brackets, like this: [\sum{k=1}n k = \frac{n(n+1)}{2}] and [ \int{-\infty}{\infty} e{-\frac{x2}{2}} \ dx = \sqrt{2\pi} ]
The code:
Inline math should come between $ dollars like$\sum_{k=1}^n k = \frac{n(n+1)}{2}$and$\frac{1+\sqrt{5}}{2} = [1; 1, 1, 1, \ldots]$. Perhaps$A=\\{a,b,c\\}$or$f(x)=x^{3n+1}$are more your style.$\forall x \in Y$. $$\int_a^b f(x)\ dx=F(b)-F(a)$$ Display math goes between \$ and \$ slash-square-brackets, like this: $\sum_{k=1}^n k = \frac{n(n+1)}{2}$ and $\int_{-\infty}^{\infty} e^{-\frac{x^2}{2}} \ dx = \sqrt{2\pi}$ To get MathJax to display, place the following in your Markdown or reStructuredText source: #!javascript <style type="text/css"> .math span { color: inherit; } </style> <script language="javascript"> if(window.MathJax===undefined){ var script = document.createElement("script"); script.type = "text/javascript"; script.src = "http://cdn.mathjax.org/mathjax/latest/MathJax.js"; var config = 'MathJax.Hub.Config({' + 'extensions: ["tex2jax.js","TeX/AMSmath.js","TeX/AMSsymbols.js"],' + 'tex2jax: { '+ 'inlineMath: [ ["$","$"]], '+ 'displayMath: [["\$","\$"]],'+ 'processEscapes: true '+ '},' + 'jax: ["input/TeX","output/HTML-CSS"]' + '});' + 'MathJax.Hub.Startup.onload();'; if (window.opera) script.innerHTML = config; else script.text = config; document.getElementsByTagName("head")[0].appendChild(script); } </script> Note: based on Christian Perfect’s code from http://christianperfect.calepin.co/mathjax-test.html and http://christianperfect.calepin.co/a-second-post.html. Markdown Note: This document is itself written using Markdown; you can see the source for it by adding ‘.text’ to the URL. ## Overview ### Philosophy Markdown is intended to be as easy-to-read and easy-to-write as is feasible. Readability, however, is emphasized above all else. A Markdown-formatted document should be publishable as-is, as plain text, without looking like it’s been marked up with tags or formatting instructions. While Markdown’s syntax has been influenced by several existing text-to-HTML filters – including Setext, atx, Textile, reStructuredText, Grutatext, and EtText – the single biggest source of inspiration for Markdown’s syntax is the format of plain text email. To this end, Markdown’s syntax is comprised entirely of punctuation characters, which punctuation characters have been carefully chosen so as to look like what they mean. E.g., asterisks around a word actually look like *emphasis*. Markdown lists look like, well, lists. Even blockquotes look like quoted passages of text, assuming you’ve ever used email. ### Inline HTML Markdown’s syntax is intended for one purpose: to be used as a format for writing for the web. Markdown is not a replacement for HTML, or even close to it. Its syntax is very small, corresponding only to a very small subset of HTML tags. The idea is not to create a syntax that makes it easier to insert HTML tags. In my opinion, HTML tags are already easy to insert. The idea for Markdown is to make it easy to read, write, and edit prose. HTML is a publishing format; Markdown is a writing format. Thus, Markdown’s formatting syntax only addresses issues that can be conveyed in plain text. For any markup that is not covered by Markdown’s syntax, you simply use HTML itself. There’s no need to preface it or delimit it to indicate that you’re switching from Markdown to HTML; you just use the tags. The only restrictions are that block-level HTML elements – e.g. <div>, <table>, <pre>, <p>, etc. – must be separated from surrounding content by blank lines, and the start and end tags of the block should not be indented with tabs or spaces. Markdown is smart enough not to add extra (unwanted) <p> tags around HTML block-level tags. For example, to add an HTML table to a Markdown article: This is a regular paragraph. <table> <tr> <td>Foo</td> </tr> </table> This is another regular paragraph. Note that Markdown formatting syntax is not processed within block-level HTML tags. E.g., you can’t use Markdown-style *emphasis* inside an HTML block. Span-level HTML tags – e.g. <span>, <cite>, or <del> – can be used anywhere in a Markdown paragraph, list item, or header. If you want, you can even use HTML tags instead of Markdown formatting; e.g. if you’d prefer to use HTML <a> or <img> tags instead of Markdown’s link or image syntax, go right ahead. Unlike block-level HTML tags, Markdown syntax is processed within span-level tags. ### Automatic Escaping for Special Characters In HTML, there are two characters that demand special treatment: < and &. Left angle brackets are used to start tags; ampersands are used to denote HTML entities. If you want to use them as literal characters, you must escape them as entities, e.g. <, and &. Ampersands in particular are bedeviling for web writers. If you want to write about ‘AT&T’, you need to write ‘AT&T’. You even need to escape ampersands within URLs. Thus, if you want to link to: http://images.google.com/images?num=30&q=larry+bird you need to encode the URL as: http://images.google.com/images?num=30&q=larry+bird in your anchor tag href attribute. Needless to say, this is easy to forget, and is probably the single most common source of HTML validation errors in otherwise well-marked-up web sites. Markdown allows you to use these characters naturally, taking care of all the necessary escaping for you. If you use an ampersand as part of an HTML entity, it remains unchanged; otherwise it will be translated into &. So, if you want to include a copyright symbol in your article, you can write: © and Markdown will leave it alone. But if you write: AT&T Markdown will translate it to: AT&T Similarly, because Markdown supports inline HTML, if you use angle brackets as delimiters for HTML tags, Markdown will treat them as such. But if you write: 4 < 5 Markdown will translate it to: 4 < 5 However, inside Markdown code spans and blocks, angle brackets and ampersands are always encoded automatically. This makes it easy to use Markdown to write about HTML code. (As opposed to raw HTML, which is a terrible format for writing about HTML syntax, because every single < and & in your example code needs to be escaped.) ## Block Elements ### Paragraphs and Line Breaks A paragraph is simply one or more consecutive lines of text, separated by one or more blank lines. (A blank line is any line that looks like a blank line – a line containing nothing but spaces or tabs is considered blank.) Normal paragraphs should not be indented with spaces or tabs. The implication of the “one or more consecutive lines of text” rule is that Markdown supports “hard-wrapped” text paragraphs. This differs significantly from most other text-to-HTML formatters (including Movable Type’s “Convert Line Breaks” option) which translate every line break character in a paragraph into a <br /> tag. When you do want to insert a <br /> break tag using Markdown, you end a line with two or more spaces, then type return. Yes, this takes a tad more effort to create a <br />, but a simplistic “every line break is a <br />” rule wouldn’t work for Markdown. Markdown’s email-style blockquoting and multi-paragraph list items work best – and look better – when you format them with hard breaks. Markdown supports two styles of headers, Setext and atx. Setext-style headers are “underlined” using equal signs (for first-level headers) and dashes (for second-level headers). For example: This is an H1 ============= This is an H2 ------------- Any number of underlining =’s or -’s will work. Atx-style headers use 1–6 hash characters at the start of the line, corresponding to header levels 1–6. For example: # This is an H1 ## This is an H2 ###### This is an H6 # This is an H1 ## This is an H2 ###### This is an H6 Optionally, you may “close” atx-style headers. This is purely cosmetic – you can use this if you think it looks better. The closing hashes don’t even need to match the number of hashes used to open the header. (The number of opening hashes determines the header level.) : # This is an H1 # ## This is an H2 ## ### This is an H3 ###### ### Blockquotes Markdown uses email-style > characters for blockquoting. If you’re familiar with quoting passages of text in an email message, then you know how to create a blockquote in Markdown. It looks best if you hard wrap the text and put a > before every line: > This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, > consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. > Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. > > Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse > id sem consectetuer libero luctus adipiscing. This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. Markdown allows you to be lazy and only put the > before the first line of a hard-wrapped paragraph: > This is a blockquote with two paragraphs. Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus. > Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing. Blockquotes can be nested (i.e. a blockquote-in-a-blockquote) by adding additional levels of >: > This is the first level of quoting. > > > This is nested blockquote. > > Back to the first level. This is the first level of quoting. This is nested blockquote. Back to the first level. Blockquotes can contain other Markdown elements, including headers, lists, and code blocks: > ## This is a header. > > 1. This is the first list item. > 2. This is the second list item. > > Here's some example code: > > return shell_exec("echo$input | $markdown_script"); ## This is a header. 1. This is the first list item. 2. This is the second list item. Here’s some example code: return shell_exec("echo$input | \$markdown_script");
Any decent text editor should make email-style quoting easy. For example, with BBEdit, you can make a selection and choose Increase Quote Level from the Text menu.
### Lists
Markdown supports ordered (numbered) and unordered (bulleted) lists.
Unordered lists use asterisks, pluses, and hyphens – interchangably – as list markers:
* Red
* Green
* Blue
is equivalent to:
+ Red
+ Green
+ Blue
and:
- Red
- Green
- Blue
• Red
• Green
• Blue
Ordered lists use numbers followed by periods:
1. Bird
2. McHale
3. Parish
It’s important to note that the actual numbers you use to mark the list have no effect on the HTML output Markdown produces. The HTML Markdown produces from the above list is:
<ol>
<li>Bird</li>
<li>McHale</li>
<li>Parish</li>
</ol>
If you instead wrote the list in Markdown like this:
1. Bird
1. McHale
1. Parish
1. Bird
2. McHale
3. Parish
or even:
3. Bird
1. McHale
8. Parish
you’d get the exact same HTML output. The point is, if you want to, you can use ordinal numbers in your ordered Markdown lists, so that the numbers in your source match the numbers in your published HTML. But if you want to be lazy, you don’t have to.
If you do use lazy list numbering, however, you should still start the list with the number 1. At some point in the future, Markdown may support starting ordered lists at an arbitrary number.
List markers typically start at the left margin, but may be indented by up to three spaces. List markers must be followed by one or more spaces or a tab.
To make lists look nice, you can wrap items with hanging indents:
* Lorem ipsum dolor sit amet, consectetuer adipiscing elit.
Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi,
viverra nec, fringilla in, laoreet vitae, risus.
* Donec sit amet nisl. Aliquam semper ipsum sit amet velit.
Suspendisse id sem consectetuer libero luctus adipiscing.
• Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus.
• Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing.
But if you want to be lazy, you don’t have to:
* Lorem ipsum dolor sit amet, consectetuer adipiscing elit.
Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi,
viverra nec, fringilla in, laoreet vitae, risus.
* Donec sit amet nisl. Aliquam semper ipsum sit amet velit.
Suspendisse id sem consectetuer libero luctus adipiscing.
• Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Aliquam hendrerit mi posuere lectus. Vestibulum enim wisi, viverra nec, fringilla in, laoreet vitae, risus.
• Donec sit amet nisl. Aliquam semper ipsum sit amet velit. Suspendisse id sem consectetuer libero luctus adipiscing.
If list items are separated by blank lines, Markdown will wrap the items in <p> tags in the HTML output. For example, this input:
* Bird
* Magic
will turn into:
<ul>
<li>Bird</li>
<li>Magic</li>
</ul>
• Bird
• Magic
But this:
* Bird
* Magic
will turn into:
<ul>
<li><p>Bird</p></li>
<li><p>Magic</p></li>
</ul>
• Bird
• Magic
List items may consist of multiple paragraphs. Each subsequent paragraph in a list item must be indented by either 4 spaces or one tab:
1. This is a list item with two paragraphs. Lorem ipsum dolor
sit amet, consectetuer adipiscing elit. Aliquam hendrerit
mi posuere lectus.
Vestibulum enim wisi, viverra nec, fringilla in, laoreet
vitae, risus. Donec sit amet nisl. Aliquam semper ipsum
sit amet velit.
2. Suspendisse id sem consectetuer libero luctus adipiscing.
It looks nice if you indent every line of the subsequent paragraphs, but here again, Markdown will allow you to be lazy:
* This is a list item with two paragraphs.
This is the second paragraph in the list item. You're
only required to indent the first line. Lorem ipsum dolor
sit amet, consectetuer adipiscing elit.
* Another item in the same list.
To put a blockquote within a list item, the blockquote’s > delimiters need to be indented:
* A list item with a blockquote:
> This is a blockquote
> inside a list item.
To put a code block within a list item, the code block needs to be indented twice – 8 spaces or two tabs:
* A list item with a code block:
<code goes here>
It’s worth noting that it’s possible to trigger an ordered list by accident, by writing something like this:
1986. What a great season.
In other words, a number-period-space sequence at the beginning of a line. To avoid this, you can backslash-escape the period:
1986\. What a great season.
### Code Blocks
Pre-formatted code blocks are used for writing about programming or markup source code. Rather than forming normal paragraphs, the lines of a code block are interpreted literally. Markdown wraps a code block in both <pre> and <code> tags.
To produce a code block in Markdown, simply indent every line of the block by at least 4 spaces or 1 tab. For example, given this input:
This is a normal paragraph:
This is a code block.
Markdown will generate:
<p>This is a normal paragraph:</p>
<pre><code>This is a code block.
</code></pre>
This is a normal paragraph:
This is a code block.
One level of indentation – 4 spaces or 1 tab – is removed from each line of the code block. For example, this:
Here is an example of AppleScript:
tell application "Foo"
beep
end tell
will turn into:
<p>Here is an example of AppleScript:</p>
<pre><code>tell application "Foo"
beep
end tell
</code></pre>
A code block continues until it reaches a line that is not indented (or the end of the article).
Within a code block, ampersands (&) and angle brackets (< and >) are automatically converted into HTML entities. This makes it very easy to include example HTML source code using Markdown – just paste it and indent it, and Markdown will handle the hassle of encoding the ampersands and angle brackets. For example, this:
<div class="footer">
© 2004 Foo Corporation
</div>
will turn into:
<pre><code><div class="footer">
&copy; 2004 Foo Corporation
</div>
</code></pre>
Regular Markdown syntax is not processed within code blocks. E.g., asterisks are just literal asterisks within a code block. This means it’s also easy to use Markdown to write about Markdown’s own syntax.
## Span Elements
### Emphasis
Markdown treats asterisks (*) and underscores (_) as indicators of emphasis. Text wrapped with one * or _ will be wrapped with an HTML <em> tag; double *’s or _’s will be wrapped with an HTML <strong> tag. E.g., this input:
*single asterisks*
_single underscores_
**double asterisks**
__double underscores__
will produce:
<em>single asterisks</em>
<em>single underscores</em>
<strong>double asterisks</strong>
<strong>double underscores</strong>
single asterisks
single underscores
double asterisks
double underscores
You can use whichever style you prefer; the lone restriction is that the same character must be used to open and close an emphasis span.
Emphasis can be used in the middle of a word:
un*frigging*believable
But if you surround an * or _ with spaces, it’ll be treated as a literal asterisk or underscore.
To produce a literal asterisk or underscore at a position where it would otherwise be used as an emphasis delimiter, you can backslash escape it:
\*this text is surrounded by literal asterisks\*
### Code
To indicate a span of code, wrap it with backtick quotes (). Unlike a pre-formatted code block, a code span indicates code within a normal paragraph. For example:
Use the printf() function.
will produce:
<p>Use the <code>printf()</code> function.</p>
Use the printf() function.
To include a literal backtick character within a code span, you can use multiple backticks as the opening and closing delimiters:
There is a literal backtick () here.
which will produce this:
<p><code>There is a literal backtick () here.</code></p>
There is a literal backtick () here.
The backtick delimiters surrounding a code span may include spaces – one after the opening, one before the closing. This allows you to place literal backtick characters at the beginning or end of a code span:
A single backtick in a code span:
A backtick-delimited string in a code span: foo
will produce:
<p>A single backtick in a code span: <code></code></p>
<p>A backtick-delimited string in a code span: <code>foo</code></p>
A single backtick in a code span:
A backtick-delimited string in a code span: foo
With a code span, ampersands and angle brackets are encoded as HTML entities automatically, which makes it easy to include example HTML tags. Markdown will turn this:
Please don't use any <blink> tags.
into:
<p>Please don't use any <code><blink></code> tags.</p>
You can write this:
— is the decimal-encoded equivalent of —.
to produce:
<p><code>&#8212;</code> is the decimal-encoded
equivalent of <code>&mdash;</code>.</p>
### Images
Admittedly, it’s fairly difficult to devise a “natural” syntax for placing images into a plain text document format.
Markdown uses an image syntax that is intended to resemble the syntax for links, allowing for two styles: inline and reference.
Inline image syntax looks like this:
That is:
• An exclamation mark: !;
• followed by a set of square brackets, containing the alt attribute text for the image;
• followed by a set of parentheses, containing the URL or path to the image, and an optional title attribute enclosed in double or single quotes.
Reference-style image syntax looks like this:
![Alt text][id]
Where “id” is the name of a defined image reference. Image references are defined using syntax identical to link references:
[id]: url/to/image "Optional title attribute"
As of this writing, Markdown has no syntax for specifying the dimensions of an image; if this is important to you, you can simply use regular HTML <img> tags.
### Backslash Escapes
Markdown allows you to use backslash escapes to generate literal characters which would otherwise have special meaning in Markdown’s formatting syntax. For example, if you wanted to surround a word with literal asterisks (instead of an HTML <em> tag), you can use backslashes before the asterisks, like this:
\*literal asterisks\*
*literal asterisks*
Markdown provides backslash escapes for the following characters:
\ backslash
backtick
* asterisk
_ underscore
{} curly braces
[] square brackets
() parentheses
# hash mark
+ plus sign
- minus sign (hyphen)
. dot
! exclamation mark
\ backslash backtick * asterisk _ underscore {} curly braces [] square brackets () parentheses
# hash mark
• plus sign
• minus sign (hyphen) . dot ! exclamation mark
# MultiMarkdown
Prototype table
Grouping
Content Long Cell
Content Cell Cell
New section More Data
And more And more
Here is some text containing a footnote.[2]
This is a statement that should be attributed to its source[p. 23, 3].
This is a statement that should be attributed to its source[3]
And following is the description of the reference to be used in the bibliography.
### Preformatted Code Blocks
Indent every line of a code block by at least 4 spaces or 1 tab.
This is a normal paragraph.
This is a preformatted
code block.
### Horizontal Rules
Three or more dashes or asterisks:
---
* * *
- - - -
### Manual Line Breaks
End a line with two or more spaces:
Roses are red,
Violets are blue.
Roses are red, Violets are blue.
1. And that’s the footnote. ↩
2. Here is the text of the footnote itself. ↩
3. John Doe. Some Big Fancy Book. Vanity Press, 2006.
|
2017-03-25 15:35:48
|
{"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": 2, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42797791957855225, "perplexity": 10509.595783851208}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218188962.14/warc/CC-MAIN-20170322212948-00635-ip-10-233-31-227.ec2.internal.warc.gz"}
|
https://asmedigitalcollection.asme.org/medicaldevices/article/3/2/027512/470804/CT-Visualization-of-Cryoablation-in-Pulmonary
|
Over 2 million adults in the United States are affected by atrial fibrillation (AF), a common cardiac arrhythmia that is associated with decreased survival, increased cardiovascular morbidities, and a decrease in quality of life. AF can be initiated by ectopic beats originating in the myocardial sleeves surrounding the pulmonary viens. Pulmonary vein (PV) isolation via radio frequency ablation is the current gold standard for treating patients with drug-refractory AF. However, cryoablation is emerging as a new minimally-invasive technique to achieve PV isolation. Cryoablation is fast gaining acceptance due to its minimal tissue disruption, decreased thrombogenicity, and reduced complications (RF can lead to low rate of stenosis). One important question in regard to this technology is whether the PV lesion is transmural and circumferential and to what extent adjacent tissues are involved in the freezing process. As ice formation lends itself to image contrast in the body, we hypothesized that intraprocedural CT visualization of the iceball formation would allow us to predict the extent of the cryolesion and provide us with a measure of the adjacent tissue damage. Cryoablation was performed using a prototype balloon catheter cryoablation system (Boston Scientific Corporation). CT visualization of iceball formation was assessed both in vitro and in vivo. Initial in vitro studies were performed in agarose gel phantoms immersed in a $37°C$ water bath. Subsequently, in vivo cryoablations were performed in 5 PV ostia in 3 crossbred farm swine. The catheters were positioned in the ostia under fluoroscopic guidance. CT scans of the thoracic region were obtained every 2.5 minutes. Animals were sacrified 6 days after the procedures. Gross pathology and histology of tissues in the region of interest were evaluated. Significant metal artifacts from the catheter and edge artifacts from the tissues surrounding the cryoballoon were observed under CT imaging both in vitro and in vivo. In vitro, it was found that the size of the iceball was comparable to that observed visually during freezing of agarose gel phantoms. In vivo, contrast change consistent with iceball formation was observed during the ablation in two out of five veins. The most clearly delineated iceball also yielded the clearest morbidity. In this case, esophageal injury on the anterior side proximal to the cryoablation site was noticed during necropsy of the animal in which the iceball was visualized. Transmural and circumferential lesions were obtained in all PVs ablated. We have shown that CT can be used to visualize iceball formation in vitro and in vivo (with limitations) using our cryoablation system. While the iceball in vitro is easily visualized, iceball growth in vivo is most evident once the iceball has grown beyond the PV into the adjacent tissues. This suggests that while CT cannot easily visualize iceball growth in the PV wall itself, it may still be an important tool to guide clinicians and reduce potential morbidities in adjacent tissues. The authors acknowledge Dan Busian (Fairview University Medical Center, Minneapolis, MN) and Dr. Erik Cressman for assistance with CT imaging.
|
2019-10-21 09:38:26
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3529656231403351, "perplexity": 5425.4054241320055}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00078.warc.gz"}
|
http://mathhelpforum.com/algebra/96499-question-help.html
|
1. ## Question Help
Hi Guys, Any help with this one would be awesome.
I think it requires some algebra or something.
If I am looking at a \$1,000,000 loan. The fixed rate is 7.34% the variable is 6.03%. If I fix half and half (half fixed, half floating) the average rate is 6.68% (7.34% + 6.03% divided by 2 = 6.68%)
What percentage (instead of 50:50) of fixed and floating do I need to get to make the average rate 6.305%
Obviously it will need to be a greater percentage in the floating (6.03%) portion as it is lower and will bring the average rate down to the desired 6.305%.
So say we had a 25:75 fixed and floating split, the average rate would be 7.34 x 0.25 + 6.03 x 0.75 = (6.3575%) --- I could probably keep trying differnt combinations until I had the correct answer, but I want to know how to actually work it out using algebra
My stab at algebra for this ended up horribly wrong, but this is what I had
6.03(x) + 7.34(1-x) = 6.305% where x = percentage of floating portion (therefore 1 - x = fixed portion)
But I dont know where to go from here (or if i am even on the right track)
|
2018-02-21 01:51:04
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8118883371353149, "perplexity": 683.5600090918798}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-09/segments/1518891813187.88/warc/CC-MAIN-20180221004620-20180221024620-00163.warc.gz"}
|
http://image.absoluteastronomy.com/topics/Arc_length
|
Arc length
Encyclopedia
Determining the length of an irregular arc segment is also called rectification of a curve
Curve
In mathematics, a curve is, generally speaking, an object similar to a line but which is not required to be straight...
. Historically, many methods were used for specific curves. The advent of infinitesimal calculus
Infinitesimal calculus
Infinitesimal calculus is the part of mathematics concerned with finding slope of curves, areas under curves, minima and maxima, and other geometric and analytic problems. It was independently developed by Gottfried Leibniz and Isaac Newton starting in the 1660s...
led to a general formula that provides closed-form solutions
Closed-form expression
In mathematics, an expression is said to be a closed-form expression if it can be expressed analytically in terms of a bounded number of certain "well-known" functions...
in some cases.
## General approach
A curve
Curve
In mathematics, a curve is, generally speaking, an object similar to a line but which is not required to be straight...
in the plane
Euclidean space
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...
can be approximated by connecting a finite number of points
Point (geometry)
In geometry, topology and related branches of mathematics a spatial point is a primitive notion upon which other concepts may be defined. In geometry, points are zero-dimensional; i.e., they do not have volume, area, length, or any other higher-dimensional analogue. In branches of mathematics...
on the curve using line segment
Line segment
In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. Examples of line segments include the sides of a triangle or square. More generally, when the end points are both vertices of a polygon, the line segment...
s to create a polygonal path
Polygonal chain
A polygonal chain, polygonal curve, polygonal path, or piecewise linear curve, is a connected series of line segments. More formally, a polygonal chain P is a curve specified by a sequence of points \scriptstyle called its vertices so that the curve consists of the line segments connecting the...
. Since it is straightforward to calculate the length
Length
In geometric measurements, length most commonly refers to the longest dimension of an object.In certain contexts, the term "length" is reserved for a certain dimension of an object along which the length is measured. For example it is possible to cut a length of a wire which is shorter than wire...
of each linear segment (using the Pythagorean theorem
Pythagorean theorem
In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...
in Euclidean space, for example), the total length of the approximation can be found by summing
Summation
Summation is the operation of adding a sequence of numbers; the result is their sum or total. If numbers are added sequentially from left to right, any intermediate result is a partial sum, prefix sum, or running total of the summation. The numbers to be summed may be integers, rational numbers,...
the lengths of each linear segment.
If the curve is not already a polygonal path, better approximations to the curve can be obtained by following the shape of the curve increasingly more closely. The approach is to use an increasingly larger number of segments of smaller lengths. The lengths of the successive approximations do not decrease and will eventually keep increasing—possibly indefinitely, but for smooth curves this will tend to a limit as the lengths of the segments get arbitrarily small
Arbitrarily large
In mathematics, the phrase arbitrarily large, arbitrarily small, arbitrarily long is used in statements such as:which is shorthand for:This should not be confused with the phrase "sufficiently large"...
.
For some curves there is a smallest number L that is an upper bound on the length of any polygonal approximation. If such a number exists, then the curve is said to be rectifiable and the curve is defined to have arc length L.
## Definition
Let C be a curve
Curve
In mathematics, a curve is, generally speaking, an object similar to a line but which is not required to be straight...
in Euclidean
Euclidean space
In mathematics, Euclidean space is the Euclidean plane and three-dimensional space of Euclidean geometry, as well as the generalizations of these notions to higher dimensions...
(or, more generally, a metric
Metric space
In mathematics, a metric space is a set where a notion of distance between elements of the set is defined.The metric space which most closely corresponds to our intuitive understanding of space is the 3-dimensional Euclidean space...
) space X = Rn, so C is the image
Image (mathematics)
In mathematics, an image is the subset of a function's codomain which is the output of the function on a subset of its domain. Precisely, evaluating the function at each element of a subset X of the domain produces a set called the image of X under or through the function...
of a continuous function
Continuous function
In mathematics, a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous".Continuity of...
f : [a, b] → X of the interval
Interval (mathematics)
In mathematics, a interval is a set of real numbers with the property that any number that lies between two numbers in the set is also included in the set. For example, the set of all numbers satisfying is an interval which contains and , as well as all numbers between them...
[a, b] into X.
From a partition
Partition of an interval
In mathematics, a partition, P of an interval [a, b] on the real line is a finite sequence of the formIn mathematics, a partition, P of an interval [a, b] on the real line is a finite sequence of the form...
a = t0 < t1 < ... < tn−1 < tn = b of the interval [a, b] we obtain a finite collection of points f(t0), f(t1), ..., f(tn−1), f(tn) on the curve C. Denote the distance
Distance
Distance is a numerical description of how far apart objects are. In physics or everyday discussion, distance may refer to a physical length, or an estimation based on other criteria . In mathematics, a distance function or metric is a generalization of the concept of physical distance...
from f(ti) to f(ti+1) by d(f(ti), f(ti+1)), which is the length of the line segment
Line segment
In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. Examples of line segments include the sides of a triangle or square. More generally, when the end points are both vertices of a polygon, the line segment...
connecting the two points.
The arc length L of C is then defined to be
where the supremum
Supremum
In mathematics, given a subset S of a totally or partially ordered set T, the supremum of S, if it exists, is the least element of T that is greater than or equal to every element of S. Consequently, the supremum is also referred to as the least upper bound . If the supremum exists, it is unique...
is taken over all possible partitions of [ab] and n is unbounded.
The arc length L is either finite or infinite. If L < ∞ then we say that C is rectifiable, and is non-rectifiable otherwise. This definition of arc length does not require that C is defined by a differentiable
Derivative
In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a...
function. In fact in general, the notion of differentiability is not defined on a metric space.
A curve may be parameterized in many ways. Suppose C also has the parameterization g : [cd] → X. Provided that f and g are injective, there is a continuous monotone function S from [ab] to [cd] so that g(S(t)) = f(t) and an inverse function S−1 from [cd] to [ab]. It is clear that any sum of the form
can be made equal to a sum of the form
by taking , and similarly a sum involving g can be made equal to a sum involving f. So the arc length is an intrinsic property of the curve, meaning that it does not depend on the choice of parameterization.
The definition of arc length for the curve is analogous to the definition of the total variation
Total variation
In mathematics, the total variation identifies several slightly different concepts, related to the structure of the codomain of a function or a measure...
of a real-valued function.
## Finding arc lengths by integrating
Consider a real function
Function (mathematics)
In mathematics, a function associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output. A function assigns exactly one output to each input. The argument and the value may be real numbers, but they can...
such that and
(its derivative with respect to x) are continuous
Continuous function
In mathematics, a continuous function is a function for which, intuitively, "small" changes in the input result in "small" changes in the output. Otherwise, a function is said to be "discontinuous". A continuous function with a continuous inverse function is called "bicontinuous".Continuity of...
on [ab]. The length s of the part of the graph of f between x = a and x = b can be found as follows:
Consider an infinitesimal part of the curve (or consider this as a limit in which the change in s approaches ). According to Pythagoras' theorem , from which:
If a curve is defined parametrically by x = X(t) and y = Y(t), then its arc length between t = a and t = b is
This is more clearly a consequence of the distance formula where instead of a and , we take the limit. A useful mnemonic is
If a function is defined in polar coordinates
Polar coordinate system
In mathematics, the polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a distance from a fixed point and an angle from a fixed direction....
by then the arc length is given by
In most cases, including even simple curves, there are no closed-form solutions of arc length and numerical integration
Numerical integration
In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. This article focuses on calculation of...
is necessary.
Curves with closed-form solution for arc length include the catenary
Catenary
In physics and geometry, the catenary is the curve that an idealised hanging chain or cable assumes when supported at its ends and acted on only by its own weight. The curve is the graph of the hyperbolic cosine function, and has a U-like shape, superficially similar in appearance to a parabola...
, circle
Circle
A circle is a simple shape of Euclidean geometry consisting of those points in a plane that are a given distance from a given point, the centre. The distance between any of the points and the centre is called the radius....
, cycloid
Cycloid
A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line.It is an example of a roulette, a curve generated by a curve rolling on another curve....
, logarithmic spiral
Logarithmic spiral
A logarithmic spiral, equiangular spiral or growth spiral is a special kind of spiral curve which often appears in nature. The logarithmic spiral was first described by Descartes and later extensively investigated by Jacob Bernoulli, who called it Spira mirabilis, "the marvelous...
, parabola
Parabola
In mathematics, the parabola is a conic section, the intersection of a right circular conical surface and a plane parallel to a generating straight line of that surface...
, semicubical parabola and (mathematically, a curve) straight line
Line (mathematics)
The notion of line or straight line was introduced by the ancient mathematicians to represent straight objects with negligible width and depth. Lines are an idealization of such objects...
. The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integral
Elliptic integral
In integral calculus, elliptic integrals originally arose in connection with the problem of giving the arc length of an ellipse. They were first studied by Giulio Fagnano and Leonhard Euler...
s.
### Derivation
In order to approximate the arc length of the curve, it is split into many linear
Linear
In mathematics, a linear map or function f is a function which satisfies the following two properties:* Additivity : f = f + f...
segments. To make the value exact, and not an approximation
Approximation
An approximation is a representation of something that is not exact, but still close enough to be useful. Although approximation is most often applied to numbers, it is also frequently applied to such things as mathematical functions, shapes, and physical laws.Approximations may be used because...
, infinitely many linear elements are needed. This means that each element is infinitely small. This fact manifests itself later on when an integral
Integral
Integration is an important concept in mathematics and, together with its inverse, differentiation, is one of the two main operations in calculus...
is used.
Begin by looking at a representative linear segment (see image) and observe that its length (element of the arc length) will be the differential
Differential (mathematics)
In mathematics, the term differential has several meanings.-Basic notions:* In calculus, the differential represents a change in the linearization of a function....
ds. We will call the horizontal element of this distance dx, and the vertical element dy.
The Pythagorean theorem
Pythagorean theorem
In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...
tells us that
Since the function is defined in time, segments (ds) are added up across infinitesimally small intervals of time (dt) yielding the integral
If y is a function of x, so that we could take t = x, then we have:
which is the arc length from x = a to x = b of the graph of the function ƒ.
For example, the curve in this figure is defined by
Subsequently, the arc length integral for values of t from -1 to 1 is
Using computational approximations, we can obtain a very accurate (but still approximate) arc length of 2.905. An expression in terms of the hypergeometric function can be obtained: it is
### Another way to obtain the integral formula
Suppose that there exists a rectifiable curve given by a function f(x). To approximate the arc length S along f between two points a and b in that curve, construct a series of right triangles whose concatenated hypotenuses "cover" the arc of curve chosen as shown in the figure. For convenience, the bases of all those triangles can be set equal to , so that for each one an associated exists. The length of any given hypotenuse is given by the Pythagorean Theorem
Pythagorean theorem
In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...
:
The summation of the lengths of the hypotenuses approximates :
Then, our previous result becomes:
As the length of these segments decreases, the approximation improves. The limit of the approximation, as goes to zero, is equal to :
### Another proof
We know that the formula for a line integral is . If we set the surface f(x,y) to 1, we will get arc length multiplied by 1, or . If x = t, and y = f(t), then y = f(x), from when x is a to when x is b. If we set these equations into our formula we get: (Note: If x = t then dt = dx). This is the arc length formula.
### Arcs of circles
The length of an arc of a circle
Circle
A circle is a simple shape of Euclidean geometry consisting of those points in a plane that are a given distance from a given point, the centre. The distance between any of the points and the centre is called the radius....
is the central angle divided by 360° multiplied by the circumference
Circumference
The circumference is the distance around a closed curve. Circumference is a special perimeter.-Circumference of a circle:The circumference of a circle is the length around it....
.
The circumference
Circumference
The circumference is the distance around a closed curve. Circumference is a special perimeter.-Circumference of a circle:The circumference of a circle is the length around it....
of a circle is , where r is the radius, or , where d is the diameter.
Arc lengths are denoted by s, since arcs "subtend" an angle.
If the angle measurement is in radians then where r is the radius and is the subtended angle. The units of s will be the same as that of the radius.
In a semicircle
Semicircle
In mathematics , a semicircle is a two-dimensional geometric shape that forms half of a circle. Being half of a circle's 360°, the arc of a semicircle always measures 180° or a half turn...
, .
### Ancient
For much of the history of mathematics
History of mathematics
The area of study known as the history of mathematics is primarily an investigation into the origin of discoveries in mathematics and, to a lesser extent, an investigation into the mathematical methods and notation of the past....
, even the greatest thinkers considered it impossible to compute the length of an irregular arc. Although Archimedes
Archimedes
Archimedes of Syracuse was a Greek mathematician, physicist, engineer, inventor, and astronomer. Although few details of his life are known, he is regarded as one of the leading scientists in classical antiquity. Among his advances in physics are the foundations of hydrostatics, statics and an...
had pioneered a way of finding the area beneath a curve with his method of exhaustion
Method of exhaustion
The method of exhaustion is a method of finding the area of a shape by inscribing inside it a sequence of polygons whose areas converge to the area of the containing shape. If the sequence is correctly constructed, the difference in area between the n-th polygon and the containing shape will...
, few believed it was even possible for curves to have definite lengths, as do straight lines. The first ground was broken in this field, as it often has been in calculus
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, differential calculus and integral calculus, which are related by the fundamental theorem...
, by approximation
Approximation
An approximation is a representation of something that is not exact, but still close enough to be useful. Although approximation is most often applied to numbers, it is also frequently applied to such things as mathematical functions, shapes, and physical laws.Approximations may be used because...
. People began to inscribe polygon
Polygon
In geometry a polygon is a flat shape consisting of straight lines that are joined to form a closed chain orcircuit.A polygon is traditionally a plane figure that is bounded by a closed path, composed of a finite sequence of straight line segments...
s within the curves and compute the length of the sides for a somewhat accurate measurement of the length. By using more segments, and by decreasing the length of each segment, they were able to obtain a more and more accurate approximation. In particular, by inscribing a polygon of many sides in a circle, they were able to find approximate values of π.
### 1600s
In the 17th century, the method of exhaustion led to the rectification by geometrical methods of several transcendental curve
Transcendental curve
In mathematics, a transcendental curve is a curve that is not an algebraic curve. Here for a curve C what matters is the point set underlying C, not a given parametrisation...
s: the logarithmic spiral
Logarithmic spiral
A logarithmic spiral, equiangular spiral or growth spiral is a special kind of spiral curve which often appears in nature. The logarithmic spiral was first described by Descartes and later extensively investigated by Jacob Bernoulli, who called it Spira mirabilis, "the marvelous...
by Evangelista Torricelli
Evangelista Torricelli
Evangelista Torricelli was an Italian physicist and mathematician, best known for his invention of the barometer.-Biography:Evangelista Torricelli was born in Faenza, part of the Papal States...
in 1645 (some sources say John Wallis in the 1650s), the cycloid
Cycloid
A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line.It is an example of a roulette, a curve generated by a curve rolling on another curve....
by Christopher Wren
Christopher Wren
Sir Christopher Wren FRS is one of the most highly acclaimed English architects in history.He used to be accorded responsibility for rebuilding 51 churches in the City of London after the Great Fire in 1666, including his masterpiece, St. Paul's Cathedral, on Ludgate Hill, completed in 1710...
in 1658, and the catenary
Catenary
In physics and geometry, the catenary is the curve that an idealised hanging chain or cable assumes when supported at its ends and acted on only by its own weight. The curve is the graph of the hyperbolic cosine function, and has a U-like shape, superficially similar in appearance to a parabola...
by Gottfried Leibniz
Gottfried Leibniz
Gottfried Wilhelm Leibniz was a German philosopher and mathematician. He wrote in different languages, primarily in Latin , French and German ....
in 1691.
In 1659, Wallis credited William Neile
William Neile
William Neile was an English mathematician and founder member of the Royal Society. His major mathematical work, the rectification of the semicubical parabola, was carried out when he was aged nineteen, and was published by John Wallis...
's discovery of the first rectification of a nontrivial algebraic curve
Algebraic curve
In algebraic geometry, an algebraic curve is an algebraic variety of dimension one. The theory of these curves in general was quite fully developed in the nineteenth century, after many particular examples had been considered, starting with circles and other conic sections.- Plane algebraic curves...
, the semicubical parabola.
### Integral form
Before the full formal development of the calculus, the basis for the modern integral form for arc length was independently discovered by Hendrik van Heuraet
Hendrik van Heuraet
Hendrik van Heuraet was a Dutch mathematician. He was noted as one of the founders of the integral. From [1653 he studied at Leiden University where he interacted with Frans van Schooten, Johannes Hudde, and Christiaan Huygens. In 1658 he and Hudde left for Saumur in France. He returned to Leiden...
and Pierre de Fermat
Pierre de Fermat
Pierre de Fermat was a French lawyer at the Parlement of Toulouse, France, and an amateur mathematician who is given credit for early developments that led to infinitesimal calculus, including his adequality...
.
In 1659 van Heuraet published a construction showing that arc length could be interpreted as the area under a curve—this integral, in effect—and applied it to the parabola
Parabola
In mathematics, the parabola is a conic section, the intersection of a right circular conical surface and a plane parallel to a generating straight line of that surface...
. In 1660, Fermat published a more general theory containing the same result in his De linearum curvarum cum lineis rectis comparatione dissertatio geometrica.
Building on his previous work with tangents, Fermat used the curve
whose tangent
Tangent
In geometry, the tangent line to a plane curve at a given point is the straight line that "just touches" the curve at that point. More precisely, a straight line is said to be a tangent of a curve at a point on the curve if the line passes through the point on the curve and has slope where f...
at x = a had a slope
Slope
In mathematics, the slope or gradient of a line describes its steepness, incline, or grade. A higher slope value indicates a steeper incline....
of
so the tangent line would have the equation
Next, he increased a by a small amount to a + ε, making segment AC a relatively good approximation for the length of the curve from A to D. To find the length of the segment AC, he used the Pythagorean theorem
Pythagorean theorem
In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle...
:
which, when solved, yields
In order to approximate the length, Fermat would sum up a sequence of short segments.
## Curves with infinite length
As mentioned above, some curves are non-rectifiable, that is, there is no upper bound on the lengths of polygonal approximations; the length can be made arbitrarily large. Informally, such curves are said to have infinite length. There are continuous curves on which every arc (other than a single-point arc) has infinite length. An example of such a curve is the Koch curve
Koch snowflake
The Koch snowflake is a mathematical curve and one of the earliest fractal curves to have been described...
. Another example of a curve with infinite length is the graph of the function defined by f(x) = x sin(1/x) for any open set with 0 as one of its delimiters and f(0) = 0. Sometimes the Hausdorff dimension
Hausdorff dimension
thumb|450px|Estimating the Hausdorff dimension of the coast of Great BritainIn mathematics, the Hausdorff dimension is an extended non-negative real number associated with any metric space. The Hausdorff dimension generalizes the notion of the dimension of a real vector space...
and Hausdorff measure
Hausdorff measure
In mathematics a Hausdorff measure is a type of outer measure, named for Felix Hausdorff, that assigns a number in [0,∞] to each set in Rn or, more generally, in any metric space. The zero dimensional Hausdorff measure is the number of points in the set or ∞ if the set is infinite...
are used to "measure" the size of infinite-length curves.
## Generalization to (pseudo-)Riemannian manifolds
Let M be a (pseudo-)Riemannian manifold
Pseudo-Riemannian manifold
In differential geometry, a pseudo-Riemannian manifold is a generalization of a Riemannian manifold. It is one of many mathematical objects named after Bernhard Riemann. The key difference between a Riemannian manifold and a pseudo-Riemannian manifold is that on a pseudo-Riemannian manifold the...
, γ : [0, 1] → M a curve in M and g the (pseudo-) metric tensor.
The length of γ is defined to be
where γ'(t)Tγ(t)M is the tangent vector of γ at t. The sign in the square root is chosen once for a given curve, to ensure that the square root is a real number. The positive sign is chosen for spacelike curves; in a pseudo-Riemannian manifold, the negative sign may be chosen for timelike curves.
In theory of relativity
Theory of relativity
The theory of relativity, or simply relativity, encompasses two theories of Albert Einstein: special relativity and general relativity. However, the word relativity is sometimes used in reference to Galilean invariance....
, arc-length of timelike curves (world line
World line
In physics, the world line of an object is the unique path of that object as it travels through 4-dimensional spacetime. The concept of "world line" is distinguished from the concept of "orbit" or "trajectory" by the time dimension, and typically encompasses a large area of spacetime wherein...
s) is the proper time
Proper time
In relativity, proper time is the elapsed time between two events as measured by a clock that passes through both events. The proper time depends not only on the events but also on the motion of the clock between the events. An accelerated clock will measure a smaller elapsed time between two...
elapsed along the world line.
• Arc (geometry)
Arc (geometry)
In geometry, an arc is a closed segment of a differentiable curve in the two-dimensional plane; for example, a circular arc is a segment of the circumference of a circle...
• Integral approximations
Numerical integration
In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. This article focuses on calculation of...
• Geodesic
Geodesic
In mathematics, a geodesic is a generalization of the notion of a "straight line" to "curved spaces". In the presence of a Riemannian metric, geodesics are defined to be the shortest path between points in the space...
s
• Meridian arc
Meridian arc
In geodesy, a meridian arc measurement is a highly accurate determination of the distance between two points with the same longitude. Two or more such determinations at different locations then specify the shape of the reference ellipsoid which best approximates the shape of the geoid. This...
|
2023-03-26 08:10:55
|
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8367513418197632, "perplexity": 489.4441949922063}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296945440.67/warc/CC-MAIN-20230326075911-20230326105911-00371.warc.gz"}
|