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Vladimir Bondarenko
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 Posted: Wed Nov 19, 2008 7:13 am Post subject: Re: Mathematica 7 is here! |
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On Nov 19, 8:26 am, "Nasser Abbasi" <n...@12000.org> wrote:
[quote]Am I the only one seeing Japanees looking fonts on this Mathematica web page
for documenation?
If you go this page below
http://reference.wolfram.com/mathematica/guide/NewIn70DynamicInteract...
and then click on the word "Manipulate" link (first function right near the
top of the page), you>ll go to page with Japanees text.
or may be it is just my brower playing tricks on me?
Nasser
[/quote]
Hello Nasser,
Yes, I confirm that I see something looking like a Japanese
text in Firefox 3.0.4 with "Manipulate" as you described.
Cheers,
Vladimir |
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Nasser Abbasi
Guest
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| Posted: Wed Nov 19, 2008 8:26 am Post subject: Re: Mathematica 7 is here! |
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<groups trimmed to sci.math.symbolic only>
"Vladimir Bondarenko" <vb@cybertester.com> wrote in message
news:845b59a5-17e9-45ec-af90-c5bb7114ceb5@a17g2000prm.googlegroups.com...
[quote]http://www.wolfram.com/products/mathematica/index.html
[/quote]
It looks like they added many new and interesting things to it. This is a
link which lists all the new functions in Mathematica 7
http://reference.wolfram.com/mathematica/guide/NewIn70AlphabeticalListing.html
I counted on that page 429 new functions (I might be off by 1 or 2). Since
the page mentions that over 500 functions are added or modified, then about
70-80 functions are modified in addition to the 430 new functions.
One thing I noticed is the addition of many more image processing functions.
Not sure if this means the functionality of the Mathematica Digital image
processing application shown here
http://www.wolfram.com/products/applications/digitalimage/ has been moved
to Mathematica 7?
I am going to try to get Mathematica 7 first thing tomorrow, can>t wait :)
Nasser |
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Nasser Abbasi
Guest
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| Posted: Wed Nov 19, 2008 8:26 am Post subject: Re: Mathematica 7 is here! |
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Am I the only one seeing Japanees looking fonts on this Mathematica web page
for documenation?
If you go this page below
http://reference.wolfram.com/mathematica/guide/NewIn70DynamicInteractivity.html
and then click on the word "Manipulate" link (first function right near the
top of the page), you>ll go to page with Japanees text.
or may be it is just my brower playing tricks on me?
Nasser |
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Vladimir Bondarenko
Guest
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| Posted: Wed Nov 19, 2008 11:08 am Post subject: Re: An exact simplification challenge - 76 (generalized Meij |
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On Nov 18, 9:40 pm, sashap <pav...@gmail.com> wrote:
[quote]The answer is 2*Pi^(3/2)*(1+I).
Oleksandr Pavlyk
In[154]:= NIntegrate[
MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x, {x, 0, Infinity}]
Out[154]= 11.1367 + 11.1367 I
In[155]:= N[MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I , 1/2]]
Out[155]= 11.1367 + 11.1367 I
In[156]:> MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x // FunctionExpand //
FullSimplify[#, x > 0] &
Out[156]= (\[Pi]^(3/2) (1 + I Sqrt[x]))/(Sqrt[x] (1 + x)^(3/2))
In[157]:= Integrate[%, {x, 0, Infinity}]
Out[157]= (2 + 2 I) \[Pi]^(3/2)
On Nov 18, 1:26 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello,
MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I, 1/2]
?
The generalized MeijerG definition is herehttp://documents.wolfram.com/mathematica/functions/MeijerG
Best wishes,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
-----------------------------------------------------------
"We must understand that technologies
like these are the way of the future."
-----------------------------------------------------------
[/quote]
Gee!
I wonder if this MeijerG product integration
formula comes from functions.wolfram.com ?
Cheers,
Vladimir |
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sashap
Guest
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| Posted: Wed Nov 19, 2008 8:40 pm Post subject: Re: An exact simplification challenge - 76 (generalized Meij |
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On Nov 19, 5:08 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
[quote]On Nov 18, 9:40 pm, sashap <pav...@gmail.com> wrote:
The answer is 2*Pi^(3/2)*(1+I).
Oleksandr Pavlyk
In[154]:= NIntegrate[
MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x, {x, 0, Infinity}]
Out[154]= 11.1367 + 11.1367 I
In[155]:= N[MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I , 1/2]]
Out[155]= 11.1367 + 11.1367 I
In[156]:> > MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x // FunctionExpand //
FullSimplify[#, x > 0] &
Out[156]= (\[Pi]^(3/2) (1 + I Sqrt[x]))/(Sqrt[x] (1 + x)^(3/2))
In[157]:= Integrate[%, {x, 0, Infinity}]
Out[157]= (2 + 2 I) \[Pi]^(3/2)
On Nov 18, 1:26 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello,
MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I, 1/2]
?
The generalized MeijerG definition is herehttp://documents.wolfram.com/mathematica/functions/MeijerG
Best wishes,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
-----------------------------------------------------------
"We must understand that technologies
like these are the way of the future."
-----------------------------------------------------------
Gee!
I wonder if this MeijerG product integration
formula comes from functions.wolfram.com ?
[/quote]
The classic convolution theorem for MeijerG was used.
This theorem is stated at functions.wolfram.com:
http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/02/02/
Oleksandr
[quote]
Cheers,
Vladimir[/quote] |
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Bhuvanesh
Guest
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| Posted: Wed Nov 19, 2008 9:32 pm Post subject: Re: Mathematica 7 is here! |
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On Nov 19, 1:13 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
[quote]On Nov 19, 8:26 am, "Nasser Abbasi" <n...@12000.org> wrote:
Am I the only one seeing Japanees looking fonts on this Mathematica web page
for documenation?
If you go this page below
http://reference.wolfram.com/mathematica/guide/NewIn70DynamicInteract...
and then click on the word "Manipulate" link (first function right near the
top of the page), you>ll go to page with Japanees text.
or may be it is just my brower playing tricks on me?
Nasser
Hello Nasser,
Yes, I confirm that I see something looking like a Japanese
text in Firefox 3.0.4 with "Manipulate" as you described.
Cheers,
Vladimir
[/quote]
I don>t see this... perhaps it>s already fixed?
Bhuvanesh |
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Vladimir Bondarenko
Guest
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| Posted: Thu Nov 20, 2008 3:44 am Post subject: Re: Mathematica 7 is here! |
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On Nov 19, 11:52 pm, "Nasser Abbasi" <n...@12000.org> wrote:
[quote]"Bhuvanesh" <BhuvaneshBh...@gmail.com> wrote in message
news:89e31ce4-0016-4d81-92ae-
"I don>t see this... perhaps it>s already fixed?"
Yes, some one at WRI must have fixed it. The page is ok now.
Bhuvanesh
[/quote]
Yes, all is OK now.
It>s great to enjoy such a speed in improvement.
Keep her steady! :) |
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Nasser Abbasi
Guest
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| Posted: Thu Nov 20, 2008 3:52 am Post subject: Re: Mathematica 7 is here! |
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"Bhuvanesh" <BhuvaneshBhatt@gmail.com> wrote in message
news:89e31ce4-0016-4d81-92ae-
"I don>t see this... perhaps it>s already fixed?"
Yes, some one at WRI must have fixed it. The page is ok now.
Bhuvanesh |
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Vladimir Bondarenko
Guest
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| Posted: Thu Nov 20, 2008 3:59 am Post subject: Re: An exact simplification challenge - 76 (generalized Meij |
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On Nov 19, 10:40 pm, sashap <pav...@gmail.com> wrote:
[quote]On Nov 19, 5:08 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
On Nov 18, 9:40 pm, sashap <pav...@gmail.com> wrote:
The answer is 2*Pi^(3/2)*(1+I).
Oleksandr Pavlyk
In[154]:= NIntegrate[
MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x, {x, 0, Infinity}]
Out[154]= 11.1367 + 11.1367 I
In[155]:= N[MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I , 1/2]]
Out[155]= 11.1367 + 11.1367 I
In[156]:> > > MeijerG[{{1/2, 1}, {}}, {{1/2, 1}, {}}, -1/
x] MeijerG[{{1}, {}}, {{1/2}, {}}, x]/x // FunctionExpand //
FullSimplify[#, x > 0] &
Out[156]= (\[Pi]^(3/2) (1 + I Sqrt[x]))/(Sqrt[x] (1 + x)^(3/2))
In[157]:= Integrate[%, {x, 0, Infinity}]
Out[157]= (2 + 2 I) \[Pi]^(3/2)
On Nov 18, 1:26 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello,
MeijerG[{{1/2, 1, 1}, {}}, {{1/2, 1/2, 1}, {}}, I, 1/2]
?
The generalized MeijerG definition is herehttp://documents.wolfram.com/mathematica/functions/MeijerG
Best wishes,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
-----------------------------------------------------------
"We must understand that technologies
like these are the way of the future."
-----------------------------------------------------------
Gee!
I wonder if this MeijerG product integration
formula comes from functions.wolfram.com ?
The classic convolution theorem for MeijerG was used.
This theorem is stated at functions.wolfram.com:
http://functions.wolfram.com/HypergeometricFunctions/MeijerG/21/02/02/
Oleksandr
Cheers,
Vladimir
[/quote]
Thank you cordially for being so much careful and insistent
in this math game.
By asking some provocative questions and sending these CAS
challenges and solving them we show to the community how to
deal with some problems which, temporarily, are not fully
automated yet in computer calculus. |
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Veikko Ekholm
Guest
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| Posted: Mon Nov 24, 2008 12:15 am Post subject: Re: Vista + keyboard layout manager? |
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Siis .. VAUDE!
_____________________
08.05.2008 jää historiaan hetkenä joka kiistatta osoitti miten
maailmanennätysvauhtia suorastaan megaromahtanut ydinvoiman kannatus
konkretisoi laajassa rintamassa myös Suomessa. Miten tällainen saavutus on
edes teoriassa mahdolista maassa, jossa ydinvastustaminen rinnastetaan
sumeilematta maanpetokselliseksi toimeksi?
No ensinnä tieto ydinvoiman petollisista vaaroista on kiistatta alkanut
saavuttaa maamme älymystöjä ja asioitten oikeista tiloista kiinostuneita.
Kun yritin lähettää tätä tietoa nettiin huomasin, miten RAIVOSSAAN
ydinhallintomme tapahtumamuutoksista ydinasioissa oli. Nettini kaadettiin
systemaattisesti sen kymmenet kerrat. Kas vaan riehuva SUPO/TVO/IAEA ei
selvästi ollut tyytyväisiä aamun uutisanteihin!))
No optimistisesti toki oletan", että vielä tulee aika jolloin TVO päästää
taas näitä materiaalejani eetteriin. .. Kenties olen liian optimisti,
kenties en, aika näyttää. Mutta kirjoitetaan nyt silti valmiiksi
off-puskuriin. Eli jysäripäänsäryn takana on YLE:n yllättäen julkaisema
ydinkannatustutkimus. Kuten hyvin muistamme vain vuosi sitten ydinvoimalla
oli maassamme kiinteä huikenteleva +80% kannatus. Ja tosiaan nyt julkaistiin
raportti siitä miten maastamme enää 20% haluaa edes yhden lisäydinvoimalan,
jota TVO/Fortum/Fennovoima himoitsee!!!! .. !!!! ..
Kerrottiin miten n.2/3 osa on ydinvastaisiksi tulleet maasamme. Ydinvoimaa
vastustaa naisista 68% joka taustoittuu hienosti taannoisen
nuorisoparlamentin Niinistön mittaamaan 80% ydinvastaisuuden kanssa
kauniisti. Ehkä kaikkein huvittavinta on se, että ydinyhtiöiden
triplarakentelu sai maanjäristystuomion, eli VAIN 4% on samaa mieltä!
Ylisanani kehumisiin loppuivat liiasta käytöstä asiassa, joten se myös
numeroista. Mutta sitten mielenkiintoisimpaan. Mistä sitten miljoonat
äänensä ydinvastaisuudelle antaneet saivat tietonsa? TV:stäkö, no hei ei
tasan! No lehdistä? Onko kukaan koskaan nähnyt yhtään ydinkriittistä lehden
omaa artikkelia ani harvojen sensuurin läpi päässeitten
yleisökirjoitususkalikkojen lisäksi, no hei EI. Radiosta ei yhtään enempää!
MISTÄ tämä massojen tieto siten on peräisin oikeesti! Naiset ja nuoret
ovat tunnetusti alkaneet löytää.. .aivan oikein nettisivuilta maailman
ainoaa vielä vaivoin SUPO:n sensorien läpi livahtelevaa faktaa siitä miten
ydinvoima on kylmä kuolema ihmiskunnalle!.. .Toki paljon olisi jatkettavaa
tästä, että ketkä ja koska, miten, mutta tosiaan miettikää miksi vain miehet
ydintöissään jossa netin käyttö on tiukasti kontrollein estetty ovat enää
merkittävämmin ydinharhapoluillaan vielä hetken?.. .. |
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Daniel Lichtblau
Guest
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| Posted: Mon Nov 24, 2008 2:52 pm Post subject: Re: An exact simplification challenge - 77 (MeijerG) |
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On Nov 24, 2:19 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
[quote]Hello,
MeijerG[{{0, 1/2, 1/2}, {}}, {{0, 1/2, 3/2}, {}}, 1]
?
Best wishes,
Vladimir Bondarenko
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester Ltd.
-----------------------------------------------------
"We must understand that technologies
like these are the way of the future."
-----------------------------------------------------
[/quote]
This is all magic to me, but...
In[5]:= InputForm[FunctionExpand[ MeijerG[{{0, 1/2, 1/2}, {}}, {{0,
1/2, 3/2}, {}}, 1]]]
Out[5]//InputForm= ((-1 + Pi)*Pi)/2
Numerical check:
In[6]:= N[MeijerG[{{0, 1/2, 1/2}, {}}, {{0, 1/2, 3/2}, {}}, 1] - ((-1
+ Pi)*Pi)/2] // InputForm
Out[6]//InputForm= 4.440892098500626*^-16 + 1.2649650455356054*^-61*I
Daniel Lichtblau
Wolfram Research |
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Daniel Lichtblau
Guest
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| Posted: Mon Nov 24, 2008 6:43 pm Post subject: Re: On recent actions of Dr. Scott Seidman of the University |
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On Nov 24, 12:00 pm, ru...@vesuvius.math.niu.edu (Dave Rusin) wrote:
[quote]In article <6c7e97d5-df18-4bf7-9880-ccb7628c3...@a12g2000yqm.googlegroups..com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
I simply want to focus on the QA challenges, and run my company rather
than to be involved into a distracting activity wasting my time.
So why are you reading and posting to USENET, the world>s biggest time-waster?
[/quote]
He apparently felt that his integrity was impugned. I mean, somebody
referred to him as a "wannabe net loon". A "wannabe"?! The fella has
his noive.
Daniel Lichtblau
(Not speaking for) Wolfram Research (or cyber-species of waterfowl,
for that matter) |
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Axel Vogt
Guest
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| Posted: Mon Nov 24, 2008 9:42 pm Post subject: Re: An exact simplification challenge - 77 (MeijerG) |
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Daniel Lichtblau wrote:
[quote]On Nov 24, 2:19 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello,
MeijerG[{{0, 1/2, 1/2}, {}}, {{0, 1/2, 3/2}, {}}, 1]
?
....
This is all magic to me, but...
In[5]:= InputForm[FunctionExpand[ MeijerG[{{0, 1/2, 1/2}, {}}, {{0,
1/2, 3/2}, {}}, 1]]]
Out[5]//InputForm= ((-1 + Pi)*Pi)/2
Numerical check:
In[6]:= N[MeijerG[{{0, 1/2, 1/2}, {}}, {{0, 1/2, 3/2}, {}}, 1] - ((-1
+ Pi)*Pi)/2] // InputForm
Out[6]//InputForm= 4.440892098500626*^-16 + 1.2649650455356054*^-61*I
Daniel Lichtblau
Wolfram Research
[/quote]
for me as well ...
Maple does not get it. However some 'experimental Math' with
Maple gives the generalization and x=1 is the given task:
1/2*Pi*(Pi*(1+3*x)+(-2+(-2+2*ln(x))*x)*x^(1/2))/(1+x)^2 |
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sashap
Guest
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| Posted: Wed Nov 26, 2008 8:15 pm Post subject: Re: An exact simplification challenge - 78 (elliptic nightma |
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In Mathematica>s conventions your input reads:
In[154]:= (EllipticK[b^2] - EllipticF[ArcSin[a], b^2] +
EllipticPi[2 b^2, b^2] - EllipticPi[2 b^2, ArcSin[a], b^2])/6
Out[154]= 1/6 (-EllipticF[ArcSin[Sqrt[5/6]], 2/5] + EllipticK[2/5] +
EllipticPi[4/5, 2/5] - EllipticPi[4/5, ArcSin[Sqrt[5/6]], 2/5])
In[155]:= N[%]
Out[155]= 0.460791
Using integral representations, and noticing that you subtract
incomplete elliptic integral from complete one the expression becomes:
In[156]:= NIntegrate[(1/6)/Sqrt[
1 - b^2 Sin[f]^2] (1 + 1/(1 - 2 b^2 Sin[f]^2)), {f, ArcSin[a], Pi/
2}]
Out[156]= 0.460791
Then we perform the change of variables mapping integration region
into
unit interval:
In[157]:= FullSimplify[
1/6 1/Sqrt[1 - b^2 Sin[f]^2] (1 + 1/(1 - 2 b^2 Sin[f]^2)) Dt[f]/
Dt[t] /. f -> ArcCos[t/Sqrt[6]]]
Out[157]= -((Sqrt[5/3] Sqrt[9 + t^2])/(Sqrt[6 - t^2] (3 + 2 t^2)))
Then perform the integration and get the answer from G.A. Edgar>s
post:
In[158]:= Integrate[%, {t, 1, 0}]
Out[158]= 1/6 Sqrt[5/3] (EllipticF[ArcCsc[Sqrt[6]], -(2/3)] -
5 EllipticPi[-4, -ArcCsc[Sqrt[6]], -(2/3)])
Check:
In[159]:= N[%, 20]
Out[159]= 0.46079057162376452952
It might be possible to reduce it further, but I have not found
that way.
Oleksandr Pavlyk
On Nov 26, 10:47 am, Vladimir Bondarenko <v...@cybertester.com> wrote:
[quote]On Nov 26, 3:57 pm, "G. A. Edgar" <ed...@math.ohio-state.edu.invalid
wrote:
In article
5ace890b-9b35-42dd-908e-297794dcf...@e12g2000yqm.googlegroups.com>,
Vladimir Bondarenko <v...@cybertester.com> wrote:
Hello,
a := sqrt(30)/6: b := sqrt(10)/5:
( + EllipticK(b)
- EllipticF(a, b)
+ EllipticPi(2*b^2, b)
- EllipticPi(a, 2*b^2, b))/6
?
Perhaps not considered simpler...
sqrt(5/108)*(EllipticF(sqrt(1/6),sqrt(-2/3))
+5*EllipticPi(sqrt(1/6),-4,sqrt(-2/3)));
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Yes, you have an amazing intuition. You just
feel the Beauty.
You are right, these 2 elliptics can be squeezed
further into.... into what? ;)
And it would be also instructive for the CAS
users if you please could give the exact steps
to get these 2 elliptics from the original 4.[/quote] |
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Axel Vogt
Guest
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| Posted: Thu Nov 27, 2008 3:48 am Post subject: Re: An exact simplification challenge - 78 (elliptic nightma |
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Vladimir Bondarenko wrote:
[quote]Hello,
a := sqrt(30)/6: b := sqrt(10)/5:
( + EllipticK(b)
- EllipticF(a, b)
+ EllipticPi(2*b^2, b)
- EllipticPi(a, 2*b^2, b))/6
?
[/quote]
My feeling is to use partial fractions, writing that as Int(A*B, z=1
... 1/6*30^(1/2))
A=-5^(1/2)*((-1+z)*(z+1))^(1/2)*(-5+2*z^2)^(1/2),
B=1/(6*(z+1))-1/(6*(-1+z))-
2/15*5^(1/2)/(-2*z+5^(1/2))-2/15*5^(1/2)/(2*z+5^(1/2)) and then caring
for z=1. |
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