text:
# Fundamental group of a Lie group

Is it true that the class of isomorphism classes of fundamental groups of Lie groups coincides with the class of isomorphism classes of finitely generated abelian groups?

• The fundamental groups of Lie groups are all abelian. So if you restrict to compact ones, they are all finitely generated abelian groups. Not sure what your question is asking about.... Mar 9, 2018 at 19:36
• I know that they are abelian, but my question is about their presentations...
– RRR
Mar 9, 2018 at 19:38
• For a finitely generated group, both "the number of generators" and "the number of relations" are not defined, simply because "generators" and "relations" are not defined. You maybe want to define something related to presentations. Such as the existence of presentations with certain properties.
– YCor
Mar 9, 2018 at 19:38
• Next, since the class of isomorphism classes of $\pi_1$ of Lie groups coincides with the class of isomorphism classes of finitely generated abelian groups, there is no point to incorporate Lie groups in the question, which (once more carefully stated) is a question about group presentations of finitely generated abelian groups.
– YCor
Mar 9, 2018 at 19:41
Yes, these two classes coincide. One direction splits into two: for a Lie group $G$,
Conversely any finitely generated abelian group is fundamental group of a Lie group. By the classification of finitely generated abelian groups, it is enough to consider cyclic groups (and then use products). Indeed $\mathbf{Z}$ is isomorphic to the fundamental group of the circle group $\mathbf{R}/\mathbf{Z}$, and $\mathbf{Z}/n\mathbf{Z}$ is isomorphic to the fundamental group of $\mathrm{PSL}_n(\mathbf{C})$.


text:
Image Management¶

Working with a directory of images¶

For the sake of argument all of the examples below assume you are working in a directory that contains FITS images.

The class ImageFileCollection is meant to make working with a directory of FITS images easier by allowing you select the files you act on based on the values of FITS keywords in their headers or based on Unix shell-style filename matching.

It is initialized with the name of a directory containing FITS images and a list of FITS keywords you want the ImageFileCollection to be aware of. An example initialization looks like:

>>> from ccdproc import ImageFileCollection
>>> from ccdproc.utils.sample_directory import sample_directory_with_files
>>> keys = ['imagetyp', 'object', 'filter', 'exposure']
>>> dir = sample_directory_with_files()
>>> ic1 = ImageFileCollection(dir, keywords=keys) # only keep track of keys

You can use the wildcard * in place of a list to indicate you want the collection to use all keywords in the headers:

>>> ic_all = ImageFileCollection(dir, keywords='*')

Normally identification of FITS files is done by looking at the file extension and including all files with the correct extension.

If the files are not compressed (e.g. not gzipped) then you can force the image collection to open each file and check from its contents whether it is FITS by using the find_fits_by_reading argument:

>> ic_from_content = ImageFileCollection(dir, find_fits_by_reading=True)

You can indicate filename patterns to include or exclude using Unix shell-style expressions. For example, to include all filenames that begin with 1d_ but not ones that include the word bad, you could do:

>>> ic_all = ImageFileCollection(dir, glob_include='1d_*',

Alternatively, you can create the collection with an explicit list of file names:

>>> ic_names = ImageFileCollection(filenames=['a.fits', '/some/path/b.fits.gz'])

Most of the useful interaction with the image collection is via its .summary property, a Table of the value of each keyword for each file in the collection:

>>> ic1.summary.colnames
['file', 'imagetyp', 'object', 'filter', 'exposure']
>>> ic_all.summary.colnames
# long list of keyword names omitted

Note that the name of the file is automatically added to the table as a column named file.

Selecting files¶

Selecting the files that match a set of criteria, for example all images in the I band with exposure time less than 60 seconds you could do:

>>> matches = (ic1.summary['filter'] == 'R') & (ic1.summary['exposure'] < 15)
>>> my_files = ic1.summary['file'][matches]

The column file is added automatically when the image collection is created.

For more simple selection, when you just want files whose keywords exactly match particular values, say all I band images with exposure time of 30 seconds, there is a convenience method .files_filtered:

>>> my_files = ic1.files_filtered(filter='R', exposure=15)

The optional arguments to files_filtered are used to filter the list of files.

Python regular expression patterns can also be used as the value if the regex_match flag is set. For example, to find all of the images whose object is in the Kelt exoplanet survey, you might do:

>>> my_files = ic1.files_filtered(regex_match=True, object='kelt.*')

To get all of the images that have image type BIAS or LIGHT you can also use a regular expression pattern:

>>> my_files = ic1.files_filtered(regex_match=True,
...                               imagetyp='bias|light')

Note that regular expression is different, and much more flexible than, file name matching (or “globbing”) at the command line. The Python documentation on the re module is useful for learning about regular expressions.

Finally, a new ImageFileCollection can be created with by providing a list of keywords. The example below makes a new collection containing the files whose imagetyp is BIAS or LIGHT:

>>> new_ic = ic1.filter(regex_match=True,
...                     imagetyp='bias|light')

Sorting files¶

Sometimes it is useful to bring the files into a specific order, e.g. if you make a plot for each object you probably want all images of the same object next to each other. To do this, the images in a collection can be sorted with the sort method using the fits header keys in the same way you would sort a Table:

>>> ic1.sort(['exposure', 'imagetyp'])

Iterating over hdus, headers, data, or ccds¶

Four methods are provided for iterating over the images in the collection, optionally filtered by keyword values.

For example, to iterate over all of the I band images with exposure of 30 seconds, performing some basic operation on the data (very contrived example):

>>> for hdu in ic1.hdus(imagetyp='LiGhT', filter='R', exposure=15):
...     new_data = hdu.data - hdu.data.mean()
15.0

Note that the names of the arguments to hdus here are the names of FITS keywords in the collection and the values are the values of those keywords you want to select. Note also that string comparisons are not case sensitive.

The other iterators are headers, data, and ccds.

All of them have the option to also provide the file name in addition to the hdu (or header or data):

>>> for hdu, fname in ic1.hdus(return_fname=True,
...                            imagetyp='LiGhT', filter='R', exposure=15):
...    hdu.header['meansub'] = True
...    hdu.data = hdu.data - hdu.data.mean()
...    hdu.writeto(fname + '.new')

That last use case, doing something to several files and saving them somewhere afterwards, is common enough that the iterators provide arguments to automate it.

Automatic saving from the iterators¶

There are three ways of triggering automatic saving.

1. One is with the argument save_with_name; it adds the value of the argument to the file name between the original base name and extension. The example below has (almost) the same effect of the example above, subtracting the mean from each image and saving to a new file:

>>> for hdu in ic1.hdus(save_with_name='_new',
...                     imagetyp='LiGhT', filter='R', exposure=15):
...    hdu.header['meansub'] = True
...    hdu.data = hdu.data - hdu.data.mean()

It saves, in the location of the image collection, a new FITS file with the mean subtracted from the data, with _new added to the name; as an example, if one of the files iterated over was intput001.fit then a new file, in the same directory, called input001_new.fit would be created.

2. You can also provide the directory to which you want to save the files with save_location; note that you do not need to actually do anything to the hdu (or header or data) to cause the copy to be made. The example below copies all of the I band images with 30 second exposure from the original location to other_dir:

>>> for hdu in ic1.hdus(save_location='other_dir',
...                     imagetyp='LiGhT', filter='I', exposure=30):
...     pass

This option can be combined with the previous one to also give the files a new name.

3. Finally, if you want to live dangerously, you can overwrite the files in the same location with the overwrite argument; use it carefully because it preserves no backup. The example below replaces each of the I band images with 30 second exposure with a file that has had the mean subtracted:

>>> for hdu in ic1.hdus(overwrite=True,
...                     imagetyp='LiGhT', filter='R', exposure=15):
...    hdu.header['meansub'] = True
...    hdu.data = hdu.data - hdu.data.mean()

Note

This functionality is not currently available on Windows.

format: plain


text:
# Bootstrap Secrets

Bootstrapping is a hard problem. With the Hashistack this problem is harder, as there are dependency loops and no clear agreement on where to break them.

For our purposes, there are two main bootstrapping loops:

• Vault depends on Consul for storage. While it is possible to use a built-in Raft database with Vault, this is still very much a tech preview, and is not yet suitable for production use. This also does not work well with the opinion of the project that Consul is the point of cluster state and topological knowledge.

• Nomad needs Consul to be available to perform automatic clustering. Because Consul is providing the backing storage to Vault, and for best-practices reasons, ACLs must be enabled. This means that now Nomad needs a token to perform automatic clustering. Ideally it would get this token from Vault, but this early in initialization that isn’t practical, so we need some way to get the token securely.

Really what we need here is an external secrets store. This secrets store could be a Vault server that is external to the cluster, or something provided by a cloud service. In the ResinStack reference implementation, this is fulfilled by the AWS Secrets Manager. This service allows us to securely escrow just enough data to break the bootstrapping loops. We still don’t want to need to go on-box to use these secrets once they become available though, so its necessary to poll for the values.

Such is the job of the emissary task on ResinStack machines. This task starts up as a service after the network has been configured and polls a configured secrets store to fill in values in templates. Once a template has been completely rendered emissary can execute a command, such as reloading a service.

The emissary task allows all-in-one control planes to be incrementally bootstrapped as secrets become available. The emissary task is simple and fairly single purpose. Its use cases are not limited to the ResinStack though, and if you have a use case extension, patches are welcome to support new secret backends.


text:
Tokyo Journal of Mathematics

Surfaces and Fronts with Harmonic-mean Curvature One in Hyperbolic Three-space

Masatoshi KOKUBU

Abstract

We show a global representation formula for a certain kind of Weingarten surface in hyperbolic three-space, which is based on the formula due to Gálvez, Martínez and Milán. As an application of the representation formula, we also investigate surfaces with harmonic-mean curvature one (HMC-1 surfaces). We allow them to have certain kinds of singularities, and discuss some global properties.

Article information

Source
Tokyo J. Math., Volume 32, Number 1 (2009), 177-200.

Dates
First available in Project Euclid: 7 August 2009

https://projecteuclid.org/euclid.tjm/1249648416

Digital Object Identifier
doi:10.3836/tjm/1249648416

Mathematical Reviews number (MathSciNet)
MR2541163

Zentralblatt MATH identifier
1211.53079

Citation

KOKUBU, Masatoshi. Surfaces and Fronts with Harmonic-mean Curvature One in Hyperbolic Three-space. Tokyo J. Math. 32 (2009), no. 1, 177--200. doi:10.3836/tjm/1249648416. https://projecteuclid.org/euclid.tjm/1249648416

References

• R. Bryant, Surfaces of mean curvature one in hyperbolic space, in Théorie des variétés minimales et applications, Astérisque, 154–155 (1988), 321–347.
• P. Collin, L. Hauswirth and H. Rosenberg, The geometry of finite topology Bryant surfaces, Ann. of Math. (2), 153 (2001), no. 3, 623–659.
• C. L. Epstein, Envelopes of Horospheres and Weingarten Surfaces in Hyperbolic $3$-Space, unpublished.,
• J. A. Gálvez, A. Martínez and F. Milán, Flat surfaces in hyperbolic $3$-space, Math. Ann., 316 (2000), no. 3, 419–435.
• J. A. Gálvez, A. Martínez and F. Milán, Complete linear Weingarten surfaces of Bryant type. A Plateau problem at infinity, Trans. Amer. Math. Soc., 356 (2004), no. 9, 3405–3428.
• M. Kokubu, W. Rossman, K. Saji, M. Umehara and K. Yamada, Singularities of flat fronts in hyperbolic space, Pacific J. Math., 221 (2005), no. 2, 303–351.
• M. Kokubu, M. Umehara and K. Yamada, An elementary proof of Small's formula for null curves in $\PSL(2,\C)$ and an analogue for Legendrian curves in $\PSL(2,\C)$, Osaka J. Math., 40 (2003), no. 3, 697–715.
• M. Kokubu, M. Umehara and K. Yamada, Flat fronts in hyperbolic $3$-space, Pacific J. Math., 216 (2004), no. 1, 149–175.
• M. Spivak, A comprehensive introduction to differential geometry, Vol. V, Publish or Perish, Inc., 1979.
• M. Umehara and K. Yamada, Complete surfaces of constant mean curvature $1$ in the hyperbolic $3$-space, Ann. of Math. (2), 137 (1993), no. 3, 611–638.


text:
1. ## financial homework help

calculating returns:
suppose you bought a 7 percent coupon bond one year ago for $893. the bond sells for$918 today.
a. assuming a \$1000 face value, what was your total dollar return on this investment over the past year.
b.what was your total nominal rate of return on this investment over the past yr.
c.if inflation rate last year was 4 percent, what was your total real rate of return on this investment?

2. ## Re: financial homework help

Please demonstrate a cashflow map and we can start.

3. ## Re: financial homework help

i'm kind of stuck to what to do because my online class instructor does not give me notes to start

4. ## Re: financial homework help

Originally Posted by dragonfly1
i'm kind of stuck to what to do because my online class instructor does not give me notes to start
Well fire him/her then!

Do you know what "a 7% bond with semiannual coupons" means?

5. ## Re: financial homework help

yes i think it means it paid twice a yr would it be 70 x 2= 140.00

6. ## Re: financial homework help

You MUST be able to draw a map. In this case, it might look like this:

Time is years.

0 -- (893.00)
½ -- +70.00
1 -- +70.00 + 918.00

7. ## Re: financial homework help

here is what i recd from the instructor
the total dollar return is the change in price plus the coupon payment
the nominal percentage return of the bond is total dollar return/purchase price
(1+R)=(1+r)(1+h)

8. ## Re: financial homework help

Originally Posted by dragonfly1
the total dollar return is the change in price plus the coupon payment
Well DFly, the dollar return is EASILY calculated as 70 + 70 + 918-893 = 165 : do you not "see" that?

9. ## Re: financial homework help

sounds like you are being smart here, if i saw it i would not ask for help
i'm new to this that's why i'm asking for help

10. ## Re: financial homework help

Originally Posted by dragonfly1
sounds like you are being smart here, if i saw it i would not ask for help
i'm new to this that's why i'm asking for help
My purpose was to try and see "where you're at", as we have no idea since you are disregarding the rules
and showing no work; if you are so easily "hurt", then I'll bow out and leave you in TK's hands

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•


text:
# Math Help - solid geometry sphere problem

1. ## solid geometry sphere problem

A sphere having a radius of 6 cm rests on 3 horizontal wires forming a plane triangle whose sides are 5 cm, 12 cm and 13 cm. What is the height of the top of the sphere above the plane of the wires?

2. Originally Posted by fishlord40
A sphere having a radius of 6 cm rests on 3 horizontal wires forming a plane triangle whose sides are 5 cm, 12 cm and 13 cm. What is the height of the top of the sphere above the plane of the wires?
1. The tangent points of the sphere and the wires are placed on the incircle of the triangle. The incircle has the radius r. The sphere has the radius R.

2. The triangle is a right triangle because $5^2+ 12^2 = 13^2$

3. Let A denote the area of the triangle. Then the radius of the incircle is calculated by:

$r = \dfrac {2A}{a+b+c} = \dfrac{ a \cdot b}{a+b+c}$

With your values you'll get $r = \dfrac{60}{30}=2$

4. The height of the sphere above the triangle is $h = R+h'$ with

$h'^2+r^2=R^2$

With your values you'll get: $h'=\sqrt{32} = 4\sqrt{2}$

and consequently $h = R+h'~\implies~h=6+4\sqrt{2}\approx 11.66$


text:
Tag

Sort by:

Parseval's Integral

The Poisson integral with ,where is a Bessel function of the first kind.

Legendre Duplication Formula

Gamma functions of argument can be expressed in terms of gamma functions of smaller arguments. From the definition of the beta function,(1)Now, let , then(2)and , so and(3)(4)(5)(6)Now, use the beta function identity(7)to write the above as(8)Solving for and using then gives(9)(10)

Trigamma Function

Min Max Re Im A special function corresponding to a polygamma function with , given by(1)An alternative function(2)is sometimes called the trigamma function, where(3)Sums and differences of for small integral and can be expressed in terms of Catalan's constant and . For example,(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)

Lanczos Approximation

An approximation for the gamma function with is given by(1)where is an arbitrary constant such that ,(2)where is a Pochhammer symbol and(3)and(4)(5)with (Lanczos 1964; Luke 1969, p. 30). satisfies(6)and if is a positive integer, then satisfies the identity(7)(Luke 1969, p. 30).A similar result is given by(8)(9)(10)where is a Pochhammer symbol, is a factorial, and(11)The first few values of are(12)(13)(14)(15)(16)(OEIS A054379 and A054380; Whittaker and Watson 1990, p. 253). Note that Whittaker and Watson incorrectly give as 227/60.Yet another related result gives(17)(Whittaker and Watson 1990, p. 261), where is a Hurwitz zeta function and is a polygamma function.

Kramp's Symbol

The symbol defined by(1)(2)(3)where is the Pochhammer symbol and is the gamma function. Note that the definition by Erdélyi et al. (1981, p. 52) incorrectly gives the prefactor of (3) as .

Dixon's Identity

where is a binomial coefficient and is a gamma function.

Thomae's Theorem

Thomae's theorem, also called Thomae's transformation, is the generalizedhypergeometric function identity(1)where is the gamma function, is a generalized hypergeometric function,(2)and (Bailey 1935, p. 14). It is a generalization of Dixon's theorem (Slater 1966, p. 52).An equivalent formulation is given by(3)(Hardy 1999, p. 104). The symmetry of this form was used by Ramanujan in his proof of the identity, which is essentially the same as Thomae's. Interestingly, this is one of the few cases in which Ramanujan gives an explicit proof of one of his propositions (Hardy 1999, p. 104).A special case of the theorem is given by(4)(J. Sondow, pers. comm., May 25, 2003).

Digamma Function

Min Max Re Im A special function which is given by the logarithmic derivative of the gamma function (or, depending on the definition, the logarithmic derivative of the factorial).Because of this ambiguity, two different notations are sometimes (but not always) used, with(1)defined as the logarithmic derivative of the gamma function , and(2)defined as the logarithmic derivative of the factorial function. The two are connected by the relationship(3)The th derivative of is called the polygamma function, denoted . The notation(4)is therefore frequently used for the digamma function itself, and Erdélyi et al. (1981) use the notation for . The digamma function is returned by the function PolyGamma[z] or PolyGamma[0, z] in the Wolfram Language, and typeset using the notation .The digamma function arises in simple sums such as(5)(6)where is a Lerch transcendent.Special cases are given by(7)(8)(9)(10)Gauss's digamma theorem states..

Incomplete Gamma Function

The "complete" gamma function can be generalized to the incomplete gamma function such that . This "upper" incomplete gamma function is given by(1)For an integer (2)(3)where is the exponential sum function. It is implemented as Gamma[a, z] in the Wolfram Language.The special case of can be expressed in terms of the subfactorial as(4)The incomplete gamma function has continued fraction(5)(Wall 1948, p. 358).The lower incomplete gamma function is given by(6)(7)(8)where is the confluent hypergeometric function of the first kind. For an integer ,(9)(10)It is implemented as Gamma[a,0, z] in the Wolfram Language.By definition, the lower and upper incomplete gamma functions satisfy(11)The exponential integral is closely related to the incomplete gamma function by(12)Therefore, for real ,(13)..

Central Beta Function

Min Max Re Im The central beta function is defined by(1)where is the beta function. It satisfies the identities(2)(3)(4)(5)With , the latter gives the Wallis formula. For , 2, ... the first few values are 1, 1/6, 1/30, 1/140, 1/630, 1/2772, ... (OEIS A002457), which have denominators .When ,(6)where(7)The central beta function satisfies(8)(9)(10)(11)For an odd positive integer, the central beta function satisfies the identity(12)

Regularized Gamma Function

The regularized gamma functions are defined by(1)(2)where and are incomplete gamma functions and is a complete gamma function. The function is implemented in the Wolfram Language as GammaRegularized[a, 0, z], and is implemented as GammaRegularized[a, z]. and satisfy the identity(3)The derivatives of and are(4)(5)and the second derivatives are(6)(7)The integrals are(8)(9)

Incomplete Beta Function

A generalization of the complete beta function definedby(1)sometimes also denoted . The so-called Chebyshev integral is given by(2)The incomplete beta function is implemented in the Wolfram Language as Beta[z, a, b].It is given in terms of hypergeometric functionsby(3)(4)It is also given by the series(5)where is a Pochhammer symbol.The incomplete beta function reduces to the usual beta function when ,(6)It has derivative(7)and indefinite integral(8)

Cauchy's Cosine Integral Formula

For ,where is the gamma function.

Regularized Beta Function

The regularized beta function is defined bywhere is the incomplete beta function and is the (complete) beta function. The regularized beta function is sometimes also denoted and is implemented in the Wolfram Language as BetaRegularized[z, a, b]. The four-argument version BetaRegularized[z1, z2, a, b] is equivalent to .

Hankel's Symbol

The symbol defined by(1)(2)where is the gamma function. If is an integer, then this simplifies to(3)given incorrectly by Erdélyi et al. (1981, p. 52).

Letwhere is a Pochhammer symbol, and let be a negative integer. Thenwhere is the gamma function.

Polygamma Function

A special function mostly commonly denoted , , or which is given by the st derivative of the logarithm of the gamma function (or, depending on the definition, of the factorial ). This is equivalent to the th normal derivative of the logarithmic derivative of (or ) and, in the former case, to the th normal derivative of the digamma function . Because of this ambiguity in definition, two different notations are sometimes (but not always) used, namely(1)(2)(3)which, for can be written as(4)(5)where is the Hurwitz zeta function.The alternate notation(6)is sometimes used, with the two notations connected by(7)Unfortunately, Morse and Feshbach (1953) adopt a notation no longer in standard use in which Morse and Feshbach's "" is equal to in the usual notation. Also note that the function is equivalent to the digamma function and is sometimes known as the trigamma function. is implemented in the Wolfram Language as PolyGamma[n, z] for positive..

Binet's Log Gamma Formulas

Binet's first formula for , where is a gamma function, is given byfor (Erdélyi et al. 1981, p. 21; Whittaker and Watson 1990, p. 249).Binet's second formula isfor (Erdélyi et al. 1981, p. 22; Whittaker and Watson 1990, pp. 250-251).

Gauss's Hypergeometric Theorem

for , where is a (Gauss) hypergeometric function. If is a negative integer , this becomeswhich is known as the Chu-Vandermonde identity.

Nu Function

(1)(2)where is the gamma function (Erdélyi et al. 1981, p. 217; Prudnikov et al. 1990, p. 799; Gradshteyn and Ryzhik 2000, p. 1109).The notation is also sometimes used to denote the divisor function giving the number of divisors or the number of distinct prime factors of a positive integer .

Mu Function

There are two functions commonly denoted , each of which is defined in terms of integrals. Another unrelated mathematical function represented using the Greek letter is the Möbius function.The two-argument -function is defined by the definite integralwhere is the gamma function (Erdélyi et al. 1981a, p. 388; Prudnikov et al. 1990, p. 798; Gradshteyn and Ryzhik 2000, p. 1109), while the three-argument -function is defined by(Prudnikov et al. 1990, p. 798; Gradshteyn and Ryzhik 2000, p. 1109).

Morley's Formula

(1)(2)where is a Pochhammer symbol and is the gamma function. This is a special case of the identity(3)

Barnes' Lemma

If a contour in the complex plane is curved such that it separates the increasing and decreasing sequences of poles, thenwhere is the gamma function (Bailey 1935, p. 7).Barnes' second lemma states thatprovided that (Bailey 1935, pp. 42-43).

Mellin's Formula

where is the digamma function and is the gamma function.

Malmstén's Formula

The integral representation of by(1)(2)where is the log gamma function and is the digamma function.

Bailey's Theorem

Let be the gamma function and denote a double factorial, thenWriting the sums explicitly, Bailey's theorem states

Exponential Integral

Min Max Re Im Let be the En-function with ,(1)(2)Then define the exponential integral by(3)where the retention of the notation is a historical artifact. Then is given by the integral(4)This function is implemented in the WolframLanguage as ExpIntegralEi[x].The exponential integral is closely related to the incomplete gamma function by(5)Therefore, for real ,(6)The exponential integral of a purely imaginarynumber can be written(7)for and where and are cosine and sine integral.Special values include(8)(OEIS A091725).The real root of the exponential integral occurs at 0.37250741078... (OEIS A091723), which is , where is Soldner's constant (Finch 2003).The quantity (OEIS A073003) is known as the Gompertz constant.The limit of the following expression can be given analytically(9)(10)(OEIS A091724), where is the Euler-Mascheroni constant.The Puiseux series of along the positive real axis is given by(11)where the denominators..

Eulerian Integral of the Second Kind

For and ,(1)(2)(3)where is the Pochhammer symbol and is the beta function.

Beta Function

The beta function is the name used by Legendre and Whittaker and Watson (1990) for the beta integral (also called the Eulerian integral of the first kind). It is defined by(1)The beta function is implemented in the Wolfram Language as Beta[a, b].To derive the integral representation of the beta function, write the product oftwo factorials as(2)Now, let , , so(3)(4)Transforming to polar coordinates with , (5)(6)(7)(8)The beta function is then defined by(9)(10)Rewriting the arguments then gives the usual form for the beta function,(11)(12)By symmetry,(13)The general trigonometric form is(14)Equation (14) can be transformed to an integral over polynomials by letting ,(15)(16)(17)(18)For any with ,(19)(Krantz 1999, p. 158).To put it in a form which can be used to derive the Legendre duplication formula, let , so and , and(20)(21)To put it in a form which can be used to develop integral representations of the Bessel functions and hypergeometric..

Log Gamma Function

Min Max Min Max Re Im The plots above show the values of the function obtained by taking the natural logarithm of the gamma function, . Note that this introduces complicated branch cut structure inherited from the logarithm function. Min Max Re Im For this reason, the logarithm of the gamma function is sometimes treated as a special function in its own right, and defined differently from . This special "log gamma" function is implemented in the Wolfram Language as LogGamma[z], plotted above. As can be seen, the two definitions have identical real parts, but differ markedly in their imaginary components. Most importantly, although the log gamma function and are equivalent as analytic multivalued functions, they have different branch cut structures and a different principal branch, and the log gamma function is analytic throughout the complex -plane except for a single branch cut discontinuity along the negative real axis. In particular,..

Pochhammer Symbol

The Pochhammer symbol(1)(2)(Abramowitz and Stegun 1972, p. 256; Spanier 1987; Koepf 1998, p. 5) for is an unfortunate notation used in the theory of special functions for the rising factorial, also known as the rising factorial power (Graham et al. 1994, p. 48) or ascending Factorial (Boros and Moll 2004, p. 16). The Pochhammer symbol is implemented in the Wolfram Language as Pochhammer[x, n].In combinatorics, the notation (Roman 1984, p. 5), (Comtet 1974, p. 6), or (Graham et al. 1994, p. 48) is used for the rising factorial, while or denotes the falling factorial (Graham et al. 1994, p. 48). Extreme caution is therefore needed in interpreting the notations and .The first few values of for nonnegative integers are(3)(4)(5)(6)(7)(OEIS A054654).In closed form, can be written(8)where is a Stirling number of the first kind.The Pochhammer symbol satisfies(9)the dimidiation formulas(10)(11)and..

Gamma Function

The (complete) gamma function is defined to be an extension of the factorial to complex and real number arguments. It is related to the factorial by(1)a slightly unfortunate notation due to Legendre which is now universally used instead of Gauss's simpler (Gauss 1812; Edwards 2001, p. 8).It is analytic everywhere except at , , , ..., and the residue at is(2)There are no points at which .The gamma function is implemented in the WolframLanguage as Gamma[z].There are a number of notational conventions in common use for indication of a power of a gamma functions. While authors such as Watson (1939) use (i.e., using a trigonometric function-like convention), it is also common to write .The gamma function can be defined as a definite integral for (Euler's integral form)(3)(4)or(5)The complete gamma function can be generalized to the upper incomplete gamma function and lower incomplete gamma function . Min Max Re Im Plots of the real and imaginary..

Stirling's Approximation

Stirling's approximation gives an approximate value for the factorial function or the gamma function for . The approximation can most simply be derived for an integer by approximating the sum over the terms of the factorial with an integral, so that(1)(2)(3)(4)(5)(6)The equation can also be derived using the integral definition of the factorial,(7)Note that the derivative of the logarithm of the integrandcan be written(8)The integrand is sharply peaked with the contribution important only near . Therefore, let where , and write(9)(10)Now,(11)(12)(13)so(14)(15)(16)Taking the exponential of each side thengives(17)(18)Plugging into the integral expression for then gives(19)(20)Evaluating the integral gives(21)(22)(Wells 1986, p. 45). Taking the logarithm of bothsides then gives(23)(24)This is Stirling's series with only the first term retained and, for large , it reduces to Stirling's approximation(25)Taking successive..

Exponential Sum Function

The exponential sum function , sometimes also denoted , is defined by(1)(2)where is the upper incomplete gamma function and is the (complete) gamma function.

Stirling's Series

The asymptotic series for the gammafunction is given by(1)(OEIS A001163 and A001164).The coefficient of can given explicitly by(2)where is the number of permutations of with permutation cycles all of which are (Comtet 1974, p. 267). Another formula for the s is given by the recurrence relation(3)with , then(4)where is the double factorial (Borwein and Corless 1999).The series for is obtained by adding an additional factor of ,(5)(6)The expansion of is what is usually called Stirling's series. It is given by the simple analytic expression(7)(8)(OEIS A046968 and A046969), where is a Bernoulli number. Interestingly, while the numerators in this expansion are the same as those of for the first several hundred terms, they differ at , 1185, 1240, 1269, 1376, 1906, 1910, ... (OEIS A090495), with the corresponding ratios being 37, 103, 37, 59, 131, 37, 67, ... (OEIS A090496)...

Weierstrass Form

There are (at least) two mathematical objects known as Weierstrass forms. The first is a general form into which an elliptic curve over any field can be transformed, given bywhere , , , , and are elements of .The second is the definition of the gamma functionaswhere is the Euler-Mascheroni constant (Krantz 1999, p. 157).

Check the price


text:
# Transactions of Kettle Corporation are as follows. 1. The company is granted a charter that...

Transactions of Kettle Corporation are as follows. 1. The company is granted a charter that authorizes the issuance of 150,000 preferred shares and 150,000 common shares without par value. 2. The founders of the corporation are issued 10,000 common shares for land valued by the board of directors at $210,000 (based on an independent valuation). 3. Sold 15,200 preferred shares for cash at$110 per share. 4. Repurchased and cancelled 3,000 shares of outstanding preferred shares for cash at $100 per share. 5. Repurchased and cancelled 4,000 preferred shares for cash at$98 per share. 6. Repurchased for cash and cancelled 500 shares of the outstanding common shares issued in item 2 above at $49 per share. 7. Issued 2,000 preferred shares at$99 per share. Instructions (a). Prepare entries in journal form to record the transactions listed above. No other transactions affecting the capital share accounts have occurred. (b). Assuming that the company has retained earnings from operations of \$1,032,000, prepare the shareholders’ equity section of its balance sheet after considering all the transactions above. (c). Why is the distinction between paid-in capital and retained earnings important? (d). How would the repurchase of the preferred shares differ if the preferred shares were retractable or callable/ redeemable? View Solution:
Transactions of Kettle Corporation are as follows 1 The company

## Plagiarism Checker

Submit your documents and get free Plagiarism report

Free Plagiarism Checker


text:
Generiloze: the difference of two squares is equal to a odd number.

For example,

3 = 2^2 - 1^2
5 = 3^2 - 2^2
7 = 4^2 - 3^2
...

Is there a general formula to explain this phenomenon?

• $(x+1)^2-x^2=x^2+2x+1-x^2=2x+1$. Sep 10, 2014 at 7:17
• Its generalize .. not generiloze :) Dec 9, 2014 at 9:10

3 Answers

The difference of two general squares doesn't need to be an odd number: $16$ and $4$ are both square numbers, but there difference $16-4=12$ is not odd.

Is there a general formula to explain this phenomenon?

Yes, if the question is, why is the difference between two consecutive square numbers odd? Two consecutive square numbers are of the form $n^2, (n+1)^2$ and their difference is $(n+1)^2 - n^2 = n^2 + 2n+ 1 - n^2 = 2n+1$, and $2n+1$ is an odd number.

If you have a number $n\geq0$ then: $$(n+1)^2 - n^2 = (n^2+2n+1)-n^2 = 2n+1,$$ which makes all odd numbers (i.e. every odd number can be written as $2n+1$, with a certain value for $n$).

The difference of an odd number and an even number (or the other way around) is an odd number.


text:
## Friday, 18 October 2019

Find here Time and Work Questions and Answers for IBPS Clerk ,IBPS PO and RRB , Bank Exams Online Practice free here

1.  A and B together can do a piece of work in 12 days and B and C finish the same work together in 16 days. After A has worked on it for 5 days and B for 7 days, C finishes it in 13 days. In how many days will C alone be able to finish the work ?
a
16
b
24
c
36
d
48
e
50
Explanation :
2.  Ramesh is twice as good a workman as Sunil and finishes a piece of work in 3 hours less than Sunil. In how many hours they together could finish the same piece of work ?
a
2$1 \over 3$
b
2
c
1$2 \over 3$
d
4
e
None of these
Explanation :
3.  A mother and a daughter working together can complete a certain work in 4 days. But if the mother worked alone she could complete the work in 6 days. Both of them worked for one day and then the mother had to leave. How long will the daughter take to complete remaining work ?
a
7 days
b
8 days
c
6 days
d
9 days
e
10 days
Explanation :

= ${12 \times 3} \over 4$ = 9 days
4.  If 15 women or 10 men can complete a project in 55 days, in how many days will 5 women and 4 men working together complete the same project ?
a
75
b
8
c
9
d
85
e
90
Explanation :
5.  24 men working 8 hours a day can finish a work in 10 days. Working at the rate of 10 hours a day, the number of men required to finish the same work in 6 days is :
a
30
b
32
c
34
d
36
e
None of these
Explanation :
6.  A can do a work in 18 days, B in 9 days and C in 6 days. A and B start working together and after 2 days C joins them. What is the total number of days taken to finish the work ?
a
4.33 day
b
4.0 day
c
4.66 day
d
5.33 day
e
None of these
Explanation :
7.  A worker is paid Rs 56 for 35 hour in a week. Up to 40 hours, he is paid at the normal rate and on overtime, 1.5 times the normal. How many hours did he work to get Rs 88 ?
a
48 hours
b
50 hours
c
58 hours
d
55 hours
e
60 hours
Explanation :
8.  A and B can do the a piece of work in 6 days. A alone can do it in 10 days. What time will B require to do it alone ?
a
20 days
b
15 days
c
25 days
d
3 days
e
2 days
Explanation :
9.  A and B weave a carpet in 10 days and 15 days, respectively. They begin to work together but B leaves after 2 days. In what time will A complete the remaining work ?
a
6$1 \over 3$ days
b
6$2 \over 3$ days
c
7 days
d
8 days
e
None of these
Explanation :
10.  A can do a piece of work in 25 days and B in 20 days. They work together for 5 days and then A goes away. In how many days will B finish the remaining work ?
a
17 days
b
11 days
c
10 days
d
12 days
e

Explanation :
11.  Two men and 7 children complete a certain piece of work in 4 days, while 4 men and 4 children complete the same work in only 3 days. The number of days required by 1 man to complete the work is :
a
60 days
b
15 days
c
6 days
d
51 days
e
52 days
Explanation :
12.  A can do a work in 9 days. If B is 50% more efficient than A, then in how many days can B do the same work alone ?
a
13.5
b
4.5
c
6
d
3
e
None of these
Explanation :
13.  A can do a piece of work in 10 days, while B alone can do it in 15 days. They work together for 5 days and the rest of the work is done by C in 2 days. If they get Rs 450 for the whole work, how should they divide the money ?
a
Rs.225,Rs.150,Rs.75
b
Rs.250,Rs.100,Rs.100
c
Rs.200,Rs.150,Rs.100
d
Rs.175,Rs.175,Rs.100
e
None of these
Explanation :
14.  A can do 50% more work as B can do in the same time. B alone can do a piece of work in 20 hours. A, with help of B, can finish the same work in how many hours ?
a
12
b
8
c
13$1 \over 3$
d
5$1 \over 2$
e
47$1 \over 2$
Explanation :
15.  If 15 men or 24 women or 36 boys can do a piece of work in 12 days, working 8 hours a day, how many men must be associated with 12 women and 6 boys to do another piece of work 2$1 \over 4$ times as great in 30 days working 6 hours a day ?
a
4
b
8
c
6
d
10
e
12
Explanation :
16.  A certain job was assigned to a group of men to do it in 20 days. But 12 men did not turn up for the job and the remaining men did the job in 32 days. The original number of men in the group was :
a
32
b
34
c
36
d
40
e
None of these
Explanation :
Let originally there were x men in the group.
Therefore, (x - 12) men did the job in 32 days.
20x = 32 (x - 12) or x = 32
17.  A daily wage worker was paid  1700 during a period of 30 days. During this period he was absent for 4 days and was fined Rs 15 per day for absence. He was paid the full salary only for 18 days as he came late on the other days. Those who came late were given only half the salary for that day. What was the total salary paid per month to the worker who came on time every day and was never absent?
a
Rs.2400
b
Rs.3000
c
Rs.2700
d
Rs.2250
e
None of these
Explanation :
18.  Rohit, Harsha and Sanjeev are three typists who working simultaneously can type 216 pages in four hour. In one hours Sanjeev can type as many pages more than Harsha as Harsha can type more than Rohit. During a period of five hours Sanjeev can type as many pages as Rohit can during seven hours. How many pages does each of them type per hour?
a
16,18,22
b
14,17,20
c
15,17,22
d
5,18,21
e
None of these
Explanation :
19.  A contractor undertakes to built a walls in 50 days. He employs 50 peoples for the same. However after 25 days he finds that only 40% of the work is complete. How many more man need to be employed to complete the work in time?
a
25
b
30
c
35
d
20
e
None of these
Explanation :
20.  A and B can finish a work in 10 days while B and C can do it in 18 days. A started the job, worked for 5 days, then B worked for 10 days and the remaining work was finished by C in 15 days. In how many days could C alone have finished the whole work ?
a
30 days
b
15 days
c
45 days
d
24 days
e
None of these
Explanation :
21.  4 men and 10 women were put on a work. They completed 1/3 of the work in 4 days. After this 2 men and 2 women were increased. They completed 2/9 more of the work in 2 days. If the remaining work is to be completed in 3 days, then how many more women must be increased ?
a
32
b
8
c
50
d
55
e
None of these
Explanation :
22.  C is twice efficient as A. B takes thrice as many days as C. A takes 12 days to finish the work alone. If they work in pairs (i.e., AB, BC, CA) starting with AB on the first day, BC on the second day and AC on the third day and so on, then how many days are required to finish the work ?
a
6$1 \over 5$ days
b
4.5 days
c
5$1 \over 9$ days
d
8 days
e
None of these
Explanation :
23.  The work done by a woman in 8 hours is equal to the work done by a man in 6 hours and by a boy in 12 hours. If working 6 hours per day 9 men can complete a work in 6 days, then in how many days can 12 men, 12 women and 12 boys together finish the same work, working 8 hours per day ?
a
3$2 \over 3$ days
b
4$1 \over 2$ days
c
3 days
d
1$1 \over 2$ days
e
None of these
Explanation :
24.  A team of workers was employed by a contractor who undertook to finish 360 pieces of an article in a certain number of days. Making four more pieces per day than was planned, they could complete the job a day ahead of schedule. How many days did they take to complete the job ?
a
10 days
b
8 days
c
9 days
d
12 days
e
None of these
Explanation :

25.  Eight children and 12 men complete a certain piece of work in 9 days. If each child takes twice the time taken by a man to finish the work, in how many days will 12 men finish the same work?
a
15
b
8
c
9
d
12
e
None of these
Explanation :
26.  A, B and C working together completed a job in 10 days. However, C only worked for the first three days when $3 \over 100$ of the job was done. Also, the work done by A in 5 days is equal to the work done by B in 4 days. How many days would be required by the fastest worker to complete the entire work ?
a
25 days
b
20 days
c
30 days
d
40 days
e
None of these
Explanation :
27.  Two men undertake to do a piece of work for Rs.600. One alone could do it in 6 days and the other in 8 days. With the assistance of a boy, they finished it in 3 days. Boy's share should be
a
Rs.75
b
Rs.225
c
Rs.300
d
Rs.100
e
None of these
Explanation :
28.  If 6 engines consume 15 metric tones of coal when each is running 9 hours a day, how much coal will be required for 8 engines, each running 12 hours a day, if being given that 3 engines of former type consume as much as 4 engines of latter type?
a
20
b
18
c
27
d
36
e
None of these
Explanation :
29.  A tank is filled in 5 hours by three pipes A, B and C. The pipe C s twice as fast as B and B is twice as fast as A. How much time will pipe A alone take to fill the tank?
a
35 hr
b
30 hr
c
28 hr
d
26 hr
e
None of these
Explanation :
30.  Kumar can do a work in 3 days while Dinesh can do same work in 2 days. Both of them finish the work together an get Rs.150. What is the share of Kumar ?
a
30
b
60
c
70
d
75
e
None of these
Ratio of 1 day work of kumar and dinesh = $1 \over 3$ : $1 \over 2$ = 2 : 3
Share of kumar in wages = $2 \over {2 + 3}$ $\times$ 150 = 2 $\times$ 30 = Rs.60


text:
# How bad does a ring have to be for a failure of “going-in-between”?

Let $A\subset B$ be an integral extension of commutative unital rings.

Let $\mathfrak{p}_0\subset\mathfrak{p}_1\subset\mathfrak{p}_2$ be a saturated chain of primes in $A$ of length $2$.

Suppose $\mathfrak{q}_0,\mathfrak{q}_2$ lie over $\mathfrak{p}_0,\mathfrak{p}_2$, and $\mathfrak{q}_0\subset\mathfrak{q}_2$.

Is there necessarily a $\mathfrak{q}_1$ satisfying $\mathfrak{q}_0\subset\mathfrak{q}_1\subset\mathfrak{q}_2$ and lying over $\mathfrak{p}_1$?

It seems to me the answer is clearly yes if the rings $A,B$ are sufficiently geometric (edit: even this is no longer clear to me, see addendum), e.g. finitely generated algebras over an algebraically closed field, since in this case, if there is no $\mathfrak{q}_1$, then $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and then $V(\mathfrak{q}_2)$ is a codimension one subvariety of $V(\mathfrak{q}_0)$, and $\operatorname{Spec}B\rightarrow\operatorname{Spec}A$ is a dimension-preserving map, so what is $V(\mathfrak{p}_1)$'s dimension?

But in general, it's not obvious to me. It seems to require that going-down holds in the integral extension of domains $A/\mathfrak{p}_0\subset B/\mathfrak{q}_0$, and because the going-down theorem requires an extra assumption of integral closure, shouldn't this fail sometimes?

So my question is:

How bad do $A,B$ have to be for $\mathfrak{q}_1$ to fail to exist? Can it happen for noetherian rings? Cohen-Macaulay rings? What's the "least pathological" example?

NB: This is crossposted from math.SE, where it hasn't gotten any answers after 2 weeks and a bounty. (Edit: it is no longer really the same question, see addendum.)

Addendum: A propos of an exchange in the comments with Jason Starr (partially deleted now), even in the geometric case it's no longer clear to me that the answer is yes. When I originally posted, I mistook the fact that a prime between $\mathfrak{q}_0$ and $\mathfrak{q}_2$ would necessarily lie over some prime strictly between $\mathfrak{p}_0$ and $\mathfrak{p}_2$ to imply it would lie over $\mathfrak{p}_1$. (Jason pointed out this mistake.) I corrected this by explicitly adding the requirement that $\mathfrak{q}_1$ lie over $\mathfrak{p}_1$, but now it is no longer true that the failure of $\mathfrak{q}_1$ to exist implies that $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated. Thus, perhaps $V(\mathfrak{q}_2)$ is codimension 2 in $V(\mathfrak{q}_0)$, it's just that none of the codimension one subvarieties of $V(\mathfrak{q}_0)$ happen to lie over $V(\mathfrak{p}_1)$. It's not at all clear to me that this can't happen.

Fred Rohrer mentions a highly relevant paper by L.J. Ratliff in the comments. A cursory scan of this paper seems to indicate that the answer might be no most of the time. Ratliff states that the condition I'm asking for here is not even guaranteed if $A$ is a complete regular local ring (bottom of p. 778), though it's not completely clear from context whether he's requiring $B$ to be integral in this case. What's an example?

One last point is that these developments distinguish this question from the one I asked at math.SE, where I asked, "In what generality does it hold that if $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated, and $\mathfrak{q}_i$ lies over $\mathfrak{p}_i$ in $A$, then $\mathfrak{p}_0\subset\mathfrak{p}_2$ is saturated?" This turns out (if I am thinking straight) to be the question addressed by Ratliff's paper, but I am asking for something stronger at present.

• You might wish to have a look at L.J.Ratliff Jr., Going between rings and contractions of saturated chains of prime ideals, Rocky Mountain J. Math. 7 (1977), 777-787. (I have not studied this in detail myself.) – Fred Rohrer Mar 13 '17 at 12:46
• @JasonStarr - you're right. I think what I was thinking was that $\mathbb{q}_1$ lies over a prime that is strictly between $\mathfrak{p}_0$ and $\mathfrak{p}_2$. I'll edit the post. – benblumsmith Mar 13 '17 at 13:10
• @JasonStarr - A propos of your correction, now it's not even completely clear to me in the geometric situation! Because the failure of $\mathfrak{q}_1$ to exist no longer implies that $\mathfrak{q}_0\subset\mathfrak{q}_2$ is saturated. – benblumsmith Mar 13 '17 at 13:36
• @FredRohrer - Thank you, this looks extremely promising. – benblumsmith Mar 13 '17 at 13:45

This is a partial answer, giving a two dimensional Noetherian counterexample. It starts from your observation that one has to consider the case where $A/\mathfrak{p}_0$ is not integrally closed. Geometrically the construction is as follows: take a normal affine surface $Y$, a curve $C$ on $Y$, a point $P_1 \in C$ and a point $P_2 \in Y \setminus C$. Construct a non-normal surface $X$ by identifying $P_1$ and $P_2$; let $P'$ be the image of $P_1$ and $P_2$ on $X$ and $C'$ be the image of $C$ on $X$. Then the chains $X \supset C' \ni P'$ on $X$ and $Y \ni P_2$ on $Y$ give a counterexample.
For an explicit example, take $Y = k^2$, where $k$ is any field, $C = x$-axis, $P_1 = (0,0)$, $P_2 = (0,1)$. So $B = k[x,y]$ and $A$ is the set of all polynomials $f \in B$ such that $f(0,0)= f(0,1)$. $B$ is integral over $A$, since e.g. $x \in B$ and $y(y-1) \in B$.
• How do you check that $A$ is Noetherian? – R. van Dobben de Bruyn Dec 13 '19 at 19:04
• $A$ is generated by $A'$ and $x$, where $A'$ is the subalgebra of $k[y]$ consisting of all polynomials $f$ such that $f(0) = f(1)$. $A'$ is finitely generated, since all subalgebras of $k[y]$ are. – auniket Dec 14 '19 at 2:09


text:
#### Category : relational-database

$posts = Post::with(‘user’)->search(‘title’,$this->search)->latest()->paginate(5); Sourc..

$posts = Post::with(‘user’) ->search(‘title’,$this->search) ->latest() ->paginate(5); Sourc..

Assume I use MySQL and there is two tables, students and parents_guardians table. Table students is used to store student’s data. While parents_guardians table is used to store student’s parents and student’s guardians. So, either a student can have father and mother as his parents, or a student can have uncle and aunt as his ..

there is a problem with the output of all data in one collection. There are 3 tables Example: categories, products_group, products. I tried to link tables via lanks but since the product table can be without a group, this is not the case. Migrations categories -id -title … products_group -id -title … products -id -title ..

I made many to many relationships for admin and school tables. "SQLSTATE[42S22]: Column not found: 1054 Unknown column ‘admin_id’ in ‘field list’ (SQL: insert into admin_school (admin_id, school_id) values (91, 2))" Table : admins schools admin_school (pivot table) Here is structure pivot table Model Admin public function schools(){ return $this->belongsToMany(‘AppModelsSchool’, ‘admin_school’, ‘admin_id’, ‘school_id’); } Model .. Read more I need some help designing my database relationships. I have these tables: documents, companies, individuals and users. A document can be internal or external. If it is external, it can be signed by a company or an individual. If it is internal it has to be signed by a user. In any case, the document .. Read more I want to do change with method by where like this User::where(〜)->with(‘post’)->orWhere(…)->with(‘post2’)->paginate(10) please help me. Sourc.. Read more I want to do change「with」 method by 「where」. like this. User::where(〜)->with(‘post’)->orWhere(…)->with(‘post2’)->paginate(10) For example(This example is not work. This is for image)$user = User::Where(function ($query) use ($id1, $id2) {$query->where(‘user_id’, $id1) ->whereIn(‘monster_id’,$id2) ->with([‘user’ => function ($query) {$query->with(‘residence’)->get(); }]); })->orWhere(function ($query) use ($id1, $id2) {$query->where(‘monster_id’, $id1) ->where(‘user_id’,$id2) ->with([‘user2’ => function ($query) { .. Read more Message Tables Column is | messages_id(pk) | user_id | target_id | message | created_at User_id has sender user’s id. Target_id has receiver user’s id And simple user table has relation to MessageTable by HasMany. public function message() { return$this->hasMany(‘AppModelTableModelMessage’, ‘user_id’, ‘user_id’); } I want to get Userdata and each last message from user_id=[1,2,3,4] to ..

I have an One-to-One relationships Between the user (penggunas) table and the role table, but when I access it in the controller it returns null and column not found Route.php Route::get(‘/test’, ‘[email protected]’); PenggunaController.php use AppPengguna; class PenggunaController extends Controller { public function index() { $pengguna = Pengguna::find(3); return response()->json([ ‘data’ =>$pengguna ]); } } ..

I’m producing an online school platform and in part of it student can see it’s classes with their informations..but the related function return this error : Undefined offset: 0 (View: C:xampphtdocsOnlineSchoolresourcesviewsadminstudentClassReport.blade.php) there are some relations between models My Models: USER(students and teachers) ، LEVEL(level of classes) ، Classroom. => there is a many to many ..

I have Book, Magazine and Song models. I want to make a Collection model that will contain models of these three types, so I could query it like this (pseudo code): Collection->get() and get a result similar to this: [ {Book_1} {Magazine_1} {Book_2} {Song_1} {Book_3} … ] Is it possible to do in Laravel, and ..

I have meals model which look like these class meals extends Model { use HasFactory; protected $appends = [ ‘image_full_path’, ]; /** * The attributes that are mass assignable. * * @var array */ protected$fillable = [ ‘name’, ‘text’,’price’,’image_url’,’rating’,’discount’,’sub_catogry_id’ ]; public function sub_catogry() { return $this->belongsTo(‘AppModelssub_catogry’); } public function order_meal_list() { return$this->hasMany(order_meal_list::class); } ..

I have a small blogging app which allows users to upload photos and audio. Users can also search for other blogs. While searching, I’d like to add a ranking to my query with the following precedence: 1) Users with most photos 2) Users with most audio User model is constructed as such: public function blog() ..

I have tow table named A and B that have n-m relation so I must have third table named C. (table C keep A id and B id !) In Laravel models, can I define relations? Sourc..

I want the permissions table to be filled in the database when selecting roles My tables: permissions -> id | name roles -> id | name permission_role -> permission_id | role_id permission_user -> permission_id | user_id role_user -> role_id | user_id I create the permissions and roles data in the admin panel and When I ..

I am trying to retrieve the number of relations using the withCount method, but using a where condition that is limiting it to a specific amount and between specific dates. for example, i want to count how much withCount results will give me 1/2/3. the methods i am using now is this, but when selecting ..

I have two Models (Store, Product) and Relationship hasMany public function products(){ return $this->hasMany(Product::class); } And I want to return response collection, in class StoresCollection extends ResourceCollection public function toArray($request) { return $this->collection->map(function ($item) { return [ ‘id’ => $item->id, ‘seller_id’ =>$item->seller_id, ‘store_product’ => \$item->products()->get(), ]; }); } But I don’t want to return ..


text:
# User Error

Gah. I’m trying to write some straightforward combinatorial code in Haskell, and I finally got to the point where I had to start using nontrivial algorithms to make it acceptably fast. When I run those algorithms the first time, I get the following gem:

    *Main> drops 100 2
*** Exception: divide by zero

The “drops” function is now 15 lines of mutually recursive Haskell, and the interactive interpreter conveniently informs me that there was a division by zero. No hint as to where it might be, of course. Come to think of it, this was probably the reason I dropped back to Python after my last Haskell stint.

This came up all the time in OCaml compiler work too. Deep in the middle of an optimization pass, I’d mess up a symbol context. The compiler would conveniently inform me that

    NotFound

OCaml exceptions are monomorphic, so the default exception thrown by map classes couldn’t even include the key, much less something like a stack trace.

The conclusion: any language that claims to be high level should be able to give the exact context in which errors occurred. This means either some form of trace reporting and/or the ability to rule out the possibility of certain errors (dependent types). Yes, both Haskell and OCaml have debuggers for this purpose, but that isn’t good enough; in my case the history function in the Haskell prints out gobs and gobs of tail recursion frames before it ever gets to the real source of the problem. Debugging in high level languages should be easy, not just possible.

Update: the problem turns out to be integer overflow, which means I’m hitting the same language flaw Guido van Rossum described here. Converting my code from Int to Integer requires adding a slew of conversion calls and type annotations because I’m using integers as both container indices and as the results of large combinatorial values. Hopefully Tim Sweeney’s idea will work out and this dichotomy can go away.

Update (16mar2009): Possible hope for the future:

http://ghcmutterings.wordpress.com/2008/12/04/explicit-stack-traces


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Teacher's Edition Go to the latest version.

# 2.4: Multiplication of Rational Numbers

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

At the end of this lesson, students will be able to:

• Multiply by $-1$.
• Multiply rational numbers.
• Identify and apply properties of multiplication.
• Solve real-world problems using multiplication.

## Vocabulary

Terms introduced in this lesson:

argument
multiplicative properties
multiplicative identity property
distributive property

## Teaching Strategies and Tips

Changing the sign of a number is equivalent to multiplying it by $-1$. In Example 1, students find the opposite of several numbers and expressions being careful to use parentheses when appropriate and simplifying the result.

Use Examples 2 and 3 to justify why, when multiplying two fractions, the numerators multiply together and the denominators multiply together.

Use the product of three or more fractions as in Examples 4c and 4d as an extension of the multiplication rule.

• Multiply the following rational numbers.

$\frac{3} {11} \cdot \frac{5} {7}$

Hint: The product of two rational numbers is the product of their numerators divided by the product of their denominators.

• Multiply the following rational numbers.

$\frac{11} {5} \cdot \frac{7} {4} \cdot \frac{3} {10}$

Hint: Multiply all the numerators and all the denominators. Do not covert the improper fraction to mixed form.

• Multiply the following rational numbers.

$\frac{5} {7} \cdot 12$

Hint: Rewrite the $12$ as $12/1$, using the “invisible $1$”.

Students first learn about the convenience of canceling before multiplying in Examples 4d and 5.

• Multiply the following rational numbers.

$\frac{24} {33} \cdot \frac{8} {27} \cdot \frac{9} {64}$

Solution:

$\frac{24} {33} \cdot \frac{8} {27} \cdot \frac{9} {64} = \frac{3 \cdot 8} {3 \cdot 11} \cdot \frac{8} {3 \cdot 9} \cdot \frac{9} {8 \cdot 8} = \frac{\cancel{3} \cdot \cancel{8}} {\cancel{3} \cdot 11} \cdot \frac{\cancel{8}} {3 \cdot \cancel{9}} \cdot \frac{\cancel{9}} {\cancel{8} \cdot \cancel{8}} = \frac{1} {33}$

Use Examples 6-8 to introduce the four properties of real numbers which involve multiplication: the commutative, associative, multiplicative identity, and distributive properties.

• A geometric interpretation of the commutative property is to consider finding the area of a rectangle. $L \times W$ is the same number no matter how you draw the rectangle or what you call $L$ and $W$; therefore, $L \times W = W \times L$. Similarly, the commutative property says that the order for multiplying any two real numbers does not matter. See Example 6.
• The associative property of multiplication concerns three or more numbers. Just as for addition, the sum is the same regardless of how they are grouped and in which pair the multiplication takes place first.
• State the rule being used in each example you do in the classroom.

## Error Troubleshooting

Example 1b: The opposite of $\pi$ is simply $-1 \cdot (\pi) = -\pi$. There is no need to use the decimal expansion.

Example 1c: Multiply both terms of the expression by $-1$. This will make more sense to students after covering the distributive law in the next lesson.

• Find the opposite of the expression

$x - 4y + 1$

Hint: multiply each of the three terms by $-1$.

The difference between absolute value and other grouping symbols is that multiplying absolute value by $-1$ will not affect the argument; that is, a negative will not distribute into the absolute value. See Example 1d.

General Tip: It is helpful to note that

• $|x|$ and $|-x|$ are always positive
• $-|x|$ is always negative.

General Tip: A common mistake is to forget to cancel like factors before multiplying the fractions, as the numbers will only get larger and thus harder to factor. Have students factor numerators and denominators first to remove any repetitions by canceling. Then carry out the remaining easier multiplication.

Feb 22, 2012

Aug 22, 2014


text:
# Tag Info

33

There are an infinite number of squares whose sides pass through exactly four circles. In fact, there are an infinite number of squares passing through the top four circles alone. Each frame of the above gif represents one possible square you can draw through those circles. In addition to the 90 squares represented in these 90 frames, there are infinitely ...

24

Yes, X can win. To simplify things I'm going to take advantage of your rules that O cannot win, so that X doesn't need to worry about O getting 1000 in a row. Just consider a 1 dimensional game, chose an origin, and label the locations with integers in order. Define the "bin $K$" as the set of spaces $x$ such that $1000\ K \le x < 1000 (K+1)$. A ...

22

I've got 20

20

This is pretty fun indeed! :) Don't forget to read @trolley813's concise proof (and upvote it :P)

19

I've done it in 21:

17

Took another approach...

15

Here is another simpler proof This generalises to all odd n. For even n, however, X. X. X..X X..X .... .XX. X..XX. X..... ..X..X ..X..X X..... X..XX. X..XX..X X......X ..X..X.. ..X..X.. X......X X..XX..X ........ .XX..XX. X..XX..XX. X......... ..X..XX..X ..X......X X...X..X.. X...X..X.. ..X......X ..X..XX..X X......... X..XX..XX.

15

I would suggest an alternate (simpler) strategy:

14

9 squares that are $1$x$1$ (trivial) 4 squares that are $\sqrt{2}$x$\sqrt{2}$ (Diamonds centered at the 4 central circles) 2 squares that are $\sqrt{5}$x$\sqrt{5}$ (Each starts at a yellow circle. Move like a knight: example 2 right 1 up, 2 up 1 left, 2 left 1 down, 2 down 1 right) 2 squares that are $\sqrt{13}$x$\sqrt{13}$ (Each starts at the orange ...

14

Thanks to @xnor presenting a solution, I was finally able to fix the bug in my Python 3 program. from collections import Counter import sys grid = [[7, 3, 1, 2, 9, 8, 5, 6, 4], [4, 9, 5, 1, 8, 6, 7, 2, 3], [9, 4, 6, 7, 5, 3, 1, 8, 2], [1, 5, 3, 8, 7, 4, 2, 9, 6], [8, 6, 2, 5, 3, 1, 4, 7, 9], [2, 8, 9, 4, 1, 5, 6, 3, ...

14

Well, not strictly an answer (since I've found exactly the same as one already posted), but rather a proof that other ways are not possible:

14

I found a different way to solve it. Note that this doesn't disprove @trolley813's proof - it only shows one part of the proof is incorrect:

13

My solution: Solution method:

10

10

Going first, you should be able to force a win. Using @kaine's terminology: box - an area on the board that can be made into a square square - a box of 4 lines three - a box of three lines which can be made into a square on the next turn safe - a move that doesn't result in a three Moves are enumerated as follows: * 1 * 2 * 3 4 5 * 6 * 7 * 8 9 ...

9

This game is called Sprouts. There is a discussion in Winning ways for your mathematical plays, Vol. 3, by Berlekamp, Conway, and Guy, on p. 598. Sprouts also has a Wikipedia page.

9

Maybe I am misunderstanding the puzzle, but it seemed straightforward to I literally plopped down circles as random as I could (but aligned to a grid, for ease of working) in a convenient drawing tool, in a 16x16 square. Then I The result is: As $a = 16$, I wonder if we are allowed to break that requirement for the bonus? Of course, circle placement will ...

9

First player wins Example:

8

I found 21 9 variations of black square 2 variations of orange square 4 variations of brown square 4 variations of gray square 2 variations of red square

8

... seems to work. (Brief) explanation of how I found it: Nice puzzle!

8

I solve it in two stages. First stage is The second stage is

8

This game is So in this case If we take "rational" to mean that when either player had a winning move they took it then Now So

8

The missing letters are T, H, U, N, and Y. Determining the first letter: Determining a second letter: And a third: And a... problem:

6

My answer is 21. It break down as follows: 9: Square from dots right next to each other 4: Squares with 1 dot in the middle 2: Squares with central 4 dots in the middle 2: Squares with previous 2 squares in the middle (The above can be more easily visualized from watching Kevin's answer) 4: From the 4 squares with 1 dot in the middle, extend out 1 more ...

6

I wouldn't call this 'elegant', but here's an argument that there can be no solution starting with Grid 1. In the first diagram, edges 1-6 must all have the same label: edges 1 and 2 because of the square they share, then edges 3 and 4 because of the vertex they share with 1 and 2, and finally 5 and 6 because of the square they share with 2 and 4. ...

6

Here's an easy "intuitive" explanation how X can win: If X can get a 998 in a row and it's her turn, she has won. So, if she can get TWO 998s in a row, and it's Os turn, she has won. {O will only be able to cover one - so it's over} So, if she can get FOUR 997s in a row, and it's O's turn, she has won. (O will only be able to cover ..two times.., so ...

6

6

Here's a way that should be faster than writing down letters. Let's say Steve was handed a piece a paper for him to draw on. Steve quickly draws an ichthus and adds some grill marks on it then rips the paper in half. He takes the bottom half, stands it up on the ground, and places some rocks around it so it stands up straight. If the answer is a destroyed ...

6

Alice wins. She starts by playing in the center square of the center board (which we'll call board C). Bob sends Alice over to another board, and she plays in the center of that one. Eventually, Bob fills up the rest of the middle board, and Alice then has the center of all but one of the outer boards. On that turn, Alice takes the "self-square" of the ...

6

It is not possible.

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Choose Language Hide Translation Bar
Highlighted
Level II

## How to plot different spec limit for different group

For example: I have a data table including 2 columns:

Value Group

12  1

65  2

13  1

67 2

Group 1 and Group 2 has different spec limits. How can I set them up and plot the corret spec limit in Graphic builder?

Thanks!

2 REPLIES 2
Highlighted
Super User

## Re: How to plot different spec limit for different group

Here is one method for doing what you want.  It envolves splitting the data into separate columns, and then setting spec limits for the split columns.  Finally the data are plotted using Graph Builder.

``````names default to here(1);
dt=open("\$SAMPLE_DATA/big class.jmp");

dtSplit = dt << Split(
Split By( :sex ),
Split( :height, :weight ),
Remaining Columns( Drop All ),
Sort by Column Property
);

dtSplit:Name("Height F") << set property("spec limits",{LSL( 57 ), USL( 65 ), Show Limits( 1 )});
dtSplit:Name("Height M") << set property("spec limits",{LSL( 60 ), USL( 70 ), Show Limits( 1 )});

Graph Builder(
Variables( X( :weight F ), Y( :height F ), Y( :height M ) ),
Elements(
Position( 1, 1 ),
Points( X, Y, Legend( 57 ) ),
Smoother( X, Y, Legend( 58 ) )
),
Elements(
Position( 1, 2 ),
Points( X, Y, Legend( 59 ) ),
Smoother( X, Y, Legend( 60 ) )
)
);``````

You could also just specify to add reference lines to the Graph Builder platform, based upon the different spec limits.

Jim
Highlighted
Staff

## Re: How to plot different spec limit for different group

If you don't want to split your data table, you can use the Process Screening platform (This platform can use your data table as it is currently set up).  I have a blog post that details how to do this.

https://community.jmp.com/t5/JMPer-Cable/Process-specification-limits-that-differ-based-on-a-groupin...

Article Labels

There are no labels assigned to this post.


text:
# How does one obtain an expression for the Dirichlet series $g(s, \theta) = \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n^{s}}$?

I would like to obtain an expression for the function $$g(s, \theta) = \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n^{s}} \qquad (\#).$$

Here is what I've tried so far: we know from the definition of the Dirichlet convolution that $$(E*I_{1})(n) = \sigma_{1}(n).$$ Here, $$E(n) = 1$$ for all $$n$$, $$I_{a}(n) = n^{a}$$ and $$\sigma_{1}$$ is the sum-of-divisors function. Now, instead of filling in numbers in these functions, I proceed with using functions as arguments. In particular, the Chebyshev polynomials of the first kind are relevant. Let $$T_{n}(s)$$ be the $$n$$'th Chebyshev polynomial of the first kind.

Then, I think: $$(E*I_{1})(T_{n}(s)) = \sigma(T_{n}(s))$$, where $$E(\cdot)$$ sends all polynomials to $$1$$, $$I_{1}(\cdot)$$ is the identity function for polynomials, and $$\sigma_{1}(\cdot)$$ is the sum of all polynomial divisors of its argument.

Question 1: is the equality $$(E*I_{1})(T_{n}(s)) = \sigma(T_{n}(s))$$ correct?

If we suppose it is, then I'd go on as follows: \begin{align*}L_{(E*I_{1})} (T_{n}(\cos(\theta))) &= L_{E} (T_{n}(\cos(\theta))) \cdot L_{I_{1}} (T_{n}(\cos(\theta))) \\ &= \zeta(s) \cdot g(s,\theta) = L_{\sigma}(\cos(n \theta)) . \end{align*}

Then I though of using an infinite product expression for $$\cos(n \theta)$$, in order to compute the sum of its polynomial divisors. An article like this one could come in handy, perhaps.

Question 2 Do you think this approach for finding $$g(\cdot)$$ as mentioned in $$(\#)$$ could work?

Question 3 Is there another method to obtain an expression for $$g(\cdot)$$ ? For instance, I think this paper could be very useful, but I'm not entirely sure how to apply their $$\bigodot$$ operator, and how it relates to the regular dirichlet convolution.

For $$\theta=a/q$$ rational it is a sum of Dirichlet L-functions modulo $$q$$, otherwise there is no Dirichlet series simplification.

For $$\Re(s) > 1, \theta\in (0,2\pi)$$ with $$C$$ a contour enclosing $$[0,\infty)$$ clockwise

$$(e^{2i \pi s}-1)\Gamma(s) \sum_{n=1}^{\infty} \frac{\exp(in \theta)}{n^{s}} = \int_C \frac{x^{s-1}}{e^{x-i\theta}-1}dx$$ The RHS is entire and for $$\Re(s) < 0$$ with the residue theorem, being careful with the branch of $$x^{s-1}$$ it is

$$= 2i\pi \sum Res(\frac{x^{s-1}}{e^{x-i\theta}-1})=2i\pi \sum_k (2i\pi k+i\theta)^{s-1}$$ $$= (2i\pi)^s (\zeta(1-s,\frac{\theta}{2\pi})+(-1)^{s-1}\zeta(1-s,1-\frac{\theta}{2\pi}))$$ (the Hurwitz zeta function)

• Thank you, @reuns. Does this mean the value of the expression is not defined at $s=1$ ? Because then you get $e^{2i\pi} -1 = 1-1=0$ on the LHS of the first equation, right? (And on the right-hand side at the bottom, one obtains an imaginary number, it seems) – Max Muller Oct 26 '19 at 17:00
• The contour integral on the RHS is zero at $s=1$, you should look there for the details of that contour integral representation – reuns Oct 26 '19 at 17:03
• Thank you for the link. Now I also see the requirement that Re(s) > 1 above and Re(s) < 0 below, sorry, should've looked better – Max Muller Oct 26 '19 at 17:07
• What do you think of this paper by the way (link: academia.edu/19563528/…) ? Do you think it formulas therein could be used to derive other identities for $g(s,\theta)$ ? – Max Muller Oct 26 '19 at 17:09

I am not sure that this could help.

Using $$\sum_{n=1}^\infty\frac {e^{i n \theta}} {n^s}=\text{Li}_s\left(e^{i \theta}\right)$$ we have $$g(s, \theta) = \sum_{n=1}^{\infty} \frac{\cos(n \theta)}{n^{s}}=\frac{1}{2} \left(\text{Li}_s\left(e^{-i \theta }\right)+\text{Li}_s\left(e^{i \theta }\right)\right)$$ where appears the polylogarithm function.

You can get the value for even $$s$$, we start with the infinite sum
$$\sum_{n=1}^{\infty}\cos(n\theta)$$ and then integrate s times to get the disired result.

• The infinite sum $\sum_{n=1}^{\infty}\cos(n\theta)$ ?? Integrating $\sum_{n=1}^{\infty}z^n (e^{in\theta}+e^{-in\theta})$ gives the value of $\sum_{n=1}^{\infty} \cos(n\theta)n^{-k}$ only for $k=1$, for $k=2$ and $\cos(\theta)\ne \pm 1$ there is no known closed-form – reuns Oct 26 '19 at 18:46


text:
# Working out the units?

I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√(R/TY)

That is A equals the square root of R divided by TY

(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?

Last edited:

Mark44
Mentor
I am having a problem finding the correct SI unitsfor the quantity A?

In the equation

A=√R/TY

That is A equals the square root of R divided by TY
Both your notation and explanation are ambiguous.

Is the expression on the right side this?
$$\sqrt{\frac{R}{TY}}$$
or this?
$$\frac{\sqrt{R}}{TY}$$
(not to good showing workings on the computer sorry)

, the SI units of the quantity R are kg m^3 s^–2, the SI units of the quantity T are kg and the SI units of the quantity Y are m s^–2. What are the correct SI units for the quantity A?

The first one R/ty all square root. Do apologise for the bad format

Then, the units of A are meters.

You have the expression
$$\sqrt{\frac{R}{TY}}$$

Simply, insert the units for each variable (in place of the variables):
$$\sqrt{\frac{\frac{kg\cdot m^3}{s^2}}{(kg)(\frac{m}{s^2})}}$$

and simplify...
$$\sqrt{\left(\frac{kg \cdot m^3}{s^2}\right) \left(\frac{s^2}{kg \cdot m}\right)}$$

kg and s2 cancel out, leaving
$$\sqrt{\frac{m^3}{m}}$$

which is
$$\sqrt{m^2}$$

or, more simply m


text:
mersenneforum.org (https://www.mersenneforum.org/index.php)
-   Number Theory Discussion Group (https://www.mersenneforum.org/forumdisplay.php?f=132)
-   -   Repeating residues in LL tests of composite Mersenne numbers (https://www.mersenneforum.org/showthread.php?t=25772)

 JeppeSN 2020-08-03 17:30

It is perhaps also interesting to note the lengths of the "pre-periods", or offsets. That is the number of terms in the LL sequence preceding the first occurrence of the period. With Batalov's data as input, that is easy (PARI/GP):
[CODE]findOffset(p,t)=s=Mod(4,2^p-1);S=s;for(i=1,t,S=S^2-2);i=0;while(s!=S,s=s^2-2;S=S^2-2;i++);i
[/CODE]
Pass the Mersenne exponent as p, and the period length (from Batalov's post; of course Batalov's technique for finding the period length is also easy to write in PARI/GP) as t. In the function, the S sequence is always t positions ahead of the s sequence, so we just check to see how long it takes until they agree.

Result:[CODE]
findOffset(11,60)
1

findOffset(23,32340)
1

findOffset(29,252)
1

findOffset(37,516924)
5

findOffset(41,822960)
1

findOffset(43,420)
3

findOffset(47,20338900)
4

findOffset(53,1309620)
2
[/CODE]
The last one, findOffset(59,345603421440), would take a long time with PARI/GP with this approach.

Not sure if the above values have any interesting interpretation. We see that for the cases I was able to resolve, the constant 120000 in Batalov's C program was on the safe side. Of course, Viliam Furik's approach (searching for "14") works exactly in the cases where findOffset gives 1.

/JeppeSN

 JeppeSN 2020-08-03 18:13

[QUOTE=Batalov;552444]Periods may (?) be different if we use the other popular seed values S0 = 10, or S0 = 2/3 (mod Mp).[/QUOTE]
That is true: This PARI/GP function imitates your program, with an optional seed argument:
[CODE]findPeriod(p,seed=4)=s=Mod(seed,2^p-1);for(i=1,120000,s=s^2-2);S=s;i=0;until(s==S,S=S^2-2;i++);i
[/CODE]
Results:
[CODE]findPeriod(11)
60

findPeriod(11,10)
10

findPeriod(11,2/3)
5
[/CODE]
So they do differ. The one for 2/3 is not even a multiple of 11 - 1 = 10. You may think PARI handles 2/3 wrong, but it does not: Evaluating findPeriod(11,683) also gives 5.

Explicitly, the LL sequence for 2^11 - 1, starting from seed 2/3 == 683, is:

683 -> 1818 -> 1264 -> 1034 -> 620 -> 1609 -> 1471 -> 160 -> 1034 -> etc.

Similarly, for 2^29 - 1, and seed 2/3, the period length is 154, which is 14 mod 28.

/JeppeSN

 Batalov 2020-08-03 18:15

You have a solid footing to retrace Tony's DiGraphs adventure, now! It is a bit of a déjà vu, but a good warm up to visit his old threads.

 jshort 2020-10-01 22:15

Let [TEX]S_{1} = 4 = (2 + \sqrt{3}) + (2 - \sqrt{3})[/TEX] and [TEX]S_{n+1} = S_{n}^{2} - 2[/TEX].

We'll first prove via induction that [TEX]S_{n} = (2 + \sqrt{3})^{2^{n-1}} + (2 - \sqrt{3})^{2^{n-1}}[/TEX].

Clearly [TEX]S_{1} = (2 + \sqrt{3})^{2^{1-1}} + (2 - \sqrt{3})^{2^{1-1} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4[/TEX]. Assume that it holds for all [TEX]k < m[/TEX].

Now [TEX]S_{m} = S_{m-1}^{2} - 2 = ((2 + \sqrt{3})^{2^{m-2}} + (2 - \sqrt{3})^{2^{m-2}})^{2} - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}} + 2(2 + \sqrt{3})^{2^{m-2}} \cdot (2 - \sqrt{3})^{2^{m-2}} - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}} + 2 - 2 = (2 + \sqrt{3})^{2^{m-1}} + (2 - \sqrt{3})^{2^{m-1}} [/TEX]

Note that in the above, I used the fact that [TEX](2 + \sqrt{3}) \cdot (2 - \sqrt{3}) = 1[/TEX]. Their inverses. Since their inverses of each other, they also have the same order over [TEX]M_{p}[/TEX].

Suppose the order of [TEX]2 + \sqrt{3}[/TEX] is [TEX]n = 2^{i} \cdot m[/TEX]. If [TEX]i=0[/TEX], the order is odd. In any case, I believe [TEX]n[/TEX] will have to be a factor of [TEX] \prod_{i} (p_{i}^{2} - 1)[/TEX] where [TEX]p_{i}[/TEX] are the prime factors of [TEX]M_{P}[/TEX].**

First assume that [TEX]n[/TEX] is odd.

Let [TEX]r[/TEX] be the smallest integer such that [TEX]2^{r} \equiv 1 mod(n)[/TEX] In this case, we have that [TEX](2 + \sqrt{3})^{2^{r} - 1} \equiv 1[/TEX] or [TEX](2 + \sqrt{3})^{2^{r}} \equiv 2 + \sqrt{3} [/TEX].

The same can be repeated with [TEX]2 - \sqrt{3}[/TEX]. The result is that [TEX]S_{r+1} = (2 + \sqrt{3})^{2^{r}} + (2 - \sqrt{3})^{2^{r}} \equiv (2 + \sqrt{3}) + (2 - \sqrt{3}) \equiv 4 \equiv S_{1}[/TEX].

However what if [TEX]n[/TEX] is even?

In this case, [TEX]i > 0[/TEX] where [TEX]n = 2^{i} \cdot m[/TEX]. We can repeat the procedure above, however we have to start at [TEX]S_{i+1}[/TEX] and instead of asking ourselves what the order of [TEX]2[/TEX] is, we're asking what the order of [TEX]2^{i}[/TEX] is over [TEX]m[/TEX].

.............................

**My guess that the order [TEX]n[/TEX] needs to be a factor of [TEX] \prod_{i} (p_{i}^{2} - 1)[/TEX] is based on the assumption that the number of invertible elements [TEX]a + b \sqrt{3} [/TEX] over [TEX]Z_{p}[/TEX] is always going to be [TEX]p^{2} - 1[/TEX], however my memory of group theory is fuzzy, so someone might want to check this!

 Viliam Furik 2020-10-20 17:23

[QUOTE=Batalov;552444]Periods may (?) be different if we use the other popular seed values S0 = 10, or S0 = 2/3 (mod Mp).[/QUOTE]

Period values for starting value 10:
M11 -> 10 (1 * 10)
M23 -> 32340 (1470 * 22)
M29 -> 252 (9 * 28)
M37 -> 516924 (14359 * 36)
M41 -> 822960 (20574 * 40)
M43 -> 420 (10 * 42)

They are all the same (at least for these exponents) as with S(0) = 4, except for the M11.

I have also looked at periods in PRP testing:
M11 -> 10 (1 * 10)
M23 -> 528 (24 * 22)
M29 -> 252 (9 * 28)
M37 -> 516924 (14359 * 36)
M41 -> x > 3077786
M43 -> 9492 (226 * 42)

Here they are more different. But still sometimes the same as for LL, and also preserve the k*(p-1) property.

 Viliam Furik 2020-10-20 23:22

I have done the PRP part because I realized that if we know the period of the PRP test of a composite exponent, we can run P-1 in a different way (which may or may not be faster, probably not), by simply computing the product of primes * 2p, finding its binary representation, and multiplying residues corresponding to powers of 1s in the representation, modulo the period, and doing the GCD.

Of course, that only pays out if the period is smaller than the log2 of the (2p * product of primes smaller than smallest B1 needed ). It depends on the smoothness of the factors of Mp.

EXAMPLE:
p = 11
B1=1
E(B1) --> product of all primes less than B1
x = E(1) * 2 * p = 22[SUB]10[/SUB] = 10110[SUB]2[/SUB]
3[SUP]x[/SUP] = 3[SUP]2^4[/SUP] * 3[SUP]2^2[/SUP] * 3[SUP]2^1[/SUP]
(I compute the

As all powers here are smaller than the period length, the method is of no use in this case.
If there was a power of 3 higher than the period length + the length of the aperiodic part, then we can find a corresponding power of 3 that is less by some multiple of the period length.

The obvious upper bound for the period is the number of quadratic residues, phi(Mp)/2[SUP]k[/SUP], where k is the number of factors of Mp, and phi(n) is the Euler totient function.

...................................
Because I realize that this method might never be possible to use, due to needed bound being much lower than necessary, I have been thinking about it for about 6 hours now. If you think it's unusable, you are most probably right.

 Viliam Furik 2020-12-04 19:47

[QUOTE=jshort;558547]...so someone might want to check this![/QUOTE]

Periods for M23 do not divide the said product.

 Dr Sardonicus 2020-12-06 15:36

[QUOTE=jshort;558547]<snip>
Suppose the order of [TEX]2 + \sqrt{3}[/TEX] is [TEX]n = 2^{i} \cdot m[/TEX]. If [TEX]i=0[/TEX], the order is odd. In any case, I believe [TEX]n[/TEX] will have to be a factor of [TEX] \prod_{i} (p_{i}^{2} - 1)[/TEX] where [TEX]p_{i}[/TEX] are the prime factors of [TEX]M_{P}[/TEX].**

First assume that [TEX]n[/TEX] is odd.
<snip>[/QUOTE]Let u = Mod(2+x,x^2-3). Then u^2 - 4*u + 1 = 0. Let R = Z[u]. If p > 3 is prime, the multiplicative order n = n(p) of u in R/pR indeed divides p^2 - 1. In fact, because norm(u) = +1 we can say more:

If p > 3, then n(p) divides p - (12/p) = p - kronecker(12,p) = p - kronecker(3,p)

which is p - 1 if p is congruent to 1 or 11 (mod 12) and p + 1 if p is congruent to 5 or 7 (mod 12).

The multiplicative order for powers of these primes, and for p = 2 and 3, are left as exercises for the reader.

For a composite modulus N, the multiplicative order n(N) of u in R/NR is the lcm of the multiplicative orders modulo the prime (power) factors of N.

[b]What you want is different, however[/b]. You want to find i and j such that

u^(2^(i+j)) == u^(2^i) (mod NR), which requires

2^(i+j) == 2^i (mod (n(N)), or 2^i*(2^j - 1) == 0 (mod n(N)).

Thus, the exponent i has to be at least as large as the exponent of the 2-power factor of n(N), and j has to be divisible by the multiplicative order of 2 modulo the odd part of n(N).

So if 2^i is the 2-power factor of n(N), m = n(N)/2^i, and j is the multiplicative order of Mod(2, m), then the repetition of u^(2^k) starts at u^(2^i) and has period j.

For N = 2^23 - 1 we find n(N) = 4105086 = 2*2052543, and znorder(Mod(2,2052543)) = 32340.

With u1 = Mod(1, 2^23 - 1)*u we find that u1^(2^32340) =/= u1, but u1^(2^(2+32340)) == u1^2.

 Viliam Furik 2020-12-06 16:05

I guess that similar thing is also possible for PRP tests, right? If so, could you write a method to work it out?

 Dr Sardonicus 2020-12-07 15:04

[QUOTE=Viliam Furik;565453]I guess that similar thing is also possible for PRP tests, right? If so, could you write a method to work it out?[/QUOTE]
Possible, sure. But it falls under "theoretically (perhaps) interesting, but computationally useless." A simple Fermat test is whether a[sup]n-1[/sup] == 1 (mod n). If instead you want to determine the smallest i and j such that

a^2[sup]i+j[/sup] == a^2[sup]i[/sup] (mod n),

[i]even assuming n is prime and a is a primitive root (mod n),[/i] 2^i is the 2-power factor of n-1, and j is the multiplicative order of 2 modulo the odd part of n-1. And to find [i]that[/i], you need to factor n-1 [i]completely[/i]. And there are [i]primality proofs[/i] that require only partial factorizations.

 LaurV 2020-12-08 06:56

[QUOTE=Dr Sardonicus;565557]the smallest i and j such that
[/QUOTE]
I had to read that few times and didn't fully got it yet. Why do you need to find both? Only j would be enough. Yet, you still may not factor n after that (you can run in a trivial factor 1 or n).

All times are UTC. The time now is 23:34.


text:
HomeEnglishClass 11MathsChapterLocus

A(2, 1) and B(1, 2) are two po...

# A(2, 1) and B(1, 2) are two points. If P is a point such that PA : PB = 2:1, then the locus of P is

Updated On: 17-04-2022

Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,

Text Solution

3x^(2)+3y^(2)+4x+14y-15=03x^(2)+3y^(2)-4x-14y+15=03x^(2)+3y^(2)+2x+7y+13=03x^(2)+3y^(2)-2x-7y+13=0


text:
One way to fix this is to replace abs(x+I*y) with sqrt(x^2+y^2).
Sage plotting routines use native Python type float under the hood for domain variables, while input functions are of the symbolic Sage type Expression. The two don't always mix well for plotting, especially when complex variables are involved.
Another way around the problem is to put lambda x,y: in front of the function containing the complex unit to force evaluation as a Python callable.