A44 === I think I'd prefer that Mathematica gave me a clue -- without beingasked explicitly in JUST the right way that only you know and hadforgotten -- that there's a precision problem.Oddly enough, when you DO ask it nicely, you get error messages, but ifyou're not aware there's a problem, it lets you go on your merry way,working with noise.Bobby-----Original Message----->> Clear[f, a, b, k]> k = $MachinePrecision;> f = SetAccuracy[333.75*b^6 +> a^2*(11*a^2*b^2 - b^6 -> 121*b^4 - 2) +> 5.5*b^8 + a/(2*b), k];> a = 77617.; b = 33096.;> a = SetAccuracy[a, k];> b = SetAccuracy[b, k];> f> Accuracy[f]> Precision[f]>> -5.517164009`0*^19>> Accuracy::mnprec:Value -23 would be inconsistent with $MinPrecision;> bounding by $MinPrecision instead.>> -20>> 0>> Accuracy::mnprec:Value !(-23) would be inconsistent with> $MinPrecision; > bounding by $MinPrecision instead.>> See that? NO precision.>> Bobby>> -----Original Message-----> Sent: Tuesday, October 01, 2002 5:53 PM> Cc: drbob@bigfoot.com>> You are of course right, I forgot that Mathematica does not try tokeep>> precision or accuracy of machine arithmetic computations. But I think> the actual precision you set need not be higher than machineprecision,>> it just has to be set explicitely (is that right?). For example:>> In[1]:=> Clear[f,a,b,k]>> In[2]:=> k = $MachinePrecision;>> In[3]:=> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/> (2*b),k];>> In[4]:=> a=77617.;b=33096.;>> In[5]:=> a=SetAccuracy[a,k];>> In[6]:=> b=SetAccuracy[b,k];> In[7]:=> f>> Out[7]=> !((-5.51716400890319`-2.8311*^19))>> In[8]:=> Accuracy[f]>> Accuracy::mnprec: Value -23 would be inconsistent with $MinPrecision;> bounding by $MinPrecision instead.>> Out[8]=> -20> Andrzej> Andrzej>> Yes, like you I took the original question to be about how to get the> result>> of using the naive rational versions in place of the inexact numbers.>> Bobby raises the question of how we might know accuracy of theresult.>> You answer this with> a = 77617.; b = 33096.;>> In[2]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +> 5.5*b^8 + a/(2*b)>> In[3]:=> f>> Out[3]=> -1.1805916207174113*^21>> In[4]:=> Accuracy[%]>> Out[4]=> -5>> However this computation is done in machine arithmetic, which means>> that>> Mathematica keeps no check on the accuracy and precision of the>> computation,>> and Mathematica gives the default accuracy value without any real>> signifcance:>> $MachinePrecision - Log[10,Abs[f]]//Round>> -5>> To get meaningful accuracy and precision values we need to force the>> computation to be in bignums (bigßoat, arbitrary precision)>> arithmetic.>> Mathematica then keeps track of the accuracy and precision that itcan>> guarantee.>> Clear[f,a,b,k]>> k=50;>> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/>> (2*b),k];>> a=77617.;b=33096.;>> a=SetAccuracy[a,k];>> b=SetAccuracy[b,k];>> f>> -0.82739605995>> Accuracy[f]>> 11>> Precision[f]>> 11>> Which tells us that the error in the computed value of f is not more> than>> 10^-11>> The above results are rounded. More accurately we get>> Accuracy[f, Round->False]>> 11.4956>> Precision[f, Round->False]>> 11.4133>> There are several related issues here:>> - is the precision (accuracy) of rhe input known?>> - how does Mathematica interpret what one gives it?>> - how does Mathematica go about the calculation?>> -->> Allan>> --------------------->> Allan Hayes>> Mathematica Training and Consulting>> Leicester UK>> www.haystack.demon.co.uk>> hay@haystack.demon.co.uk>> Voice: +44 (0)116 271 4198> It seems clear to me that what Allan and what you mean by succeeds> here refer to quite different things and your objection is therefore> beside the point. There are obviously two ways in which one can> interpret the original posting. The first interpretation, whichAllan> and myself adopted, was that the question concerned purely the> computational mechanism of Mathematica. Or, to put it in otherwords,> it was why are the results of these two computations not thesame?.> In this sense success means no more than making Mathematica return> the same answer using the two different routes the original poster> used.> You on the other hand choose to assume that the posting shows that> its> author does not understand not just the mechanism of significance> arithmetic used by Mathematica but also computations with inexact> numbers in general. I do not think this is necessarily the correct> assumption. I also don't see that Mathematica is doing anything> wrong.> After all, one can always check the accuracy of your answer:>> In[1]:=> a = 77617.; b = 33096.;>> In[2]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +> 5.5*b^8 + a/(2*b)>> In[3]:=> f>> Out[3]=> -1.1805916207174113*^21>> In[4]:=> Accuracy[%]>> Out[4]=> -5>> which tells you that it can't be very reliable. What more do you> demand?>> Andrzej> Andrzej Kozlowski> Yokohama, Japan> http://www.mimuw.edu.pl/~akoz/> http://platon.c.u-tokyo.ac.jp/andrzej/>> Actually, we don't know whether SetAccuracy succeeds, because we>> don't>> know how inexact those numbers really are. If they are known to> more>> digits than shown in the original post, they should be entered with> as>> much precision as they deserve. If not, there's no trick or>> algorithm>> that will give the result precision, because the value of f really> is>> in the noise. That is, we have no idea what the value of fshould>> be.>> Mathematica's failing is in returning a value without pointing out>> that>> it has no precision.>> Bobby>> -----Original Message----->> Sent: Monday, September 30, 2002 11:59 AM>> Andrzej, Bobby, Peter>> It looks as if using SetAccuracy succeeds here because the inexact>> numbers>> that occur have finite binary representations. If we change them>> slightly to>> avoid this then we have to use Rationalize:>> 1) Using SetAccuracy>> Clear[a,b,f]>> f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/>> (2*b),>> Infinity];>> a=77617.1;>> b=33096.1;>> a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity ];>> f>>- 15640321149084868351974949239896188679725401538739519428131155 14949>> 3891236234> 52500771916869370459119776018798804630436149786919912931962574 3010292>> 36>> 3>> 1246>> 75>> /> 10867106143970760551000357827554793888198143135975649579607989 8677435>> 72>> 8240>> 16>> 0653953612982932181371232436367739737604096>> 2) Rewriting as fractions>> a=776171/10;>> b=330961/10;>>f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121* b^4-2)+54/10*b^8+a/(2*b)>> -(5954133808997234115690303589909929091649391296257/>> 41370125000000)>> 3) Using Rationalize>> Clear[a,b,f]>> f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/>> (2*b),>> 0];>> a=77617.1;>> b=33096.1;>> a=Rationalize[a,0];b=Rationalize[b,0];>> f>> -(5954133808997234115690303589909929091649391296257/>> 41370125000000)>> I use Rationalize[. , 0] besause of results like>> Rationalize[3.1415959]>> 3.1416>> Rationalize[3.1415959,0]>> 31415959/10000000>> -->> Allan>> --------------------->> Allan Hayes>> Mathematica Training and Consulting>> Leicester UK>> www.haystack.demon.co.uk>> hay@haystack.demon.co.uk>> Voice: +44 (0)116 271 4198> Well, first of of all, your using SetAccuracy and SetPrecisiondoes> nothing at all here, since they do not change the value of a or b.>> You> should use a = SetAccuracy[a, Infinity] etc. But even then you> won't> get the same answer as when you use exact numbers because of the> way> you evaluate f. Here is the order of evaluation that will give you>> the> same answer, and should explain what is going on:>> f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*> b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity];>> a = 77617.;> b = 33096.;>> a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity];>> f>> 54767> -(-----)> 66192>> Andrzej Kozlowski> Toyama International University> JAPAN>> Could someone explain what is going on here, please?>> In[1]:=>> a = 77617.; b = 33096.;>> In[2]:=>> SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity];>> SetPrecision[a, Infinity]; SetPrecision[b, Infinity];>> In[4]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8+>> a/(2*b)>> In[5]:=>> SetAccuracy[f, Infinity]; SetPrecision[f, Infinity];>> In[6]:=>> f>> Out[6]=>> -1.1805916207174113*^21>> In[7]:=>> a = 77617; b = 33096;>> In[8]:=>> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> (55/10)*b^8 + a/(2*b)>> In[9]:=>> g>> Out[9]=>> -(54767/66192)>> In[10]:=>> N[%]>> Out[10]=>> -0.8273960599468214>> PK>> >> Andrzej Kozlowski> Yokohama, Japan> http://www.mimuw.edu.pl/~akoz/> http://platon.c.u-tokyo.ac.jp/andrzej/>> ==== I've been looking over the file and directory manipulation functions in Mathematica 4.1, and I'm not finding good examples of how to test for the existence of a file or directory before I attempt to create, access, or delete it. Are there any good examples of this somewhere?STH ==== FileNames[] does what you want.Steve Luttrell> I've been looking over the file and directory manipulation functions inMathematica> 4.1, and I'm not finding good examples of how to test for the existence of> a file or directory before I attempt to create, access, or delete it. Are> there any good examples of this somewhere?>> STH>Reply-To: kuska@informatik.uni-leipzig.de ==== FileNames[] returns an empty list if there are nofiles. SoFileNames[blub.txt]will return {} if the file does not exist. Jens> I've been looking over the file and directory manipulation functions in Mathematica> 4.1, and I'm not finding good examples of how to test for the existence of> a file or directory before I attempt to create, access, or delete it. Are> there any good examples of this somewhere?> STH ==== If you don't wont error messages all you need to do is to set MinPrecision low enough:In[1]:=$MinPrecision = -3;In[2]:=k = $MachinePrecision;In[3]:=f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), k];In[4]:=a = 77617.; b = 33096.;In[5]:=a = SetAccuracy[a, k];In[6]:=b = SetAccuracy[b, k];In[7]:=fOut[7]=-5.517164009`-2.8311*^19In[8]:= Accuracy[f]Out[8]=-23In[9]:=Precision[f]Out[9]=-3In any case the answer you get is meaningless.> I think I'd prefer that Mathematica gave me a clue -- without being> asked explicitly in JUST the right way that only you know and had> forgotten -- that there's a precision problem.>> Oddly enough, when you DO ask it nicely, you get error messages, but if> you're not aware there's a problem, it lets you go on your merry way,> working with noise.>> Bobby>> -----Original Message-----> Sent: Tuesday, October 01, 2002 8:05 PM> Cc: ÔAllan Hayes'; mathgroup@smc.vnet.net>> So? It's just as it should be, isn't it?>> Andrzej>> Andrzej Kozlowski> Toyama International University> JAPAN> http://sigma.tuins.ac.jp/~andrzej/>> Go one step further:>> Clear[f, a, b, k]>> k = $MachinePrecision;>> f = SetAccuracy[333.75*b^6 +>> a^2*(11*a^2*b^2 - b^6 ->> 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b), k];>> a = 77617.; b = 33096.;>> a = SetAccuracy[a, k];>> b = SetAccuracy[b, k];>> f>> Accuracy[f]>> Precision[f]>> -5.517164009`0*^19>> Accuracy::mnprec:Value -23 would be inconsistent with $MinPrecision;>> bounding by $MinPrecision instead.>> -20>> 0>> Accuracy::mnprec:Value !(-23) would be inconsistent with>> $MinPrecision; >> bounding by $MinPrecision instead.>> See that? NO precision.>> Bobby>> -----Original Message----->> Sent: Tuesday, October 01, 2002 5:53 PM>> Cc: drbob@bigfoot.com>> You are of course right, I forgot that Mathematica does not try to> keep>> precision or accuracy of machine arithmetic computations. But I think>> the actual precision you set need not be higher than machine> precision,>> it just has to be set explicitely (is that right?). For example:>> In[1]:=>> Clear[f,a,b,k]>> In[2]:=>> k = $MachinePrecision;>> In[3]:=>> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/>> (2*b),k];>> In[4]:=>> a=77617.;b=33096.;>> In[5]:=>> a=SetAccuracy[a,k];>> In[6]:=>> b=SetAccuracy[b,k];>> In[7]:=>> f>> Out[7]=>> !((-5.51716400890319`-2.8311*^19))>> In[8]:=>> Accuracy[f]>> Accuracy::mnprec: Value -23 would be inconsistent with $MinPrecision;>> bounding by $MinPrecision instead.>> Out[8]=>> -20>> Andrzej> Andrzej>> Yes, like you I took the original question to be about how to get the> result> of using the naive rational versions in place of the inexact numbers.> Bobby raises the question of how we might know accuracy of the> result.>> You answer this with>> a = 77617.; b = 33096.;>> In[2]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b)>> In[3]:=>> f>> Out[3]=>> -1.1805916207174113*^21>> In[4]:=>> Accuracy[%]>> Out[4]=>> -5>> However this computation is done in machine arithmetic, which means> that> Mathematica keeps no check on the accuracy and precision of the> computation,> and Mathematica gives the default accuracy value without any real> signifcance:>> $MachinePrecision - Log[10,Abs[f]]//Round>> -5>> To get meaningful accuracy and precision values we need to force the> computation to be in bignums (bigßoat, arbitrary precision)> arithmetic.> Mathematica then keeps track of the accuracy and precision that it> can> guarantee.>> Clear[f,a,b,k]> k=50;>> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/> (2*b),k];> a=77617.;b=33096.;> a=SetAccuracy[a,k];> b=SetAccuracy[b,k];> f>> -0.82739605995>> Accuracy[f]>> 11>> Precision[f]>> 11>> Which tells us that the error in the computed value of f is not more> than>> 10^-11> The above results are rounded. More accurately we get>> Accuracy[f, Round->False]>> 11.4956>> Precision[f, Round->False]>> 11.4133>> There are several related issues here:> - is the precision (accuracy) of rhe input known?> - how does Mathematica interpret what one gives it?> - how does Mathematica go about the calculation?>> --> Allan>> ---------------------> Allan Hayes> Mathematica Training and Consulting> Leicester UK> www.haystack.demon.co.uk> hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198>> It seems clear to me that what Allan and what you mean by succeeds>> here refer to quite different things and your objection is therefore>> beside the point. There are obviously two ways in which one can>> interpret the original posting. The first interpretation, which> Allan>> and myself adopted, was that the question concerned purely the>> computational mechanism of Mathematica. Or, to put it in other> words,>> it was why are the results of these two computations not the> same?.>> In this sense success means no more than making Mathematica return>> the same answer using the two different routes the original poster>> used.>> You on the other hand choose to assume that the posting shows that>> its>> author does not understand not just the mechanism of significance>> arithmetic used by Mathematica but also computations with inexact>> numbers in general. I do not think this is necessarily the correct>> assumption. I also don't see that Mathematica is doing anything>> wrong.>> After all, one can always check the accuracy of your answer:>> In[1]:=>> a = 77617.; b = 33096.;>> In[2]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b)>> In[3]:=>> f>> Out[3]=>> -1.1805916207174113*^21>> In[4]:=>> Accuracy[%]>> Out[4]=>> -5>> which tells you that it can't be very reliable. What more do you>> demand?>> Andrzej>> Andrzej Kozlowski>> Yokohama, Japan>> http://www.mimuw.edu.pl/~akoz/>> http://platon.c.u-tokyo.ac.jp/andrzej/> Actually, we don't know whether SetAccuracy succeeds, because we> don't> know how inexact those numbers really are. If they are known to>> more> digits than shown in the original post, they should be entered with> as> much precision as they deserve. If not, there's no trick or> algorithm> that will give the result precision, because the value of f really>> is> in the noise. That is, we have no idea what the value of f> should> be.>> Mathematica's failing is in returning a value without pointing out> that> it has no precision.>> Bobby>> -----Original Message-----> Sent: Monday, September 30, 2002 11:59 AM>> Andrzej, Bobby, Peter>> It looks as if using SetAccuracy succeeds here because the inexact> numbers> that occur have finite binary representations. If we change them> slightly to> avoid this then we have to use Rationalize:>> 1) Using SetAccuracy>> Clear[a,b,f]> f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/> (2*b),> Infinity];>> a=77617.1;> b=33096.1;>> a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity ];>> f>> - 15640321149084868351974949239896188679725401538739519428131155 14949> 3891236234>> 52500771916869370459119776018798804630436149786919912931962574 3010292> 36> 3> 1246> 75>> /> 108671061439707605510003578275547938881981431359756495796079898 677435> 72> 8240> 16> 0653953612982932181371232436367739737604096>> 2) Rewriting as fractions>> a=776171/10;> b=330961/10;> f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+54/10*b^8+a/(2 *b)>> -(5954133808997234115690303589909929091649391296257/> 41370125000000)>> 3) Using Rationalize>> Clear[a,b,f]> f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/> (2*b),> 0];>> a=77617.1;> b=33096.1;>> a=Rationalize[a,0];b=Rationalize[b,0];>> f>> -(5954133808997234115690303589909929091649391296257/> 41370125000000)> I use Rationalize[. , 0] besause of results like>> Rationalize[3.1415959]>> 3.1416>> Rationalize[3.1415959,0]>> 31415959/10000000> --> Allan>> ---------------------> Allan Hayes> Mathematica Training and Consulting> Leicester UK> www.haystack.demon.co.uk> hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198>> Well, first of of all, your using SetAccuracy and SetPrecision> does>> nothing at all here, since they do not change the value of a or b.>> You>> should use a = SetAccuracy[a, Infinity] etc. But even then you>> won't>> get the same answer as when you use exact numbers because of the>> way>> you evaluate f. Here is the order of evaluation that will give you>> the>> same answer, and should explain what is going on:>> f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*>> b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity];>> a = 77617.;>> b = 33096.;>> a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity];>> f>> 54767>> -(-----)>> 66192>> Andrzej Kozlowski>> Toyama International University>> JAPAN> Could someone explain what is going on here, please?>> In[1]:=> a = 77617.; b = 33096.;>> In[2]:=> SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity];> SetPrecision[a, Infinity]; SetPrecision[b, Infinity];>> In[4]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8> +> a/(2*b)>> In[5]:=> SetAccuracy[f, Infinity]; SetPrecision[f, Infinity];>> In[6]:=> f>> Out[6]=> -1.1805916207174113*^21>> In[7]:=> a = 77617; b = 33096;>> In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +> (55/10)*b^8 + a/(2*b)>> In[9]:=> g>> Out[9]=> -(54767/66192)>> In[10]:=> N[%]>> Out[10]=> -0.8273960599468214> PK>> > Andrzej Kozlowski>> Yokohama, Japan>> http://www.mimuw.edu.pl/~akoz/>> http://platon.c.u-tokyo.ac.jp/andrzej/>Andrzej KozlowskiYokohama, Japanhttp://www.mimuw.edu.pl/~akoz/http:// platon.c.u-tokyo.ac.jp/andrzej/ ==== >>In any case the answer you get is meaningless.Precisely.Bobby-----Original Message----- 121*b^4 - 2) + 5.5*b^8 + a/(2*b), k];In[4]:=a = 77617.; b = 33096.;In[5]:=a = SetAccuracy[a, k];In[6]:=b = SetAccuracy[b, k];In[7]:=fOut[7]=-5.517164009`-2.8311*^19In[8]:= Accuracy[f]Out[8]=-23In[9]:=Precision[f]Out[9]=-3In any case the answer you get is meaningless.> I think I'd prefer that Mathematica gave me a clue -- without being> asked explicitly in JUST the right way that only you know and had> forgotten -- that there's a precision problem.>> Oddly enough, when you DO ask it nicely, you get error messages, butif> you're not aware there's a problem, it lets you go on your merry way,> working with noise.>> Bobby>> -----Original Message-----> Sent: Tuesday, October 01, 2002 8:05 PM> Cc: ÔAllan Hayes'; mathgroup@smc.vnet.net>> So? It's just as it should be, isn't it?>> Andrzej>> Andrzej Kozlowski> Toyama International University> JAPAN> http://sigma.tuins.ac.jp/~andrzej/>> Go one step further:>> Clear[f, a, b, k]>> k = $MachinePrecision;>> f = SetAccuracy[333.75*b^6 +>> a^2*(11*a^2*b^2 - b^6 ->> 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b), k];>> a = 77617.; b = 33096.;>> a = SetAccuracy[a, k];>> b = SetAccuracy[b, k];>> f>> Accuracy[f]>> Precision[f]>> -5.517164009`0*^19>> Accuracy::mnprec:Value -23 would be inconsistent with $MinPrecision;>> bounding by $MinPrecision instead.>> -20>> 0>> Accuracy::mnprec:Value !(-23) would be inconsistent with>> $MinPrecision; >> bounding by $MinPrecision instead.>> See that? NO precision.>> Bobby>> -----Original Message----->> Sent: Tuesday, October 01, 2002 5:53 PM>> Cc: drbob@bigfoot.com>> You are of course right, I forgot that Mathematica does not try to> keep>> precision or accuracy of machine arithmetic computations. But I think>> the actual precision you set need not be higher than machine> precision,>> it just has to be set explicitely (is that right?). For example:>> In[1]:=>> Clear[f,a,b,k]>> In[2]:=>> k = $MachinePrecision;>> In[3]:=>> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/>> (2*b),k];>> In[4]:=>> a=77617.;b=33096.;>> In[5]:=>> a=SetAccuracy[a,k];>> In[6]:=>> b=SetAccuracy[b,k];>> In[7]:=>> f>> Out[7]=>> !((-5.51716400890319`-2.8311*^19))>> In[8]:=>> Accuracy[f]>> Accuracy::mnprec: Value -23 would be inconsistent with $MinPrecision;>> bounding by $MinPrecision instead.>> Out[8]=>> -20>> Andrzej> Andrzej>> Yes, like you I took the original question to be about how to getthe> result> of using the naive rational versions in place of the inexactnumbers.> Bobby raises the question of how we might know accuracy of the> result.>> You answer this with>> a = 77617.; b = 33096.;>> In[2]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b)>> In[3]:=>> f>> Out[3]=>> -1.1805916207174113*^21>> In[4]:=>> Accuracy[%]>> Out[4]=>> -5>> However this computation is done in machine arithmetic, which means> that> Mathematica keeps no check on the accuracy and precision of the> computation,> and Mathematica gives the default accuracy value without any real> signifcance:>> $MachinePrecision - Log[10,Abs[f]]//Round>> -5>> To get meaningful accuracy and precision values we need to force the> computation to be in bignums (bigßoat, arbitrary precision)> arithmetic.> Mathematica then keeps track of the accuracy and precision that it> can> guarantee.>> Clear[f,a,b,k]> k=50;>> f=SetAccuracy[333.75*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.5*b ^8+a/> (2*b),k];> a=77617.;b=33096.;> a=SetAccuracy[a,k];> b=SetAccuracy[b,k];> f>> -0.82739605995>> Accuracy[f]>> 11>> Precision[f]>> 11>> Which tells us that the error in the computed value of f is notmore> than>> 10^-11> The above results are rounded. More accurately we get>> Accuracy[f, Round->False]>> 11.4956>> Precision[f, Round->False]>> 11.4133>> There are several related issues here:> - is the precision (accuracy) of rhe input known?> - how does Mathematica interpret what one gives it?> - how does Mathematica go about the calculation?>> --> Allan>> ---------------------> Allan Hayes> Mathematica Training and Consulting> Leicester UK> www.haystack.demon.co.uk> hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198>>message>> It seems clear to me that what Allan and what you mean bysucceeds>> here refer to quite different things and your objection istherefore>> beside the point. There are obviously two ways in which one can>> interpret the original posting. The first interpretation, which> Allan>> and myself adopted, was that the question concerned purely the>> computational mechanism of Mathematica. Or, to put it in other> words,>> it was why are the results of these two computations not the> same?.>> In this sense success means no more than making Mathematicareturn>> the same answer using the two different routes the original poster>> used.>> You on the other hand choose to assume that the posting shows that>> its>> author does not understand not just the mechanism of significance>> arithmetic used by Mathematica but also computations with inexact>> numbers in general. I do not think this is necessarily the correct>> assumption. I also don't see that Mathematica is doing anything>> wrong.>> After all, one can always check the accuracy of your answer:>> In[1]:=>> a = 77617.; b = 33096.;>> In[2]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b)>> In[3]:=>> f>> Out[3]=>> -1.1805916207174113*^21>> In[4]:=>> Accuracy[%]>> Out[4]=>> -5>> which tells you that it can't be very reliable. What more do you>> demand?>> Andrzej>> Andrzej Kozlowski>> Yokohama, Japan>> http://www.mimuw.edu.pl/~akoz/>> http://platon.c.u-tokyo.ac.jp/andrzej/> Actually, we don't know whether SetAccuracy succeeds, because we> don't> know how inexact those numbers really are. If they are known to>> more> digits than shown in the original post, they should be enteredwith> as> much precision as they deserve. If not, there's no trick or> algorithm> that will give the result precision, because the value of f really>> is> in the noise. That is, we have no idea what the value of f> should> be.>> Mathematica's failing is in returning a value without pointing out> that> it has no precision.>> Bobby>> -----Original Message-----> Sent: Monday, September 30, 2002 11:59 AM>> Andrzej, Bobby, Peter>> It looks as if using SetAccuracy succeeds here because the inexact> numbers> that occur have finite binary representations. If we change them> slightly to> avoid this then we have to use Rationalize:>> 1) Using SetAccuracy>> Clear[a,b,f]> f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/> (2*b),> Infinity];>> a=77617.1;> b=33096.1;>> a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity ];>> f>> - 15640321149084868351974949239896188679725401538739519428131155 14949> 3891236234>> 52500771916869370459119776018798804630436149786919912931962574 3010292> 36> 3> 1246> 75>> /> 108671061439707605510003578275547938881981431359756495796079898 677435> 72> 8240> 16> 0653953612982932181371232436367739737604096>> 2) Rewriting as fractions>> a=776171/10;> b=330961/10;> f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+54/10*b^8+a/(2 *b)>> -(5954133808997234115690303589909929091649391296257/> 41370125000000)>> 3) Using Rationalize>> Clear[a,b,f]> f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4*b ^8+a/> (2*b),> 0];>> a=77617.1;> b=33096.1;>> a=Rationalize[a,0];b=Rationalize[b,0];>> f>> -(5954133808997234115690303589909929091649391296257/> 41370125000000)> I use Rationalize[. , 0] besause of results like>> Rationalize[3.1415959]>> 3.1416>> Rationalize[3.1415959,0]>> 31415959/10000000> --> Allan>> ---------------------> Allan Hayes> Mathematica Training and Consulting> Leicester UK> www.haystack.demon.co.uk> hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198>> Well, first of of all, your using SetAccuracy and SetPrecision> does>> nothing at all here, since they do not change the value of a orb.>> You>> should use a = SetAccuracy[a, Infinity] etc. But even then you>> won't>> get the same answer as when you use exact numbers because of the>> way>> you evaluate f. Here is the order of evaluation that will giveyou>> the>> same answer, and should explain what is going on:>> f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*>> b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity];>> a = 77617.;>> b = 33096.;>> a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity];>> f>> 54767>> -(-----)>> 66192>> Andrzej Kozlowski>> Toyama International University>> JAPAN> Could someone explain what is going on here, please?>> In[1]:=> a = 77617.; b = 33096.;>> In[2]:=> SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity];> SetPrecision[a, Infinity]; SetPrecision[b, Infinity];>> In[4]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8> +> a/(2*b)>> In[5]:=> SetAccuracy[f, Infinity]; SetPrecision[f, Infinity];>> In[6]:=> f>> Out[6]=> -1.1805916207174113*^21>> In[7]:=> a = 77617; b = 33096;>> In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +> (55/10)*b^8 + a/(2*b)>> In[9]:=> g>> Out[9]=> -(54767/66192)>> In[10]:=> N[%]>> Out[10]=> -0.8273960599468214> PK>> > Andrzej Kozlowski>> Yokohama, Japan>> http://www.mimuw.edu.pl/~akoz/>> http://platon.c.u-tokyo.ac.jp/andrzej/>Andrzej KozlowskiYokohama, Japanhttp://www.mimuw.edu.pl/~akoz/http:// platon.c.u-tokyo.ac.jp/andrzej/ ==== >Many thanks to all who replied.>>The original problem was as follows:>>In a presentation I wish to use Plot to generate a sequence of frames and>then animate them. The problem is that the audience sees the animation>twice. Once when the frames are being generated and then again after I have>closed the group and double clicked on the top graphic. However, the first>showing during generation is enough (but uncontrolled).>>Is it possible to tidy up the generation of the graphic so that it becomes>the animation?>There were two main solutions which I now apply to my real problem and not>the simple, generic, problem in the original post.>The real problem was how to build up a probably density function (PDF) in>real time. In the examples below I redraw the PDF every time I add 10>points.>>Initialise>><<<<< JLink`;>wb=WeibullDistribution[ 2.101349094155377,22.58126779173235`];>midpts=Table[i,{i, 0.5,50,1}];>>Solution from Omega Consulting>>GraphicCell[graphics_] :=> Cell[GraphicsData[PostScript, DisplayString[graphics]],Graphics]>>Block[{$ DisplayFunction=Identity, graphs},> data={};> graphs => Table[data=Flatten[Join[data,RandomArray[wb,10]]];> freq=BinCounts[data,{0,50,1}];> GraphicCell[> BarChart[Transpose[{freq,midpts}],ImageSize ->500] ], {500}];> NotebookWrite[EvaluationNotebook[],Cell[CellGroupData[graphs ,Closed]]];> SelectionMove[EvaluationNotebook[], All, GeneratedCell];> FrontEndExecute[{FrontEndToken[EvaluationNotebook[],> SelectionAnimate]}]> ]>>This solution works but it generates 500 frames and sometimes exceeds the>memory.>The paradigm here is generate all frames, then animate all frames. We>really need a loop that does:>> generate next frame, delete last frame, show next frame>>Is it possible to do this?Yes, however, I thought you didn't want to see the selection bar moving around during the animation. That's why I chose to generate the whole animation in one shot. Also, when you write the new cell over the old cell there is a ßash between frames as the old frame is deleted. Here's an example of a frame-by-frame method.GraphicCell[graphics_] := Cell[GraphicsData[PostScript, DisplayString[graphics]],Graphics]CellPrint[Cell[, Graphics]];Block[{$DisplayFunction=Identity}, Do[ SelectionMove[EvaluationNotebook[], All, GeneratedCell]; NotebookWrite[EvaluationNotebook[], GraphicCell[Plot[x y,{x,0,1}, PlotRange->{0,50}]]], {y,50} ] ]Also, if you add ShowCellBracket->False to GraphicCell and a Pause to the loop, then things get much better visually.---------------------------------------------------- ----------Omega ConsultingThe final answer to your Mathematica needsSpend less time searching and more time finding.http://www.wz.com/internet/Mathematica.html ==== I am trying to use FindRoot in mathematica 4.0 to find the zeros of a complex-valued function of one complex variable. In particular, I am looking for the one root with a positive imaginary part. I have a rough approximation for where the root should be, and this is good enough to give a reasonable guess.However, there are always two other roots near the one I want - one with 0 imaginary part and another with negative imag part. (For those who are interested, the zeros are the roots of the dispersion relation for a plasma interacting with a laser). Sometimes FindRoot picks up one of these instead of the one I want.So, I'd like to tell mathematica to look for a root only in a certain rectangular region of the complex plane. Well, if I could tell it, Ôlook for roots with imag. part > something', I'd be happy too.I tried specifying complex values for the start and stop points of an interval, hoping mathematica would interpret these as the corners of a rectangle. No such luck.Any help would be greatly appreciated.I'd also like to point out that this and other issues about complex roots are not clearly addressed in the built-in help files. ==== Just a shot: Try using the DampingFactor option to force FindRoot to take smaller steps.---Selwyn Hollis ==== > I am trying to use FindRoot in mathematica 4.0 to find the zeros of a> complex-valued function of one complex variable. In particular, I am> looking for the one root with a positive imaginary part. I have a rough> approximation for where the root should be, and this is good enough to> give a reasonable guess.> However, there are always two other roots near the one I want - one with> 0 imaginary part and another with negative imag part. (For those who> are interested, the zeros are the roots of the dispersion relation for a> plasma interacting with a laser). Sometimes FindRoot picks up one of> these instead of the one I want.> So, I'd like to tell mathematica to look for a root only in a certain> rectangular region of the complex plane. Well, if I could tell it,> Ôlook for roots with imag. part > something', I'd be happy too.> I tried specifying complex values for the start and stop points of an> interval, hoping mathematica would interpret these as the corners of a> rectangle. No such luck.> Any help would be greatly appreciated.> I'd also like to point out that this and other issues about complex> roots are not clearly addressed in the built-in help files.> I believe it is difficult to restrict FindRoot. Below are severalsuggestions. (1) You might tryrigging the function, say withg[x_?NumberQ] := f[x] + If[Im[x]<=0,1000,0](2) If you have version 4.2 a related approach would be to takeadvantage of constrained optimization and doNMinimize[{Re[f[x,y]]^2 + Im[f[x,y]]^2, rectangle range constraints},{x,y}]or something along those lines.(3) If proximity of the other roots is the main problem it might bealleviated by rescaling your variable.(4) Alternatively you could write your function as a pair of real valuedfunctions of two real valued arguments. Then you can use Interval tosplit the rectangle into parts, keeping subrectangles for which {0,0}lives in f[Interval[x],Interval[y]] where Interval[x], for example, is ashorthand for an interval for the x part of the rectangle being split.Daniel LichtblauWolfram ResearchReply-To: kuska@informatik.uni-leipzig.de ==== and solving the real and imaginary part for for z->x+I*y withFindRoot[Evaluate[({Re[#] == 0, Im[#] == 0} &[ z^3 - 1]) /. z -> x + I y], {x, -3/2, 0}, {y, 0.5, 0.9}]helps not ? Jens> I am trying to use FindRoot in mathematica 4.0 to find the zeros of a> complex-valued function of one complex variable. In particular, I am> looking for the one root with a positive imaginary part. I have a rough> approximation for where the root should be, and this is good enough to> give a reasonable guess.> However, there are always two other roots near the one I want - one with> 0 imaginary part and another with negative imag part. (For those who> are interested, the zeros are the roots of the dispersion relation for a> plasma interacting with a laser). Sometimes FindRoot picks up one of> these instead of the one I want.> So, I'd like to tell mathematica to look for a root only in a certain> rectangular region of the complex plane. Well, if I could tell it,> Ôlook for roots with imag. part > something', I'd be happy too.> I tried specifying complex values for the start and stop points of an> interval, hoping mathematica would interpret these as the corners of a> rectangle. No such luck.> Any help would be greatly appreciated.> I'd also like to point out that this and other issues about complex> roots are not clearly addressed in the built-in help files.> ==== The last part of my message you are quoting was completely wrong, as was pointed out by Allan Hayes. Mathematica does not track precision of machine arithmetic computations. In order for Mathematica to give reliable information about the precision of a computation you have to explicitly set the precision of all the numerical quantities.Your own example at the bottom simply shows you have not understood the evaluation mechanism of Mathematica. What you are doing is this:In[1]:=a = SetAccuracy[77617., Infinity];b = SetAccuracy[33096., Infinity];At this point you have converted a and be to have the following exact values:In[3]:=aOut[3]=77617In[4]:=bOut[4]=33096Next you evaluate:f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity]but this is a two step process (which is what you seem not to have grasped). First Mathematica computes:In[5]:=333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b)Out[5]=1.1805916207174113*^21Now, because there were machine ßoats in the formula (333.75 and 5.5 ) the whole expression was computed using machine arithmetic without keeping track of precision. The answer is therefore completely inacurate. Mathematica returns the purely formal accuracy:In[6]:=Accuracy[%]Out[6]=-5But the second part of your evaluation tells it to take this inaccurate answer and convert it to an exact number.In[7]:=SetAccuracy[%%, Infinity]Out[7]=1180591620717411303424In[8]:=Accuracy[%] Out[8]=InfinityBut of course doing this is meaningless, after converting an inexact answer to an exact number does not make it a an exact answer!Of course had you evaluated f before you assigned the values to a and b you would not have encountered the problem because no machine reals would have appeared in the computation. ALternatively you might have used:f = SetAccuracy[333.75, Infinity]*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + SetAccuracy[5.5, Infinity]*b^8 + a/(2*b)Andrzej KozlowskiYokohama, Japanhttp://www.mimuw.edu.pl/~akoz/http:// platon.c.u-tokyo.ac.jp/andrzej/>> It seems clear to me that what Allan and what you mean by succeeds>> here refer to quite different things and your objection is therefore>> beside the point. There are obviously two ways in which one can>> interpret the original posting. The first interpretation, which Allan>> and myself adopted, was that the question concerned purely the>> computational mechanism of Mathematica. Or, to put it in other words,>> it was why are the results of these two computations not the same?.>> In this sense success means no more than making Mathematica return>> the same answer using the two different routes the original poster >> used.>> You on the other hand choose to assume that the posting shows that its>> author does not understand not just the mechanism of significance>> arithmetic used by Mathematica but also computations with inexact>> numbers in general. I do not think this is necessarily the correct>> assumption. I also don't see that Mathematica is doing anything wrong.>> After all, one can always check the accuracy of your answer:>> In[1]:=>> a = 77617.; b = 33096.;>> In[2]:=>> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) +>> 5.5*b^8 + a/(2*b)>> In[3]:=>> f>> Out[3]=>> -1.1805916207174113*^21>> In[4]:=>> Accuracy[%]>> Out[4]=>> -5>> which tells you that it can't be very reliable. What more do you >> demand?>> As you are dealing here only with machine-precision numbers, your>> When you do calculations with arbitrary-precision numbers, as> discussed in the previous section, Mathematica always keeps track of> the precision of your results, and gives only those digits which are> known to be correct, given the precision of your input. When you do> calculations with machine-precision numbers, however, Mathematica> always gives you a machine-ç.8dprecision result, whether or not all the> digits in the result can, in fact, be determined to be correct on the> basis of your input. >> In practice, to rely on a numerical result, you ALWAYS have to check> its accuracy. How reliable is Accuracy anyway?>> In[1]:=> a = SetAccuracy[77617., Infinity];> b = SetAccuracy[33096., Infinity];>> In[3]:=> f = SetAccuracy[333.75*b^6 +> a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 +> a/(2*b), Infinity]>> Out[3]=> 1180591620717411303424>> In[4]:=> Accuracy[f]>> Out[4]=> Infinity>> We got completely wrong result with Infinite accuracy. In conclusion,> one can not rely on Accuracy. Checking numerical results in> Mathematica sounds like a tough task.:-)>> --PK> Andrzej>> Andrzej Kozlowski>> Yokohama, Japan>> http://www.mimuw.edu.pl/~akoz/>> http://platon.c.u-tokyo.ac.jp/andrzej/> [...]> ==== leap to MathGroup so I'll respond there as well to some of the issuesraised herein.> Daniel,> >>The precision/accuracy tracking mechanism will generally let you know,>>in some fashion, that you have no trustworthy digits. But it is up to>>the user to check that sort of thing.> In this case Mathematica did NOT let us know, in any fashion, that we> had no trustworthy digits. Precision and Accuracy outputs were> completely misleading. (16 and -5 respectively.) Precision, in this case, is itself a bit misleading. As I recall machinearithmetic was in use. For that domain just regard 16 as the definitionof precision. Moreover no tracking is done. It is only when bignums areused that significance arithmetic will be at play.By the way, the distinction between precision for machine numbers vs.bignums will be more clear with our next big release. At present onedoes not know if 16 refers to a machine number or a bignum of that sameprecision.> Even Andrzej> Kozlowski, who's adept in Mathematica, thought that would be meaningful,> and never came up with a better way to check (other than using infinite> precision for numbers that probably aren't known that exactly). Peter> Kosta demonstrated that he could get a completely erroneous answer with> Infinite precision.> I blame the problem primarily, and I don't think there's any way to make> the answer meaningful. That's not Mathematica's fault at all, and users> need to be aware of that old maxim: garbage in, garbage out.Right. One cannot artificially raise precision or accuracy after thefact and expect a meaningful result. Again, had significance arithmeticbeen used, I think there would have been adequate information to assessthe problem.> comes up with a 22-digit result, it doesn't take much sophistication to> realize the answer can't have 16-digit precision.> Here's an even more extreme result:> f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 -> 121*b^4 - 2) + 5.5*b^8 + a/(2*b), 50];> a = 77617.; b = 33096.;> f> Precision[f]> -1.180591620717411303424`71.0721*^21> 71> 71.0721 digits of precision? I don't think so!! It's correct, albeit the number is garbage. You start with a number thatis, in your words, all noise. (I assume you had set a and b before f,because otherwise I get a different result for Precision[f]). Nowf ~ 10^21, and so has accuracy of around 71 digits. Garbagenotwithstanding, this behavior is entirely correct and as it should be.> We can do the following instead:> x = Interval[333.75];> y = Interval[5.5];> a = Interval[77617.];> b = Interval[33096.];> x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8 + a/(2*b)> Interval[{-4.486248158726164*^22, 4.2501298345826815*^22}]> and that looks like the right answer, finally!! I like that method.Yes, it's a useful way to show that the result has no digits to trust.Actually you can achieve a similar effect using significance arithmetic,as below.x = SetPrecision[33375/100, 16];y = SetPrecision[11/2, 16];a = SetPrecision[77617, 16];b = SetPrecision[33096, 16];In[18]:= InputForm[x*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + y*b^8+ a/(2*b)]Out[18]//InputForm= -1.1339143158169`0*^14This tells you there are NO reliable digits. The advantage to this isthat is is more ßexible than interval arithmetic in Mathematica whichwill not, for example, deal well with complex-valued functions orcomplex domains. Also significance arithmetic (aka arbitrary precisionarithmetic) is much faster than interval arithmetic for large problems.> However, that doesn't change the fact that Accuracy, Precision, and> SetAccuracy appear to be completely useless.My own experience is that they are quite useful. But only if used withcare and in appropriate ways. Raising accuracy after the fact does notfall into that category.> I haven't seen an example> in which they did what anyone (but you) thought they should do.> BobbyMaybe. But, curiously enough, I am a user rather than developer when itcomes to that stuff. I used the precision mechanism alot in polynomialNSolve, and it did indeed work to my liking. Which counts for somethingfrom the ordinary users' perspective, because now NSolve works betterthan it otherwise might.> -----Original Message----->> I think I'd prefer that Mathematica gave me a clue -- without being> > asked explicitly in JUST the right way that only you know and had> forgotten -- that there's a precision problem.> >> Oddly enough, when you DO ask it nicely, you get error messages, but> if> you're not aware there's a problem, it lets you go on your merry way,> working with noise.>> Bobby> Mathematica is not a mind reader. But the evaluation sequence, while> complicated, is reasonably well documented. If you perform machine> arithmetic, or for that matter significance arithmetic, and there is> massive cancellation error, no use of SetAccuracy after the fact will> fix it.> The precision/accuracy tracking mechanism will generally let you know,> in some fashion, that you have no trustworthy digits. But it is up to> the user to check that sort of thing. It is not obvious to me what sort> of error the software might notice to report. If you have a concise> example of input, and expected output, I can look further. I've not seen> anything in this thread that struck me as a failure of the software to> warn the user, but maybe I missed something.> DanielDaniel LichtblauWolfram Research ==== > The more I play with the example the more depressing it gets. Start> with ßoating point numbers but explicitely arbitrary-precision ones.> In[1]:=> a=77617.00000000000000000000000000000;> b=33095.00000000000000000000000000000;> In[3]:=> !(333.7500000000000000000000000000000 b^6 + a^2 ((11 a^2> b^2 - > b^6 - 121 b^4 - 2)) + 5.500000000000000000000000000000 b^8 +> a/(2> b))> Out[3]=> !((-4.78339168666055402578083604864320577443814`26.6715 *^32))> In[4]:=> Accuracy[%]> Out[4]=> -6> Due to the manual section 3.1.6:> When you do calculations with arbitrary-precision numbers, as> discussed in the previous section, Mathematica always keeps track of> the precision of your results, and gives only those digits which are> known to be correct, given the precision of your input. When you do> calculations with machine-precision numbers, however, Mathematica> always gives you a machine[CapitalEth]precision result, whether or not all the> digits in the result can, in fact, be determined to be correct on the> basis of your input. > Because I started with arbitrary-precision numbers Mathematica should display> only those digits that are correct, that is none.No, 26 digits are correct (check Precision instead of Accuracy to seethis).You appear to be showing output in InputForm. If you use OutputForm orStandardForm only 26 digits will be shown. 32Out[3]= -4.7833916866605540257808360 10InputForm is showing more because it exposes bad digits as well asgood ones.> To relax a bit, set a new input cell to StandardForm and type> 77617.000000000000000000000000000000000> Convert it to InputForm. You get> 77616.999999999999999999999999999999999999999999952771` 37.9031> Convert back to StandardForm> 77616.99999999999999999999999999999999999999999976637`37.9031 > Again to InputForm> 77616.99999999999999999999999999999999999999999963735`37.9031 > Back to StandardForm> 77616.99999999999999999999999999999999999999999951376`37.9031 > See what you can get if you have enough patience or a small program.> PKAgreed, it's not very pretty. I am uncertain as to whether thisindicates a bug in StandardForm or elsewhere in the underlying numericscode, and will defer to our numerics experts on that issue. My guess isit is a bug if only because it violates the spirit of IEEE arithmeticwherein ßoats that have integer values should be representable as such(or something to that effect). I will point out, however, that the twonumbers in question are equal to the specified precision. Also itappears to be improved in our development kernel.Daniel LichtblauWolfram Research ==== The more I play with the example the more depressing it gets. Startwith ßoating point numbers but explicitely arbitrary-precision ones.In[1]:=a=77617.00000000000000000000000000000;b= 33095.00000000000000000000000000000;In[3]:=!( 333.7500000000000000000000000000000 b^6 + a^2 ((11 a^2b^2 - b^6 - 121 b^4 - 2)) + 5.500000000000000000000000000000 b^8 +a/(2 b))Out[3]=!((- 4.78339168666055402578083604864320577443814`26.6715*^32)) In[4]:=Accuracy[%]Out[4]=-6Due to the manual section 3.1.6:When you do calculations with arbitrary-precision numbers, asdiscussed in the previous section, Mathematica always keeps track ofthe precision of your results, and gives only those digits which areknown to be correct, given the precision of your input. When you docalculations with machine-precision numbers, however, Mathematicaalways gives you a machine[CapitalEth]precision result, whether or not all thedigits in the result can, in fact, be determined to be correct on thebasis of your input. Because I started with arbitrary-precision numbers Mathematica should displayonly those digits that are correct, that is none.To relax a bit, set a new input cell to StandardForm and type 77617.000000000000000000000000000000000Convert it to InputForm. You get77616.999999999999999999999999999999999999999999952771` 37.9031Convert back to StandardForm 77616.99999999999999999999999999999999999999999976637`37.9031 Again to InputForm 77616.99999999999999999999999999999999999999999963735`37.9031 Back to StandardForm 77616.99999999999999999999999999999999999999999951376`37.9031 See what you can get if you have enough patience or a small program.PK ==== > The more I play with the example the more depressing it gets. Start> with ßoating point numbers but explicitely arbitrary-precision ones.> In[1]:=> a=77617.00000000000000000000000000000;> b=33095.00000000000000000000000000000;> In[3]:=> !(333.7500000000000000000000000000000 b^6 + a^2 ((11 a^2> b^2 - > b^6 - 121 b^4 - 2)) + 5.500000000000000000000000000000 b^8 +> a/(2> b))> Out[3]=> !((-4.78339168666055402578083604864320577443814`26.6715 *^32))> In[4]:=> Accuracy[%]> Out[4]=> -6> Due to the manual section 3.1.6:> When you do calculations with arbitrary-precision numbers, as> discussed in the previous section, Mathematica always keeps track of> the precision of your results, and gives only those digits which are> known to be correct, given the precision of your input. When you do> calculations with machine-precision numbers, however, Mathematica> always gives you a machine[CapitalEth]precision result, whether or not all the> digits in the result can, in fact, be determined to be correct on the> basis of your input. > Because I started with arbitrary-precision numbers Mathematica should display> only those digits that are correct, that is none.I retract the above comment. I did not notice that was an error in the input. b=33095.00000000000000000000000000000intstead of intended b=33096.00000000000000000000000000000I am sorry for the mistake.PK> To relax a bit, set a new input cell to StandardForm and type > 77617.000000000000000000000000000000000> Convert it to InputForm. You get> 77616.999999999999999999999999999999999999999999952771` 37.9031> Convert back to StandardForm> 77616.99999999999999999999999999999999999999999976637`37.9031 > Again to InputForm> 77616.99999999999999999999999999999999999999999963735`37.9031 > Back to StandardForm> 77616.99999999999999999999999999999999999999999951376`37.9031 > See what you can get if you have enough patience or a small program.> PK ==== > The last part of my message you are quoting was completely wrong, as > was pointed out by Allan Hayes. Mathematica does not track precision of > machine arithmetic computations. In order for Mathematica to give > reliable information about the precision of a computation you have to > explicitly set the precision of all the numerical quantities.> Your own example at the bottom simply shows you have not understood the > evaluation mechanism of Mathematica. Just opposite, thanks to you and other participants, I completelyunderstood it. SetAccuracy just takes anything and calls it accurate.This behavior is useless if not stupid. It was apparently intended byMathematica developers but that doesn't make it right.On a side note, I hate the argument It is descibed in the manual,therefore it is correct. Legal doesn't mean right. Besides there isno supreme court here to overrule some stupid law. :-)PK> [...] ==== >Dear Colleagues,>>I calculated:>>Sum[1/Prime[n], {n, 15000}] // N>>Result: 2.74716>>Now I wonder if this sum will converge or keep on growing, albeit very>slowly.>Matthias Bode>Sal. Oppenheim jr. & Cie. KGaA>Koenigsberger Strasse 29>D-60487 Frankfurt am Main>GERMANY>Mobile: +49(0)172 6 74 95 77>Internet: http://www.oppenheim.de>The serie is divergent!I suggest you to look at this beautiful introduction to primes distributionhttp://www.maths.ex.ac.uk/%7Emwatkins/zeta/ vardi.htmlBye, robRoberto BrambillaCESIVia Rubattino 5420134 Milanotel +39.02.2125.5875fax +39.02.2125.5492rlbrambilla@cesi.it ==== > I've been looking over the file and directory manipulation functions in Mathematica> 4.1, and I'm not finding good examples of how to test for the existence of> a file or directory before I attempt to create, access, or delete it. Are> there any good examples of this somewhere?>> STH>>Steven,Directory[]returns the current directory$Pathreturns the directories in your path where files can be found.SetDirectory[]changes to the specified directory and makes it the current directory.FileNames[]returns a list of all the file names (text documents, picture documentsetc) and subdirectory (folder) names in the current directory.Since FileNames returns a list of elements, you can use all sorts of teststo see if a file you're looking for is in a said directory or not orwhether it is a subdirectory or not.I hope that this will get you started.Yas ==== has anybody tried to modify the classes.m package in the way, it wouldbe possible to have several superclasses?Oleg. ==== What does this page say, auf Englisch? I only pretend to understand German.http://www.mertig.com/neu/HTMLLinks/index_6.htmlSTH ==== The sum diverges. See Elementary Number Theory, 5th edition by David M.Burton, pages 355-356.-----Original Message-----Sal. Oppenheim jr. & Cie. KGaAKoenigsberger Strasse 29D-60487 Frankfurt am MainGERMANYMobile: +49(0)172 6 74 95 77Internet: http://www.oppenheim.de ==== Technical support has responded to my query. They say that one of thesimplifications that Mathematica does when using Simplify is to Factor theexpression, and factoring an expression consisting of inexact numbers leadsto the type of behavior I observed. They stated that changing the behaviorof Simplify to avoid the problems I noticed would make Simplify lesscapable, and that the majority of users would prefer to have Simplify behavethe way it currently does.Of course, that doesn't answer the question about why the number of terms inthe expression would cause the coefficients to go from rational to inexactto rational again.At any rate, I have another lesson learned. When using Simplify, check theresult if you are mixing infinite precision and inexact numbers, as theresult may not be what you wanted. In the past I always assumed that theresult of using Simplify on an expression produced a result equivalent tothe original expression, but now I discover that this is not true.Carl WollPhysics DeptU of Washington----- Original Message -----> {i, > n}]], x[1]]>> Now lets try CW for the first 100 values of n:> In[30]:=> Table[CW[n], {n, 100}]>> Out[30]=> {1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,> 0.5`12.3085, 0.5`12.0074, 0.5`11.7064, 0.5`11.4054,> 0.5`11.1043, 0.5`10.8033, 0.5`10.5023, 0.5`10.2012,> 0.5`9.9002, 0.5`9.5992, 0.5`9.2982, 0.5`8.9971,> 0.5`8.6961, 0.5`8.3951, 0.5`8.094, 0.5`7.793, 0.5`7.492,> 0.5`7.1909, 0.5`6.8899, 0.5`6.5889, 0.5`6.2879,> 0.5`5.9868, 0.5`5.6858, 0.5`5.3848, 0.5`5.0837,> 0.5`4.7827, 0.5`4.4817, 0.5`4.1806, 0.5`3.8796,> 0.5`3.5786, 0.5`3.2776, 0.5`2.9765, 0.5`2.6755,> 0.5`2.3745, 0.5`2.3745, 0.5`1.7724, 0.5`1.7724,> 0.5`1.1703, 0.5`1.1703, 0.5`0.5683, 0``0.1423, 0``0.1423,> 0``0.1423, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,> 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,> 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,> 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,> 1/2, 1/2, 1/2}>> In[31]:= Map[Length,Split[#]]>> Out[31]=> {12,43,45}>> This is indeed curious.The problem seems to occur for values between 13> and 55 and then go away (for good???) At first I thought it maybe in> some fascinating way related to some properties of the integer n, but> now I am not sure. It certainly worth a careful examination. I hope> whoever discovers the cause of this will let us know.>> Andrzej>> Andrzej Kozlowski> Yokohama, Japan> http://www.mimuw.edu.pl/~akoz/> http://platon.c.u-tokyo.ac.jp/andrzej/>> To Technical Support and the Mathematica User community,>> I'm writing to report what I consider to be a bug. First, I want to> show a> simplified example of the problem. Consider the following expression:>> expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16;>> When simplified I expected to get some real number plus> x[0]+3x[1]/4+9x[2]/16, but instead I get the following:>> Simplify[expr]> 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2]> >> As you can see, for some reason Mathematica converted the fractions> 3/4 and> 9/16 to real machine numbers. I consider this to be a bug.>> Now, for an example more representative of the situation that I've been> coming across.>> expr12 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 12}];> expr13 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 13}];> expr55 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 55}];>> As you can see, I have replaced the real numbers by extended precision> numbers. The simplified example above demonstrates that the problem> exists> when using machine numbers. Now, we'll see what happens when we use> arbitrary precision numbers. First, let's simplify the expression with> 12> terms.>> Simplify[expr12]> (1.0010000000000 + 2048 x[1] + 1024 x[2] + 512 x[3] + 256 x[4] + 128> x[5] +>> 64 x[6] + 32 x[7] + 16 x[8] + 8 x[9] + 4 x[10] + 2 x[11] + x[12])> / 4096>> As you can see, a sum with 12 terms upon simplification has> coefficients> which are still integers as they should be. However, increasing the> number> of terms to 13 yields>> Simplify[expr13]> 0.0001221923828125 + 0.500000000000 x[1] + 0.250000000000 x[2] +>> 0.1250000000000 x[3] + 0.0625000000000 x[4] + 0.0312500000000 x[5] +>> 0.01562500000000 x[6] + 0.00781250000000 x[7] + 0.00390625000000> x[8] +>> 0.001953125000000 x[9] + 0.000976562500000 x[10] + 0.000488281250000> x[11]> +>> > 0.000244140625000 x[12] + 0.0001220703125000 x[13]>> Now, all of the coefficients are converted to real numbers,> replicating the> bug from the simplified example. Finally, let's see what happens when> we> have 55 terms. Rather than putting the resulting expression here, I> will> just leave it at the end of the post. The result though is somewhat> surprising. Each of the coefficients of the x[i] are again real> numbers, but> now their precision is only 0! The proper result of course is the sum> of> some real number (with a precision close to 0 due to numerical> cancellation)> and an expression consisting of rational numbers multiplied by x[i].> The> loss of precision of the coefficients of the x[i] is precisely what> > occurred> to me. Of course, in this simple example, simply using Expand instead> of> Simplify produces the expected result, and is my workaround. I hope you> agree with me that this is a bug, and one that Wolfram needs to> correct.>> Carl Woll> Physics Dept> U of Washington>> Simplify[expr55]> -16 -1 -1 -1> 0. 10 + 0. x[1] + 0. x[2] + 0. 10 x[3] + 0. 10 x[4] + 0. 10> x[5] +> >> -2 -2 -2 -3 -3> 0. 10 x[6] + 0. 10 x[7] + 0. 10 x[8] + 0. 10 x[9] + 0. 10> x[10]> +>> -3 -3 -4 -4> -4> 0. 10 x[11] + 0. 10 x[12] + 0. 10 x[13] + 0. 10 x[14] + 0. 10> x[15] +>> -5 -5 -5 -6> -6> 0. 10 x[16] + 0. 10 x[17] + 0. 10 x[18] + 0. 10 x[19] + 0. 10> x[20] +>> -6 -6 -7 -7> -7> 0. 10 x[21] + 0. 10 x[22] + 0. 10 x[23] + 0. 10 x[24] + 0. 10> x[25] +>> -8 -8 -8 -9> -9> 0. 10 x[26] + 0. 10 x[27] + 0. 10 x[28] + 0. 10 x[29] + 0. 10> x[30] +>> -9 -9 -10 -10> 0. 10 x[31] + 0. 10 x[32] + 0. 10 x[33] + 0. 10 x[34] +>> -10 -11 -11 -11> 0. 10 x[35] + 0. 10 x[36] + 0. 10 x[37] + 0. 10 x[38] +>> -12 -12 -12 -12> 0. 10 x[39] + 0. 10 x[40] + 0. 10 x[41] + 0. 10 x[42] +>> -13 -13 -13 -14> 0. 10 x[43] + 0. 10 x[44] + 0. 10 x[45] + 0. 10 x[46] +>> -14 -14 -15 -15> 0. 10 x[47] + 0. 10 x[48] + 0. 10 x[49] + 0. 10 x[50] +>> -15 -15 -16 -16> -> 16> 0. 10 x[51] + 0. 10 x[52] + 0. 10 x[53] + 0. 10 x[54] +> 0. 10> x[55]> >> ==== Dear CarlYou have discovered what is perhaps a bug but maybe something even mor einteresting . However, I think you stopped your investigation a little prematurely. Here is a function that just computes the coefficient of x[1] in your Simplified expression for various values of n:CW[n_] := Coefficient[Simplify[1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, n}]], x[1]]Now lets try CW for the first 100 values of n:In[30]:=Table[CW[n], {n, 100}]Out[30]={1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 0.5`12.3085, 0.5`12.0074, 0.5`11.7064, 0.5`11.4054, 0.5`11.1043, 0.5`10.8033, 0.5`10.5023, 0.5`10.2012, 0.5`9.9002, 0.5`9.5992, 0.5`9.2982, 0.5`8.9971, 0.5`8.6961, 0.5`8.3951, 0.5`8.094, 0.5`7.793, 0.5`7.492, 0.5`7.1909, 0.5`6.8899, 0.5`6.5889, 0.5`6.2879, 0.5`5.9868, 0.5`5.6858, 0.5`5.3848, 0.5`5.0837, 0.5`4.7827, 0.5`4.4817, 0.5`4.1806, 0.5`3.8796, 0.5`3.5786, 0.5`3.2776, 0.5`2.9765, 0.5`2.6755, 0.5`2.3745, 0.5`2.3745, 0.5`1.7724, 0.5`1.7724, 0.5`1.1703, 0.5`1.1703, 0.5`0.5683, 0``0.1423, 0``0.1423, 0``0.1423, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2}In[31]:= Map[Length,Split[#]]Out[31]={12,43,45}This is indeed curious.The problem seems to occur for values between 13 and 55 and then go away (for good???) At first I thought it maybe in some fascinating way related to some properties of the integer n, but now I am not sure. It certainly worth a careful examination. I hope whoever discovers the cause of this will let us know.AndrzejAndrzej KozlowskiYokohama, Japanhttp://www.mimuw.edu.pl/~akoz/http:// platon.c.u-tokyo.ac.jp/andrzej/> To Technical Support and the Mathematica User community,>> I'm writing to report what I consider to be a bug. First, I want to > show a> simplified example of the problem. Consider the following expression:>> expr=0.22 + x[0] + (3*(-0.16+ x[1]))/4 + (9*(0.546 + x[2]))/16;>> When simplified I expected to get some real number plus> x[0]+3x[1]/4+9x[2]/16, but instead I get the following:>> Simplify[expr]> 0.407125 + x[0] + 0.75 x[1] + 0.5625 x[2]>> As you can see, for some reason Mathematica converted the fractions > 3/4 and> 9/16 to real machine numbers. I consider this to be a bug.>> Now, for an example more representative of the situation that I've been> coming across.>> expr12 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 12}];> expr13 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 13}];> expr55 = 1.001`17 + Sum[(x[i] - 1.001`17)/2^i, {i, 55}];>> As you can see, I have replaced the real numbers by extended precision> numbers. The simplified example above demonstrates that the problem > exists> when using machine numbers. Now, we'll see what happens when we use> arbitrary precision numbers. First, let's simplify the expression with > 12> terms.>> Simplify[expr12]> (1.0010000000000 + 2048 x[1] + 1024 x[2] + 512 x[3] + 256 x[4] + 128 > x[5] +>> 64 x[6] + 32 x[7] + 16 x[8] + 8 x[9] + 4 x[10] + 2 x[11] + x[12]) > / 4096>> As you can see, a sum with 12 terms upon simplification has > coefficients> which are still integers as they should be. However, increasing the > number> of terms to 13 yields>> Simplify[expr13]> 0.0001221923828125 + 0.500000000000 x[1] + 0.250000000000 x[2] +>> 0.1250000000000 x[3] + 0.0625000000000 x[4] + 0.0312500000000 x[5] +>> 0.01562500000000 x[6] + 0.00781250000000 x[7] + 0.00390625000000 > x[8] +>> 0.001953125000000 x[9] + 0.000976562500000 x[10] + 0.000488281250000 > x[11]> +>> 0.000244140625000 x[12] + 0.0001220703125000 x[13]>> Now, all of the coefficients are converted to real numbers, > replicating the> bug from the simplified example. Finally, let's see what happens when > we> have 55 terms. Rather than putting the resulting expression here, I > will> just leave it at the end of the post. The result though is somewhat> surprising. Each of the coefficients of the x[i] are again real > numbers, but> now their precision is only 0! The proper result of course is the sum > of> some real number (with a precision close to 0 due to numerical > cancellation)> and an expression consisting of rational numbers multiplied by x[i]. > The> loss of precision of the coefficients of the x[i] is precisely what > occurred> to me. Of course, in this simple example, simply using Expand instead > of> Simplify produces the expected result, and is my workaround. I hope you> agree with me that this is a bug, and one that Wolfram needs to > correct.>> Carl Woll> Physics Dept> U of Washington>> Simplify[expr55]> -16 -1 -1 -1> 0. 10 + 0. x[1] + 0. x[2] + 0. 10 x[3] + 0. 10 x[4] + 0. 10 > x[5] +>> -2 -2 -2 -3 -3> 0. 10 x[6] + 0. 10 x[7] + 0. 10 x[8] + 0. 10 x[9] + 0. 10 > x[10]> +>> -3 -3 -4 -4 > -4> 0. 10 x[11] + 0. 10 x[12] + 0. 10 x[13] + 0. 10 x[14] + 0. 10> x[15] +>> -5 -5 -5 -6 > -6> 0. 10 x[16] + 0. 10 x[17] + 0. 10 x[18] + 0. 10 x[19] + 0. 10> x[20] +>> -6 -6 -7 -7 > -7> 0. 10 x[21] + 0. 10 x[22] + 0. 10 x[23] + 0. 10 x[24] + 0. 10> x[25] +>> -8 -8 -8 -9 > -9> 0. 10 x[26] + 0. 10 x[27] + 0. 10 x[28] + 0. 10 x[29] + 0. 10> x[30] +>> -9 -9 -10 -10> 0. 10 x[31] + 0. 10 x[32] + 0. 10 x[33] + 0. 10 x[34] +>> -10 -11 -11 -11> 0. 10 x[35] + 0. 10 x[36] + 0. 10 x[37] + 0. 10 x[38] +>> -12 -12 -12 -12> 0. 10 x[39] + 0. 10 x[40] + 0. 10 x[41] + 0. 10 x[42] +>> -13 -13 -13 -14> 0. 10 x[43] + 0. 10 x[44] + 0. 10 x[45] + 0. 10 x[46] +>> -14 -14 -15 -15> 0. 10 x[47] + 0. 10 x[48] + 0. 10 x[49] + 0. 10 x[50] +>> -15 -15 -16 -16 > -> 16> 0. 10 x[51] + 0. 10 x[52] + 0. 10 x[53] + 0. 10 x[54] + > 0. 10> x[55]> ==== I think, for input of that many numbers (and other inputs), I might usean input file that has textual names or descriptions in the first ten ortwenty columns, followed by values starting at a fixed column afterthat. Mathematica or Java could easily read inputs from that, and ahuman could read it as well. If you're concerned that a human mightjumble the file format -- accidentally deleting lines, etc. -- a programcould key on the names or descriptions rather than trusting them to bein correct order, and point out missing values. A strategy like thatallows you to start with a previous input file (not from scratch) andchange only what needs to change.Also, Mathematica 4.2 adds XML support to the picture, and that might beuseful.Bobby-----Original Message----->> I agree with Mr. Kuska, that the system Mr Nagesh describes is not> userfriendly. But I think, the suggestions of Mr. Kuska do not makeitmore> userfriendly, rather the opposite is true.>> Mr. Nagesh asks>> Is any body here have expertise or information about the capabilityof> > Mathematica as a system simulation tool?> Mr. Kuska answers:> Since the most system simulation tools are simply solving asystem of> ordinary differntial equations it is simple to do this withNDSolve[].>> My comment:>> That is: He sees the simulation system merely as a set ofdifferential> equations.>> hmm, since the original poster write>> My refrigeration system simulation package is likely to have> approximately 60 First order Differential equations.>> it seems not completly wrong to assume that the system consists of> of ode's ..>Yout should not ignore the word merely.It is not enough to have a set 60 differential equations and a set of200-250 numbers. That is not simulation system, which can be used byuserswith the exception, perhaps, of the programmer of the system himself.How does e.g. the user know what meaning a number in the set has, oughthecount the numbers from the beginning?Your nice command shows only, if there is an input, which is not anumber.But I think the user would like to know, which of the 200 elements arenotnumbers.The only good of your command is, that it looks nice and shows yourknowledge!>> The question of Mr. Nagesh:>> My 4th Objective:- How can I program the check for correctness ofthe> input values supplied by the package user ?> The answer of Mr. Kuska is:> And @@ (NumericQ /@ {aListOfAllYourNumericParameters})>> My comment:>> This is a nice command and shows the knowledge of Mr.Kuska. But doesMr.> Nagesh understand it and is it sufficient to check, if all inputsare> numerical?>> It seems you have a deeper knowlege about the things that Mr. Nagesh> understand. You know him personally ? It is not very polite to> make speculations *what* a other person understand.>> And no it is not sufficent to check that all parameters are numbers.> Typical paramters described by intervals, where the assumptions of> a model are valid. But for this checks one needs more knowlege> about the meaning of the parameters. And one needs the knowlege about> the differntial equations, to find out the eigenvalues of the jacobian> ...> Additionally I think, it is not userfriendly to see the input merelyasa> set of 200-250 numbers.>> My suggestion is, that JLink is used, a suggestion Mr. Kusk takesinto> consideration, too.>> That will be fast as lightning !> But further I suggest, that classes are defined in Java, whichrepresentthe> parts of the system.>> That is notable nonsense! When the differntial equations should be> solved with Mathematica, the parts can't be Java classes.Mathematica's> NDSolve[] need a explicit expression to integrate the equations.That's not nonsense, the Mathematica program does not fetch the classesbutthe numbers in the classes (or better in the objects).inthe Java classes in textform. They can then be fetched from theMathematicaprogram and transformed into expressions by the Command ToExpression.The aim is, to have a clear separation of the system into components,whichare manageable and understandable.>> OK you can call a Java class member from Mathematica but this will> be incredible slow when the right hand side is evaluated 200-300> times and every evaluation make several callbacks to the Java source.> Event handler of the Java main program (without some modification in> the event loop) while it is evaluating an other command.My idea is to fetch the values once from the Java objects at thebeginning.200-250 numbers is not so much..> Constructors of the classes should build objects with defaultvalues.>> That's a great idea. If a simulation run should be documented, one> always> need the full listing of the Java source to find the actual parameter> settings.That is not my opinion. I think not every user of the simulation systemshould have to know Java and Mathematica.The user must look for the values in the objects. And the values are intheobjects, if they come from the constructor or from the graphical userinterface.I think, there should be listings of the objects including names of thevariables. In the objects the values are in an meaningful environment.>> Graphical user interfaces> should give the opportunity to change the data fields in the objectsand> check the input for correctness.>> *and* what has a GUI for 200-250 variables to do with Mathematica ?> BTW one has typical much less variables because many parameters> are fixed and it make no sense to change, for example, material> constants of materials that can't exchanged> The system should give the opportunity, to store the objects onharddisk> (serialization).>> accessed directly.> Can you be so kind, to explain *how* your posting help a person that> ask How can I build a simulation system with Mathematica ?>> You *can* say Don't use Mathematica, use Java! but this has nothing> to do with the question or with my reply.It is you, who proposes to solve the problem with C/C++ and not to useMathematica (see below!)My point of view is:Use Mathematica, for what Mathemtica is good, and Java, for what Java isgood.Mathematica is not so good as Java for data entry and Java is betterthanMathematica to represent the simulation system (by objects).>> But I still would suggest to use C/C++ and a numerical> ode-solver, make a fancy GUI/Script> interface and don't use Mathematica for such a simple task.> The ode-solver is the smallest part of such a simulation system.>> Jens> My name is Nagesh and pursuing research studies inRefrigeration. At> > present I am writing a Dynamic Refrigeration System SimulationPackage.> I> am using Mathematica as a programming language for the samesincelast> one> > year. I don't have any programming experience before this. Ihave> following> querries:-> 1. Is any body here have expertise or information about thecapability> of> Mathematica as a system simulation tool?>> Since the most system simulation tools are simply solving asystemof> ordinary differntial equations it is simple to do this withNDSolve[].>> 2. Is is possible to program a user friendly interface for mysystem> simulation package with Mathematica or I have to use some other> software?>> Write a MathLink or J/Link frontend that launch the kernel. Butyou> should keep> in mind that the user interface is typical 80-90 % of your code.> If you just whant to solve some ode's it is probably easyer to> include one of the excelent ode-solvers from netlib in your C-code> than to call Mathematica to do that. As long as you dont wish tochange> the ode's very often (than Mathematica is more ßexible) youshould> not use Mathematica.>> 3. My refrigeration system simulation package is likely to have> approximately 60 First order Differential equations. Is ispossibleto> > solve these in Mathematica ?>> Sure.> If yes then can anybody here guide me about> this further.> >> Write down the equations and call NDSolve[].>> >> I am explaining below in short about the objectives I want tofulfill> from> coding out of my main input file>> 1. Example from Main Input File ( this will contain about200-250> variables> which will be entered by the user of this package)>> This sounds like a *very* userfiendly interface ;-)> > Below is examples of two variables entered into this file, whichwill be> used in other analysis files for further evaluation.>> 2. Example from other analysis file ( there will be about 20-25other> such> component analysis files ) where the above mentioned variablesfrommain> input file will be used for further evaluations:->> Below is one example from this file explaining how the variablesfrom> main> input file will be used in other files.>> I hope that this short information will be useful for guiding meto> solve> the following problems that I am facing. I am facing follwingproblems> or> objectives:->> 1. My 1st Objective:- The user of this package must be able tochange> only> the value of the variable in the main input file but he must notbeable> to> change the name of the variable itself. For example he must beableto> change the value of the variable but he must not be able tochange> the> name of this variable itself.> > Here our problem is how to achieve or program it so that ourobjective> will> be fullfilled.>> Options with defaulf values ? or something like>> {aParam,bParam}={ODEParameter1,ODEParameter2} /.> userRules /.> {ODEParameter1->1,ODEParameter2->2}>> >> 2. My 2nd Objective:- How I can program the main input file sothatit> will> be user friendly in terms of its visuals and satisfying theconstraint> mentioned above in objective1.>> What is *userfiendly* in a file with 250 variables ???> 3. My 3rd Objective:- How can I program the optional values foreach> > variable in the main input file ? so that there will be always avalue> assigned to each variable listed in main input file whenever theuser> opens> up this file. If user want to change the values of somevariablesthen> he> can change them and run the simulation otherwise the simulationrunwill> be> > done with optional values assigned to each variable in the inputfile.>> See above.> 4. My 4th Objective:- How can I program the check forcorrectness ofthe> input values supplied by the package user ?> >> And @@ (NumericQ /@ {aListOfAllYourNumericParameters})> Jens> >> ==== hi> AFAIK, Mac OS is now BSD or something like that. That makes it almost> certain that it could support QT. As I pointed out in another post, I can> run the KDE on Windows XP. I haven't been in the trenches with the Qt> impression is, it really is Ôcode once, run everywhere'.i've checked the trolltech hp. qt fully supports the following OS:MS Windows 95/98/Me, NT4, 2000 and XP. Mac OS X. > This is one of the reasons I am such a Mozilla fan. Konqueror works quite> like. But Mozilla runs everywhere with more or less a uniform look and> feel. Yes, Mozilla is written with Gtk and not with Qt, but that just> shows that WRI has options.i'm not sure if it is allowed to create a closed-source product like the mathematica fronend based on a gpl'ed library like gtk. with qt you have the possibility to buy licenses for the commercial version of qt, allowing you to create closed-source apps. > I'm a KDE fan. I've used the KDE since it was in alpha 0.4. I remember> back when it would compile in a few minutes on a pentium II. Now it takes> several hours on a P4. i started with beta1 (if i remember correctly) - sure it is really big now, but i think kde is still far away from being bloated..... imagine the mathematica frontend being seamlessly integrated into the linux ui's look and feel.... > I've always hated motif. The file chooser simply stinks. And that's just a> start.the whole motif ui stinks....:-)gerald************************************* Gerald RothM@th Desktop Development************************************* ==== > hi,> moving the frontend over to QT would have some neat side effects:> consistent look & feel with the modern linux gui, themeability, source> antialiased> truetype fonts as QT supports Xrender and Xft (looks great - see KDE3). i> think all of those points are of value, but the most important might be> source compatibility. ONE frontend for MOST (or ALL) platforms - sounds> like a dream :-))AFAIK, Mac OS is now BSD or something like that. That makes it almost certain that it could support QT. As I pointed out in another post, I can run the KDE on Windows XP. I haven't been in the trenches with the Qt impression is, it really is Ôcode once, run everywhere'.This is one of the reasons I am such a Mozilla fan. Konqueror works quite like. But Mozilla runs everywhere with more or less a uniform look and feel. Yes, Mozilla is written with Gtk and not with Qt, but that just shows that WRI has options. I'm a KDE fan. I've used the KDE since it was in alpha 0.4. I remember back when it would compile in a few minutes on a pentium II. Now it takes several hours on a P4. But if WRI wanted to go the Gtk route, they could achieve the same ends.I've always hated motif. The file chooser simply stinks. And that's just a start.> gerald> STH ==== Can I solve this inequality with Mathematica?Log[x,a]+Log[a x,a]>0I've tried all know, but I get get any good result.CeZaR ==== > First thanks to all, and in particular Bobby Treat, for your help with> this question.> The best solution was as follows:> lst = ReadList[c:data.txt, {Number, Number}]> adjacenceMatrix[> x:{{_, _}..}] := Module[{actors, events},> {actors, events} = Union /@ Transpose[x];> Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & ,> {Length[actors], Length[events]}]]> a = adjacenceMatrix[lst];> b = a . Transpose[a];> c = b (1 - IdentityMatrix[Length[b]])> C is the desired symmetric matrix with off diagonal values of >=0,> indicating the number of times two actors participate in the same event.> The diagonal is set to 0.> A few items in response to Bobby's message, below. While c is, in fact,> a huge matrix with lots of cells equal to zero, that is exactly how we> need it structured for our analysis and research question (not relevant> to the list, but I'd be happy to discuss off list). Processing time is> actually not too bad!! I'm running a PIII 900 with 512 SDRAM, and the> code ran a 177 x 3669 matrix in under 90 seconds. MatrixForm [c]> presented no problems in viewing in the front end, but then it's only> 177 x 177.> Tom> **********************************************> Thomas P. Moliterno> Graduate School of Management> University of California, Irvine> tmoliter@uci.edu> **********************************************> [...]There are several ways to go about this and which is best will varybased on relative number of events vs. number of actors. Below I showthree variations. The first is a minor recoding of the one above. Thesecond iterates over all pairs of actors. The third looks at all eventsfor common actors. I then show three examples. The first two methodshave the advantage that they do not require that events be positiveintegers. With some extra work the third method could also get aroundthis restriction.toAdjacency0[data:{{_, _}..}] := Module[ {actors, events, mat1, mat2}, {actors, events} = Union /@ Transpose[data]; mat1 = Array[If[MemberQ[data, {actors[[#1]], events[[#2]]}], 1, 0] &, {Length[actors], Length[events]}]; mat2 = mat1 . Transpose[mat1]; mat2 * (1-IdentityMatrix[Length[mat2]]) ]toAdjacency1[origdata_] := Module[ {data=Union[origdata], mat}, data = Map[Last, Split[data,#1[[1]]===#2[[1]]&], {2}]; mat = Table [If [j>k, Length[Intersection[data[[j]],data[[k]]]], 0], {j,Length[data]}, {k,Length[data]}]; mat+Transpose[mat] ]toAdjacency2[origdata_] := Module[ {data=Sort[Map[Reverse,Union[origdata]]], mat, len, event}, data = Map[Last, Split[data,#1[[1]]===#2[[1]]&], {2}]; dim = Length[Union[Flatten[data]]]; len = Length[data]; mat = Table[0, {dim}, {dim}]; Do [ event = data[[j]]; Do [ Do [ mat[[event[[m]],event[[k]]]] += 1; mat[[event[[k]],event[[m]]]] += 1, {m,k-1}], {k,Length[event]}], {j,len}]; mat ]data1 = Table[{Random[Integer,{1,1000}], Random[Integer,50]}, {10000}];data2 = Table[{Random[Integer,{1,1000}], Random[Integer,100]}, {10000}];data3 = Table[{Random[Integer,{1,1000}], Random[Integer,200]}, {10000}];Timings are on a 1.5 GHz machine running the Mathematica 4.2 kernelIn[107]:= Timing[m0 = toAdjacency0[data1];]Out[107]= {5.44 Second, Null}In[108]:= Timing[m1 = toAdjacency1[data1];]Out[108]= {10.5 Second, Null}In[109]:= Timing[m2 = toAdjacency2[data1];]Out[109]= {16.24 Second, Null}In[110]:= m0 === m1 === m2Out[110]= TrueNote that for this example the result is not terrible sparse (less than20%).In[112]:= Count[Flatten[m0], 0]Out[112]= 191374In[115]:= Timing[m0 = toAdjacency0[data2];]Out[115]= {11.51 Second, Null}In[116]:= Timing[m1 = toAdjacency1[data2];]Out[116]= {10.92 Second, Null}In[117]:= Timing[m2 = toAdjacency2[data2];]Out[117]= {9.07 Second, Null}Curiously this was the first example I tried, and all three methodsperform about the same in this case. The result, not suprisingly, ismore sparse (40%) because we have the same number of actors and pairs aspreviously, but now with more events to spread out over the pairs.In[118]:= Count[Flatten[m0], 0]Out[118]= 403232When we get sparser still, the third method begins to dominate and thefirst is relatively slower.In[119]:= Timing[m0 = toAdjacency0[data3];]Out[119]= {22.73 Second, Null}In[120]:= Timing[m1 = toAdjacency1[data3];]Out[120]= {10.88 Second, Null}In[121]:= Timing[m2 = toAdjacency2[data3];]Out[121]= {4.96 Second, Null}Now sparsity is over 60%.In[122]:= Count[Flatten[m0], 0]Out[122]= 624350The relative speed of this last method, in this instance, is derivedfrom the fact that individual event lists are on average half the sizeof the previous case. Hence the main loop is expected to improve onaverage by a factor of 2 (you get a factor of 4 for iterating over allpairs in each event, but lose a factor of 2 because there are twice asmany event lists).My guess is that a preprocessor that assesses number of actors vs.number of events would be the best way to choose between the first andthird methods (which, inexplicably, are labelled as methods 0 and 2). Itis not clear to me whether the middle approach will ever dominate. Ihave not given much thought to concocting examples where it wouldbecause offhand I suspect they would be pathological as in dense andwith large intersections.As a last remark I'll note that these might run significantly faster ifcoded with Compile. Whether that is viable depends on the form of thedata. In the above example where everything is a machine integer thatapproach would certainly work.Daniel LichtblauWolfram Research ==== First thanks to all, and in particular Bobby Treat, for your help withthis question.The best solution was as follows:lst = ReadList[c:data.txt, {Number, Number}]adjacenceMatrix[ x:{{_, _}..}] := Module[{actors, events}, {actors, events} = Union /@ Transpose[x]; Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & , {Length[actors], Length[events]}]]a = adjacenceMatrix[lst];b = a . Transpose[a];c = b (1 - IdentityMatrix[Length[b]])C is the desired symmetric matrix with off diagonal values of >=0,indicating the number of times two actors participate in the same event.The diagonal is set to 0. A few items in response to Bobby's message, below. While c is, in fact,a huge matrix with lots of cells equal to zero, that is exactly how weneed it structured for our analysis and research question (not relevantto the list, but I'd be happy to discuss off list). Processing time isactually not too bad!! I'm running a PIII 900 with 512 SDRAM, and thecode ran a 177 x 3669 matrix in under 90 seconds. MatrixForm [c]presented no problems in viewing in the front end, but then it's only177 x 177.Tom********************************************** Thomas P. MoliternoGraduate School of ManagementUniversity of California, Irvinetmoliter@uci.edu************************************ **********-----Original Message-----columns will be numbered from 1 to the number of observed actors orevents, and will correspond to actors and events in sorted order.That said, you're asking for a VERY large matrix, and most of itsentries will be zero. I'll suggest another way, later.The following indicates AT MOST 13.4% of the entries could be non-zero:lst = ReadList[moliterno-test1996.txt, {Number, Number}]; {actors, events} = Union /@ Transpose[lst]; N[Length[lst]/(Length[actors]*Length[actors])] 0.13350994338800987However, a random sample shows that less than 1% will be non-zero:Timing[ Count[(MemberQ[lst, {actors[[Random[Integer, Length[actors]]]], events[[Random[Integer, Length[events]]]]}] & ) /@ Range[10000], True]/ 10000.]{7.515999999999998*Second, 0.008}Nevertheless, the following code should build the matrices you want.I'm using a 2.2GHz P4 and 1024MB RDRAM, so if you have a slower machine,be warned.adjacenceMatrix[ x:{{_, _}..}] := Module[{actors, events}, {actors, events} = Union /@ Transpose[x]; Array[If[MemberQ[x, {actors[[#1]], events[[#2]]}], 1, 0] & , {Length[actors], Length[events]}]]Timing[a = adjacenceMatrix[lst]; ]Dimensions[a]{5.671999999999997*Second, Null}{166, 1778}Timing[b = a . Transpose[a]; ]{0.5309999999999988*Second, Null}You don't want to display a or b in MatrixForm. It will crash youranything at all from the result, and use something likeb[[Range[20],Range[4]]]//MatrixForm{{47, 0, 0, 0}, {0, 3, 0, 0}, {0, 0, 7, 1}, {0, 0, 1, 59}, {0, 0, 0, 3}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 0}, {0, 0, 0, 0}, {2, 0, 0, 1}, {0, 0, 0, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}, {0, 0, 0, 2}, {0, 0, 0, 0}, {0, 0, 0, 0}}to get a glimpse at some of it. It's the Ôa' matrix that's terriblysparse -- the Ôb' matrix isn't unreasonable. Elements of the Ôa' matrixcan be quickly computed by the functionaFunction = If[MemberQ[lst, {actors[[#1]], events[[#2]]}], 1, 0] &;That stores a line of code rather than all those ones and zeroes. The bmatrix (call it bb this time) can be computed as:Timing[bb = (#1 . Transpose[#1] & ) [Array[aFunction, {Length[actors], Length[events]}]]; ]bb == b{6.1569999999999965*Second, Null}TrueBobby Treat-----Original Message-----posting a this was the most helpful solution message to the list, butfirst I hoped to ask you a follow up question, if I may (and I'llcapture your off-line response here in my final posting to the list).I've run the code (copied from you) below, and get the correct outputfor made-up data, but when I import in real data, I get an errormessage. Here's the input I'm running:lst = ReadList[c:test1996.txt, {Number, Number}]AdjacenceMatrix[lst : {{_, _} ..}] := Module[{actors, events, adj}, {actors, events} = Union /@ Transpose[lst]; adj = Table[0, {Length[actors]}, {Length[events]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, events]]; adj]MatrixForm[a = AdjacenceMatrix[lst]]MatrixForm[b = a.Transpose[a]]And here's what I get for output:Set::partw : Part 300007 of <<1>> does not exist.Set::partw : Part 300007 of <<1>> does not exist.Set::partw : Part 300007 of <<1>> does not exist.General::stop : Further output of Set::partw will be suppressed duringthis calculation.Then I get the two matrices (a & b as per your code), but they are justfilled with zeros. So it gets to about the 4th line of your code, butthen doesn't fill-in from my data. Finally, I should note that30007 is one of the actors in the data that I've read in. In caseyou want to run this yourself, I've attached the raw data file. Thereare 166 actors and 1778 events: both actors and events are coded with6-digit numbers, actors begin with 3's, events with 2's.I'm sure this is a silly question, and that there is an easy answer... But I sure can't find it. So I really appreciate your help andinterest!!!!Tom*************************************** *******Thomas P. MoliternoGraduate School of ManagementUniversity of California, Irvinetmoliter@uci.edu************************************ **********-----Original Message-----If you multiply it by its transpose, you get something else that'suseful:lst = {{1, A}, {1, B}, {2, B}, {3, C}, {3, D}, {1, D}, {1, C}};AdjacenceMatrix[lst : {{_, _} ..}] := Module[{actors, events, adj}, {actors, events} = Union /@ Transpose[lst]; adj = Table[0, {Length[actors]}, {Length[events]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, events]]; adj]MatrixForm[a = AdjacenceMatrix[lst]]MatrixForm[b = a.Transpose[a]]Matrix Ôb' records how many events two actors have in common. On thediagonal, it shows the total number of events each actor is connectedto.It's easy to put zeroes on the diagonal:MatrixForm[c = b (1 - IdentityMatrix[Length[b]])]To get the originally intended incidence matrix, this works:d = c /. {_?Positive -> 1}However, I think matrices Ôa' and Ôb' are actually more useful, and Ôa'easily leads to all the others.Bobby-----Original Message-----AdjacenceMatrix[lst : {{_, _} ..}] := Module[ {actors,events adj}, {events, actors} = Union /@ Transpose[lst]; adj = Table[0, {Length[events]}, {Length[actors]}]; Scan[(Part[adj, Sequence @@ #] = 1) &, lst /. MapIndexed[Rule[#1, First[#2]] &, actors]]; adj ]you getIn[]:=AdjacenceMatrix[lst]Out[]={{1, 1, 0, 1}, {0, 1, 0, 0}, {0, 0, 1, 1}} Jens> I need to create an adjacency matrix from my data, which is currentlyin> the form of a .txt file and is basically a two column incidence list.> For example:> 1 A> 1 B> 2 B> 3 C> . .> . .> . .> m n> Where 1 to m represent actors and A to n represent events. My goal isto> have an (m x m) matrix where cell i,j equals 1 if two actors are> incident to the same event (in the sample above, 1 and 2 are both> incident to B) and 0 otherwise (w/ zeros on the diagonal).> I'm new to Mathmatica, and so I'm on the steep part of the learning> curve ... All I've been able to figure out so far is how to get my> incidence list into the program using Import[filename.txt]. But then> what? How do I convert to the adjacency matrix? I've found the> ToAdjacencyMatrix[] command in DiscreteMath`Combinatorica`, but Ican't> seem to get it to work ...> Tom> **********************************************> Thomas P. Moliterno> Graduate School of Management> University of California, Irvine> tmoliter@uci.edu> ********************************************** ==== ====Awk! Legends!Basically, the answer to your question is that the PlotLegend option worksONLY for the Plot command and does not work for other types of plots. Forother types of plots you have to use ShowLegend. And ShowLegend is not allthat easy to use, especially since WRI does not give an example for multiplecurves in the Help.Needs[Graphics`Graphics`]Needs[Graphics`Legend`] {q1[t_], q2[t_], q3[t_]} = {0.1 Exp[-0.02 t], 0.2 Exp[-0.025 t], 0.4 Exp[-0.028 t]};Let's look at your first plot.Plot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> {{AbsoluteThickness[0.5], AbsoluteDashing[{4, 4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1, 8}]}}, AxesLabel -> {Y, X}, PlotLabel -> Title, PlotLegend -> {1, 3, 5}, LegendPosition -> {0.5, 0}, ImageSize -> 500];The legend is almost as big as the plot. It distracts from the realinformation you are trying to convey. Furthermore, the order of the curvesin the legend is the reverse of their order in the plot.The following shows how to put the legend in a LogLogPlot, or other types ofplots. I defined the plot styles independently because they are used inseveral places. I made the legend much smaller and put it in an empty areaof the plot. I also reversed the order of the keys so they would match theorder of the curves in the plot.styles={{AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]},{AbsoluteThickness[1.5]},{ AbsoluteThickness[ 2],AbsoluteDashing[{1,8}]}};ShowLegend[ LogLogPlot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> styles, AxesLabel -> {Y, X}, PlotLabel -> Title, ImageSize -> 500, DisplayFunction -> Identity], {MapThread[{Graphics[{Sequence @@ #1, Line[{{0, 0}, {1, 0}}]}], #2} &, {styles, {1, 3, 5}}] // Reverse, LegendPosition -> {-0.7, -0.4}, LegendSize -> {0.2, 0.3}, LegendShadow -> {0.02, -0.02}, LegendSpacing -> 0.5} ];But why use a legend at all? After all, a legend is nothing but another plotin which you have put labels on the curves. Why not put the labels directlyon the curves in the real plot in the first place?LogLogPlot[{q1[t], q2[t], q3[t]}, {t, 0, 100}, PlotStyle -> styles, AxesLabel -> {Y, X}, PlotLabel -> Title, ImageSize -> 500, Epilog -> MapThread[ Text[SequenceForm[Case , #1], {Log[10, 0.01], Log[10, #2[0.01]]}, {0, -1}] &, {{1, 2, 3}, {q1, q2, q3}}]];In the legend you have keyed the curves to numbers 1, 3 and 5. (Perhaps youjust used these as examples and meant to use something different in the realplots?) But these don't seem to have any obvious relation to your functions.I suppose the reader will have to look at another table or look into thetext of your paper or notebook to find out what 1, 3 and 5 mean. So thereader has to go from the graph to the legend then to the text and thenmentally transfer the meaning of the curve back to the main plot. It is somuch nicer to put the meaning right on the curve if you can.For the most part, legends are just computer junk and not even easy tonicely construct. When the legend urge comes over you - try to resist.David Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/ AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}](*However with LogPlot or LogLogPlot the legend desappear*)LogLogPlot[{q1[t],q2[t],q3[t]}, {t, 0, 100},PlotStyle[Rule]{ {AbsoluteThickness[0.5], AbsoluteDashing[{4,4}]}, AbsoluteThickness[1.5], {AbsoluteThickness[2], AbsoluteDashing[{1,8}]}}, AxesLabel[Rule]{Y, X}, PlotLabel[Rule]Title, PlotLegend[Rule]{1,3,5}, LegendPosition[Rule] {0.5,0}]I have shown a particular case, but I has this problem always with LegendandLogPlot and LogPlotPlot. I will appreciate any help.GuillermoSanchez------------------------------------ ---------This message was sent using Endymion MailMan. ==== Using the Front End as a interface with the kernel I was running some calulations when suddenly pressing Shift+Enter causes the contents of the cell being evaluated to transform to the next text underlined with a red line: NotebookObject[FrontEndObject[LinkObject[dd8,1,1]],8] foollowed by the next messages:An unknown box name (NotebookObject) was sent as the BoxForm for the expression. Check the format rules for the expression.An unknown box name (FrontEndObject) was sent as the BoxForm for the expression. Check the format rules for the expression.An unknown box name (LinkObject) was sent as the BoxForm for the expression. Check the format rules for the expression.An invalid typeset structure was generated: Missing BoxFormData.Any suggestions will be very aprreciated.Cesar ==== I have an odd problem. I need to use and simplify functions that havebeen provided by a piece of software that insists on outputing thefunctional results of a data mining proceedure, using e whenoutputing numbers in scientific notation.I'm having difficultly using Replace, Hold, etc. to correctly evaluatethese types of function formats. For example, y = 5e+5x1+2e-1x2,should be transcribed into 5 10^5 x1 + 0.2 x2.ChuckReply-To: kuska@informatik.uni-leipzig.de ==== str = 5e+5x1+2e-1x2;StringJoin[Characters[str] /. e -> *10^] // ToExpression??Work fine for me.But this type of output is typical generated by a C/FORTRANProgram and you should rewrite the formating rules inthe code that produce this output. Jens> I have an odd problem. I need to use and simplify functions that have> been provided by a piece of software that insists on outputing the> functional results of a data mining proceedure, using e when> outputing numbers in scientific notation.> I'm having difficultly using Replace, Hold, etc. to correctly evaluate> these types of function formats. For example, y = 5e+5x1+2e-1x2,> should be transcribed into 5 10^5 x1 + 0.2 x2.> Chuck ==== RB> I am considering the following integralRB> W[m_,n_]:=Integrate[BesselJ[m, x]*BesselJ[n, x], {x, 0, Infinity}]RB> where m,n are reals >=0. With Mathematica 4.1 I obtain:RB> If[Re[m+n]>-1, -Cos[(m-n)Pi/2]/(2 Pi)*RB> (2 EulerGamma + Log[4] +RB> PolyGamma[0, 1/2(1 + m - n)] +RB> PolyGamma[0, 1/2(1 - m + n)] +RB> 2PolyGamma[0, 1/2(1 + m + n)])RB> Any explanation about the analytical expression will beRB> gratefully accepteed.The expression for W[m_,n_] returned by Mathematica is wrong.To prove, just substitute m = n = 0 which is exactly what you had doneand observe that the output you had hadW[0,0]=-(2 EulerGamma + Log[4] + 4 PolyGamma[0, 1/2])/(2 Pi)= 0.84564was incorrect. The correct answer is 1/2.Mathematica can handle the numeric integration successfullyIn[1] := NIntegrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}, Method -> Oscillatory] (* The warnings are skipped *)Out[1] = 0.5Using NIntegrate[BesselJ[0, x]*BesselJ[0, x], {x, 0, Infinity}]without Method -> Oscillatory is not the optimal choice asthe integrand oscillates fairly rapidly over the integrationregion.RB> I suspect that these integrals are divergent (*).In fact, not exactly.Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}]is equal to 1/2, and Mathematica 4.1 for Microsoft Windows(November 2, 2000) does it correctly, while Mathematica 4.2for Microsoft Windows (February 28, 2002) concocts a strangemixture of a wrong divergence message and the warning thatit cannot check the convergence [should I trust to the secondwarning? or the first?]As a matter of fact,Integrate[BesselJ[1, x]*BesselJ[0, x], {x, 0, Infinity}]converges because the integrand is regular at x=0, bounded overthe whole right semi-axis, and decays as2*Cos[Pi/4 - x]*Cos[(3*Pi)/4 - x]/(Pi*x) + o(1/x)at x -> Infinity .Say, calculateNormal[Series[BesselJ[1, x], {x, Infinity, 1}]] Normal[Series[BesselJ[0, x], {x, Infinity, 1}]] // InputForm->(2*(Cos[Pi/4 - x] - Sin[Pi/4 - x]/(8*x))*(Cos[(3*Pi)/4 - x] +(3*Sin[(3*Pi)/4 - x])/(8*x)))/(Pi*x)then Plot[%,{x,1,10}]and Plot[BesselJ[1,x]*BesselJ[0,x],{x,1,10}]and you could hardly see the difference.Generally, to get to the convergence domain for W in terms ofm and n is easy via the asymtotics of the Bessel functions(use something likeExpand[Normal[Series[BesselJ[m, x], {x, Infinity, 1}]]Normal[Series[BesselJ[n, x], {x, Infinity, 1}]]]then analyze the main term).Best wishes,Vladimir BondarenkoMathematical DirectorSymbolic Testing GroupWeb : http://www.CAS-testing.org/ http://maple.bug-list.org/VER2/ (under tuning) http://maple.bug-list.org/VER3/ (under tuning) http://maple.bug-list.org/VER1/ (under tuning) http://www.beautyriot.com/ (teamwork) http://www.ohaha.com/ (teamwork) Voice: (380)-652-447325 Mon-Fri 9 a.m. - 6 p.m.Mail : 76 Zalesskaya Str, Simferopol, Crimea, Ukraine ==== > inside a program I need to solve this linear equation in terms of p1.> However something odds happens. Sometimes the solution is computed and> sometimes the result is empty [I mean no output...]. Is this a bug of the> solve command or am I doing something wrong? The problem is robust to:> changing name to the variables and other makeups..> DavidThat's the weirdest bug I've seen in weeks. As it happens, it's mine. Atleast the inconsistent behavior, that is. I'll fix it, and maybe alsotry to address the issue of how to handle approximate numbers in testingsubexpressions for zero.I've excised your code and put in place a substantially smaller examplethat I believe is responsible. The table will tend to give erraticresults.zz = (-1.*x^7*(-1. + p - 7.*x^6 + p*x^6 + 6.*x^7)* (7.000000000000002 - 7.000000000000002*x + 14.000000000000004*x^6 - 14.000000000000004*x^7 + 7.000000000000002*x^12 -6.999999999999998*x^13))/ (p^2*(1. + 0.9*x^6)^2*(1. + x^6)^5);One workaround would be to use exact input, say by preprocessing withRationalize.Daniel LichtblauWolfram Research ==== >inside a program I need to solve this linear equation in terms of p1.>However something odds happens. Sometimes the solution is computed and>sometimes the result is empty [I mean no output...]. Is this a bug of the>solve command or am I doing something wrong? The problem is robust to:>changing name to the variables and other makeups..>David>>ps: Sorry for the stupid way in which I copied the command...>>Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 - 14.010000000000002*x^12 -> 18.04*x^18 - 13.06*x^24 -> 5.040000000000001*x^30 - 0.81*x^36) +> x*(7.777777777777779 - 9.074074074074076*x +> 30.333333333333336*x^6 -> 21.51851851851852*x^7 -> 16.333333333333336*x^8 +> 44.33333333333334*x^12 +> 3.188888888888883*x^13 -> 65.68333333333332*x^14 +> 28.777777777777786*x^18 +> 47.937037037037044*x^19 -> 100.10000000000002*x^20 + 7.*x^24 +> 45.6037037037037*x^25 -> 69.53333333333333*x^26 +> 13.299999999999999*x^31 -> 19.833333333333332*x^32 -> 1.0499999999999996*x^38) +> p*(-6 + 8.296296296296296*x -> 28.799999999999997*x^6 +> 32.785185185185185*x^7 +> 9.333333333333336*x^8 -> 55.260000000000005*x^12 +> 49.04777777777776*x^13 +> 38.38333333333334*x^14 -> 52.980000000000004*x^18 +> 34.20518518518518*x^19 +> 60.20000000000001*x^20 -> 25.380000000000003*x^24 +> 11.736296296296294*x^25 +> 43.63333333333334*x^26 - 4.86*x^30 +> 2.8999999999999986*x^31 +> 13.533333333333333*x^32 + 0.81*x^37 +> 1.0499999999999996*x^38)))/(x + 1.9*x^7 +> 0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 +> 6*x^7)*(1.2962962962962965 - 3.111111111111112*x^6>+> 9.333333333333336*x^7 - 10.111111111111114*x^12>+> 22.05*x^13 - 5.703703703703705*x^18 + 17.15*x^19>+> 5.483333333333331*x^25 + 1.0499999999999996*x^31>+> p1*x^5*(7.000000000000002 - 7.000000000000002*x>+> 14.000000000000004*x^6 -> 14.000000000000004*x^7 +> 7.000000000000002*x^12 -> 6.999999999999998*x^13) -> 1.166666666666667*p*x^4*x1 -> 3.500000000000001*p*x^10*x1 -> 1.0500000000000003*p*x^11*x1 -> 3.500000000000001*p*x^16*x1 -> 3.150000000000001*p*x^17*x1 -> 1.166666666666667*p*x^22*x1 -> 3.150000000000001*p*x^23*x1 -> 1.0500000000000003*p*x^29*x1))/((1 + 0.9*x^6)^2*(1>+> x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1]>You might find it more robust (and the results cleaner) if you Simplify the equation prior to using Solve. Such asSolve[eqn // Rationalize // Simplify, p1]However, if you are assigning values to p or x prior to using Solve, there may not be a solution. That is, for whenever the numerator of the expression for p1 would be zero, e.g., p = (-6 x^7 + 7 x^6 +1)/(x^6 + 1).Bob Hanlon ==== > Could someone calculate the number Pi to 67,108,864 (2^26) decimal places> I made the calculation in another program and would like to verify itsDoes it really matter what program is used to verify it? If not,here's the digits (computed with the fastest pi crunching program fora PC):33863220896223409803 ==== >>I believe the complexity is O(n log n), so this should be good enough.Umm ...good enough? I understand the words individually, but thephrase makes no sense to me.Bobby Treat-----Original Message-----> crash the Front End. I was thinking about the fact that I calculated> all those digits and then threw them away. I could save them withSave> or DumpSave, and read them in with Get the next time I wanted any of> them, although the file would be close to 70 MB (if not more). I maydo> that, in fact -- I have plenty of disk space.>> The next step would be to somehow reuse the stored digits if I wanted> MORE digits. But how?>> The Bailey-Borwein-Plouffe Pi algorithm is an avenue of attack, sinceit> can calculate digits far from the decimal point, without calculating> those in between. Unfortunately, it calculates hexadecimal digits in> that way, not decimal digits. (That's true for the version I've seen,> anyway.) Still, I could take the stored digits, convert tohexadecimal,> add more hexadecimal digits with the B-B-P algorithm, and then convert> back to decimal. In both conversions, I'd have to be very cognizantof> how much precision I end up with, but that shouldn't be too difficult.> It might go faster if I store hexadecimal digits, as well as decimal> digits, to eliminate one of those conversions at each increase in the> number of digits.>> The next step would be to set up an application that allowed anyone to> ping for digits across the Internet, and would return them if they're> stored.>> Hasn't someone already done that? It seems as if someone would have.>> Bobby TreatIf you're interested in decimal digits, I don't think the BBP algorithmis theway to go. In order to get the nth decimal digit of Pi you need tocompute theprevious n-1 digits, since base conversion is global, not local. ThealgorithmMathematica uses for computing Pi is quite fast - I believe thecomplexity is O(nlog n), so this should be good enough.David> -----Original Message-----calculation in>> So would it take about the same amont of time for the completeprintout> of digits? Of course it would take a few additional seconds to format> the output...>> Or does Mathematica take alot less time when it truncates the output?> Could you tell me the CPU you used and its speed etc...i amcurious,performance> to> other programs out there.>> I used one processor of a dual 1GH Mac and got the same answer with> the> following speed:>> 4.2 for Mac OS X (June 4, 2002)> oldmax = $MaxPrecision> 6> 1. 10> $MaxPrecision = Infinity> Infinity> With[{n = 2^26}, Timing[> pd = RealDigits[N[Pi, n + 1], 10, 20,> 19 - n]; ]]> {28794.1 Second, Null}> MaxMemoryUsed[]> 512055204> pd> {{3, 3, 8, 6, 3, 2, 2, 0, 8, 9, 6, 2, 2, 3,>> 4, 0, 9, 8, 0, 3}, -67108844}>> Tom Burton ==== I prefer to delete all output and then Copy As>Notebook expression. ItBobby-----Original Message-----andoutput --- I try to use one tab indent for input and two tabs indent foroutput, plus some blank line adjustment.I wonder if anyone has a way of automatically achieving thisreformating.--Allan---------------------Allan HayesMathematica Training and ConsultingLeicester UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice : +44 (0)116 271 4198> Often posters to MathGroup copy and paste in the complete cellexpression,> including the In and Out numbers, when posting to MathGroup.>> I wonder if this is the best method because one can't then just copyoutall> the statements and paste them into a Mathematica notebook. All thestatement> numbers have to be edited out and if there are many statementdefinitions> this is an extended task for any responder. This, of course, decreasesthe> chances for a response. A better method is for the poster to just copyand> paste the CONTENTS of each cell. This is more work for the poster, butit> may pay off in better responses.>> David Park> djmp@earthlink.net> http://home.earthlink.net/~djmp/>> ==== Could someone explain what is going on here, please?In[1]:= a = 77617.; b = 33096.; In[2]:=SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity]; SetPrecision[a, Infinity]; SetPrecision[b, Infinity]; In[4]:=f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b)In[5]:=SetAccuracy[f, Infinity]; SetPrecision[f, Infinity]; In[6]:=fOut[6]=-1.1805916207174113*^21In[7]:=a = 77617; b = 33096; In[8]:=g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + (55/10)*b^8 + a/(2*b)In[9]:=gOut[9]=-(54767/66192)In[10]:=N[%]Out[ 10]=-0.8273960599468214PK ==== Peter,I hope that the following example will help - it is a matter or when thingsevaluate.The a in SetAccuracy[a, Infinity], below, evaluates before SetAccuracy acts,so we getSetAccuracy[2.3, Infinity]. The value of a is not changed. a = 2.3; aa = SetAccuracy[a, Infinity] 2589569785738035/1125899906842624But, a 2.3Whereas aa 2589569785738035/1125899906842624--Allan------------------- --Allan HayesMathematica Training and ConsultingLeicester UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice : +44 (0)116 271 4198> Could someone explain what is going on here, please?>> In[1]:=> a = 77617.; b = 33096.;>> In[2]:=> SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity];> SetPrecision[a, Infinity]; SetPrecision[b, Infinity];>> In[4]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + a/(2*b)>> In[5]:=> SetAccuracy[f, Infinity]; SetPrecision[f, Infinity];>> In[6]:=> f>> Out[6]=> -1.1805916207174113*^21>> In[7]:=> a = 77617; b = 33096;>> In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + (55/10)*b^8+ a/(2*b)>> In[9]:=> g>> Out[9]=> -(54767/66192)>> In[10]:=> N[%]>> Out[10]=> -0.8273960599468214> PK> ==== In receiving notebooks from many different people I have noticed thatbeginners often do not know how to use Text cells and write all of theircomments as Input cells. I have even run across some extremely advancedusers who did not know the easy method for entering Text cells. A goodnotebook is usually a blend of Text cells, Input/Output cells and graphicscells. Text cells are very useful for documenting what you are doing andpassing information to other people. Since many people do not know how touse Text cells, I thought I would write a little explanation for beginnerswho are followers of MathGroup.The very easiest method for entering a Text cell is to put the insertionpoint where you want the new cell to be (at the end of the notebook orbetween two existing cells) and then type Alt-7. Then just start typing andyou will have a Text cell.Alternatively you can use MenuFormatStyleText to start a new Text cell.Often, it is useful to put the ToolBar at the top of the notebook. UseMenuFormatShow ToolBar. The drop-down menu on the ToolBar has the variouskinds of cells available for the current style of the notebook. You canselect Text (or any other style) from there.Some users may hesitate to use Text cells because they want to include amathematical expression in the comments. However, that is also very easy.Just use an Inline cell within the text cell. At the point within the textcell where you want to include a mathematical expression, start an Inlinecell by typing Ctrl-(. A selection placeholder will appear on a pinkbackground. You can type a Mathematica expression there just as in an Inputcell. Use Ctrl-) to complete the Inline cell, or Shift-Space. You can evenselect an Inline cell and evaluate it with Shift-Ctrl-Enter.Putting comments in Text cells is far better than using Input cells (or cellgroup header cells). Mathematica won't try to evaluate Text cells, the textwill wrap properly and adjust better to the notebook width if you change it.You can also check the spelling of words by putting the cursor after a wordand using Ctrl-K. (In an Input cell Mathematica doesn't use the dictionary,but uses the table of symbols instead and hence it won't check spelling.)David Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/=== =This is because inline cells are not in StandardForm by default, but TraditionalForm.Use the menu item Cell -> Default Inline Format Type -> StandardForm.>David Park's posting reminded me of a frequent annoyance when I am>trying to include some Mathematica expressions within text cells -- a>Mathematica input expression in Standard Form that involves use of a>Control-key combination to form a superscript, square-root, or built-up>fraction:>>For example, suppose I want to include within a text cell a Standard>Form expression for the square of x, with the exponent 2 raised. If I>type the x first, even if I immediately highlight it and change it to>Courier (to match the default font for Input cells in Standard Form), as>soon as I press the Control-^ key combination, an Inline cell is created>beginning with the x, and then when I type the exponent 2 everything in>that Inline cell is now in Times, and the x is Italic. To change both>characters to Courier is not so easy: it seems to require separately>the entire Inline cell and selecting Courier does not change the exponent!)>>So to avoid this annoyance I normally must first type the desired>expression in a separate Input cell, then copy the contents of that cell>to the desired point in the Text cell.>>Any suggestions on a more efficient method for handling this?> In receiving notebooks from many different people I have noticed that> beginners often do not know how to use Text cells ....>> Some users may hesitate to use Text cells because they want to include a> mathematical expression in the comments....> Just use an Inline cell within the text cell....>>-->Murray Eisenberg murray@math.umass.edu>Mathematics & Statistics Dept.>Lederle Graduate Research Tower phone 413 549-1020 (H)>University of Massachusetts 413 545-2859 (W)>710 North Pleasant Street>Amherst, MA 01375-------------------------------------------------------- ------Omega ConsultingThe final answer to your Mathematica needsSpend less time searching and more time finding.http://www.wz.com/internet/Mathematica.htmlReply-To : murray@math.umass.edu ==== David Park's posting reminded me of a frequent annoyance when I am trying to include some Mathematica expressions within text cells -- a Mathematica input expression in Standard Form that involves use of a Control-key combination to form a superscript, square-root, or built-up fraction:For example, suppose I want to include within a text cell a Standard Form expression for the square of x, with the exponent 2 raised. If I type the x first, even if I immediately highlight it and change it to Courier (to match the default font for Input cells in Standard Form), as soon as I press the Control-^ key combination, an Inline cell is created beginning with the x, and then when I type the exponent 2 everything in that Inline cell is now in Times, and the x is Italic. To change both characters to Courier is not so easy: it seems to require separately the entire Inline cell and selecting Courier does not change the exponent!)So to avoid this annoyance I normally must first type the desired expression in a separate Input cell, then copy the contents of that cell to the desired point in the Text cell.Any suggestions on a more efficient method for handling this?> In receiving notebooks from many different people I have noticed that> beginners often do not know how to use Text cells ....> Some users may hesitate to use Text cells because they want to include a> mathematical expression in the comments....> Just use an Inline cell within the text cell....-- Murray Eisenberg murray@math.umass.eduMathematics & Statistics Dept.Lederle Graduate Research Tower phone 413 549-1020 (H)University of Massachusetts 413 545-2859 (W)710 North Pleasant StreetAmherst, MA 01375 ==== correctly, all you need to do is to make sure the default inline cell formatis StandardForm. Go to the menu item Cell, select Default Inline FormatType,and change it to StandardForm.Carl WollPhysics DeptU of Washington----- Original Message -----> type the x first, even if I immediately highlight it and change it to> Courier (to match the default font for Input cells in Standard Form), as> soon as I press the Control-^ key combination, an Inline cell is created> beginning with the x, and then when I type the exponent 2 everything in> that Inline cell is now in Times, and the x is Italic. To change both> characters to Courier is not so easy: it seems to require separately> the entire Inline cell and selecting Courier does not change theexponent!)>> So to avoid this annoyance I normally must first type the desired> expression in a separate Input cell, then copy the contents of that cell> to the desired point in the Text cell.>> Any suggestions on a more efficient method for handling this?> In receiving notebooks from many different people I have noticed that> beginners often do not know how to use Text cells ....>> Some users may hesitate to use Text cells because they want to include a> mathematical expression in the comments....> Just use an Inline cell within the text cell....>> --> Murray Eisenberg murray@math.umass.edu> Mathematics & Statistics Dept.> Lederle Graduate Research Tower phone 413 549-1020 (H)> University of Massachusetts 413 545-2859 (W)> 710 North Pleasant Street> Amherst, MA 01375>>Reply-To: kuska@informatik.uni-leipzig.de ==== just one comment: the meaning of the Alt-7 key depend on the style sheet that is in use.The TMJ style use Alt-8 for text and one hasto learn new key short-cuts for every style sheet ! Jens> In receiving notebooks from many different people I have noticed that> beginners often do not know how to use Text cells and write all of their> comments as Input cells. I have even run across some extremely advanced> users who did not know the easy method for entering Text cells. A good> notebook is usually a blend of Text cells, Input/Output cells and graphics> cells. Text cells are very useful for documenting what you are doing and> passing information to other people. Since many people do not know how to> use Text cells, I thought I would write a little explanation for beginners> who are followers of MathGroup.> > The very easiest method for entering a Text cell is to put the insertion> point where you want the new cell to be (at the end of the notebook or> between two existing cells) and then type Alt-7. Then just start typing and> you will have a Text cell.> Alternatively you can use MenuFormatStyleText to start a new Text cell.> Often, it is useful to put the ToolBar at the top of the notebook. Use> MenuFormatShow ToolBar. The drop-down menu on the ToolBar has the various> kinds of cells available for the current style of the notebook. You can> select Text (or any other style) from there.> Some users may hesitate to use Text cells because they want to include a> mathematical expression in the comments. However, that is also very easy.> Just use an Inline cell within the text cell. At the point within the text> cell where you want to include a mathematical expression, start an Inline> cell by typing Ctrl-(. A selection placeholder will appear on a pink> background. You can type a Mathematica expression there just as in an Input> cell. Use Ctrl-) to complete the Inline cell, or Shift-Space. You can even> select an Inline cell and evaluate it with Shift-Ctrl-Enter.> Putting comments in Text cells is far better than using Input cells (or cell> group header cells). Mathematica won't try to evaluate Text cells, the text> will wrap properly and adjust better to the notebook width if you change it.> You can also check the spelling of words by putting the cursor after a word> and using Ctrl-K. (In an Input cell Mathematica doesn't use the dictionary,> but uses the table of symbols instead and hence it won't check spelling.)> David Park> djmp@earthlink.net> http://home.earthlink.net/~djmp/ ==== Solve[youre equation, p1, VerifySolutions->True] will return a solution. So willSolve[Rationalize[your equation],p1].Andrzej KozlowskiToyama International UniversityJAPAN>> inside a program I need to solve this linear equation in terms of p1.> However something odds happens. Sometimes the solution is computed and> sometimes the result is empty [I mean no output...]. Is this a bug of > the> solve command or am I doing something wrong? The problem is robust to:> changing name to the variables and other makeups..> David>> ps: Sorry for the stupid way in which I copied the command...>> Solve[(x^2*((-0.9*x^7*(p^2*(-1 - 5.8*x^6 - 14.010000000000002*x^12 -> 18.04*x^18 - 13.06*x^24 -> 5.040000000000001*x^30 - 0.81*x^36) +> x*(7.777777777777779 - 9.074074074074076*x +> 30.333333333333336*x^6 -> 21.51851851851852*x^7 -> 16.333333333333336*x^8 +> 44.33333333333334*x^12 +> 3.188888888888883*x^13 -> 65.68333333333332*x^14 +> 28.777777777777786*x^18 +> 47.937037037037044*x^19 -> 100.10000000000002*x^20 + 7.*x^24 +> 45.6037037037037*x^25 -> 69.53333333333333*x^26 +> 13.299999999999999*x^31 -> 19.833333333333332*x^32 -> 1.0499999999999996*x^38) +> p*(-6 + 8.296296296296296*x -> 28.799999999999997*x^6 +> 32.785185185185185*x^7 +> 9.333333333333336*x^8 -> 55.260000000000005*x^12 +> 49.04777777777776*x^13 +> 38.38333333333334*x^14 -> 52.980000000000004*x^18 +> 34.20518518518518*x^19 +> 60.20000000000001*x^20 -> 25.380000000000003*x^24 +> 11.736296296296294*x^25 +> 43.63333333333334*x^26 - 4.86*x^30 +> 2.8999999999999986*x^31 +> 13.533333333333333*x^32 + 0.81*x^37 +> 1.0499999999999996*x^38)))/(x + 1.9*x^7 +> 0.9*x^13)^2 - ((-1 + p - 7*x^6 + p*x^6 +> 6*x^7)*(1.2962962962962965 - > 3.111111111111112*x^6 +> 9.333333333333336*x^7 - > 10.111111111111114*x^12 +> 22.05*x^13 - 5.703703703703705*x^18 + > 17.15*x^19 +> 5.483333333333331*x^25 + > 1.0499999999999996*x^31 +> p1*x^5*(7.000000000000002 - > 7.000000000000002*x +> 14.000000000000004*x^6 -> 14.000000000000004*x^7 +> 7.000000000000002*x^12 -> 6.999999999999998*x^13) -> 1.166666666666667*p*x^4*x1 -> 3.500000000000001*p*x^10*x1 -> 1.0500000000000003*p*x^11*x1 -> 3.500000000000001*p*x^16*x1 -> 3.150000000000001*p*x^17*x1 -> 1.166666666666667*p*x^22*x1 -> 3.150000000000001*p*x^23*x1 -> 1.0500000000000003*p*x^29*x1))/((1 + > 0.9*x^6)^2*(1 +> x^6)^2)))/(p^2*(1 + x^6)^3) == 0, p1]>Reply-To: murray@math.umass.edu ==== For all names, perhaps: Names[*`*]For names you defined at a normal session (without changing to some other context than the default Global`): Names[Global`*]> IIRC, there is a way to get a list of all the symbols defined in the > currently running session. I can't seem to find the reference to that > command. Could somone point me in the direction of documentation which > will tell me how to get information about the current session?> TIA,> -- Murray Eisenberg murray@math.umass.eduMathematics & Statistics Dept.Lederle Graduate Research Tower phone 413 549-1020 (H)University of Massachusetts 413 545-2859 (W)710 North Pleasant StreetAmherst, MA 01375 ==== > For all names, perhaps:> Names[*`*]> For names you defined at a normal session (without changing to some> other context than the default Global`):> Names[Global`*]>> IIRC, there is a way to get a list of all the symbols defined in the>> currently running session. I can't seem to find the reference to that>> command. Could somone point me in the direction of documentation which>> will tell me how to get information about the current session?>> TIA,>> I think I asked the wrong question. I'll have to look at things some more. What you gave me resulted in far more than I was looking for.http://public.globalsymmetry.com/proprietary/com/wri/ system-symbols.htmlhttp://66.92.149.152/proprietary/com/wri /system-symbols.htmlI think I really hosed the code for generating that table. I used 5 lines. I probably didn't need more than two, but I'm too tired right now to think about it. Mathematica is totaly awesome when it comes to what it was intended for. They really need to rent a Troll for a few months and fix this interface. Qt will run on just about anything. Heck, my Win-XP partition runs XFree86, with the KDE, or it did when I booted into XP a month ago.STH ==== ?*does the trick. You can limit it to Global variables with?Global`*Bobby-----Original Message-----Sender: steve@smc.vnet.netApproved: Steven M. Christensen , Moderator ==== > Has anyone had any luck getting Mathematica 4.2 to work> on a Sun running Solaris 6?> I get the following error message when starting Mathematica:> ./Mathematica> ld.so.1: ./Mathematica: fatal: librt.so.1: open failed: No such file or> directory> Killed> But math works.> Any thoughts on this would be appreciated - even if it's just> Solaris 6 sucks - Reinstall with Solaris 8/9. I'd rather not go to the> trouble of reinstalling the box unless absolutely necessary (It is a> 270mhz Ultra 5)I'd be tempted to run Ôldd' and see what libraries it needs. On my Solaris 9box, with Mathematica 4.2 I see:wren /usr/local/stow/Mathematica-4.2/SystemFiles/FrontEnd/Binaries /Solaris % ldd../Mathematica libXt.so.4 => /usr/lib/libXt.so.4 libXext.so.0 => /usr/lib/libXext.so.0 libXmu.so.4 => /usr/lib/libXmu.so.4 libX11.so.4 => /usr/lib/libX11.so.4 libnsl.so.1 => /usr/lib/libnsl.so.1 libsocket.so.1 => /usr/lib/libsocket.so.1 libc.so.1 => /usr/lib/libc.so.1 libucb.so.1 => /usr/ucblib/libucb.so.1 librt.so.1 => /usr/lib/librt.so.1 libpthread.so.1 => /usr/lib/libpthread.so.1 libSM.so.6 => /usr/lib/libSM.so.6 libICE.so.6 => /usr/lib/libICE.so.6 libm.so.1 => /usr/lib/libm.so.1 libdl.so.1 => /usr/lib/libdl.so.1 libmp.so.2 => /usr/lib/libmp.so.2 libelf.so.1 => /usr/lib/libelf.so.1 libaio.so.1 => /usr/lib/libaio.so.1 libmd5.so.1 => /usr/lib/libmd5.so.1 /usr/platform/SUNW,Ultra-60/lib/libc_psr.so.1 libthread.so.1 => /usr/lib/libthread.so.1 /usr/platform/SUNW,Ultra-60/lib/libmd5_psr.so.1You should be able to find the libraries it's using, and hopefully it might justbe looking in the wrong place, in which case you may be able to create asymbolic link. However, if Solaris 2.6 is not supported (I don't know if it is) it is quitepossible it wants a library you don't have, in which case you are stuck unlessyou re-install a later OS. -- Dr. David Kirkby,Senior Research Fellow,Department of Medical Physics,University College London,11-20 Capper St, London, WC1E 6JA.Internal telephone: ext 46408 ==== Well, first of of all, your using SetAccuracy and SetPrecision does nothing at all here, since they do not change the value of a or b. You should use a = SetAccuracy[a, Infinity] etc. But even then you won't get the same answer as when you use exact numbers because of the way you evaluate f. Here is the order of evaluation that will give you the same answer, and should explain what is going on:f = SetAccuracy[333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121* b^4 - 2) + 5.5*b^8 + a/(2*b), Infinity];a = 77617.; b = 33096.;a = SetAccuracy[a, Infinity]; b = SetAccuracy[b, Infinity];f 54767-(-----) 66192Andrzej KozlowskiToyama International UniversityJAPAN> Could someone explain what is going on here, please?>> In[1]:=> a = 77617.; b = 33096.;>> In[2]:=> SetAccuracy[a, Infinity]; SetAccuracy[b, Infinity];> SetPrecision[a, Infinity]; SetPrecision[b, Infinity];>> In[4]:=> f := 333.75*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + 5.5*b^8 + > a/(2*b)>> In[5]:=> SetAccuracy[f, Infinity]; SetPrecision[f, Infinity];>> In[6]:=> f>> Out[6]=> -1.1805916207174113*^21>> In[7]:=> a = 77617; b = 33096;>> In[8]:=> g := (33375/100)*b^6 + a^2*(11*a^2*b^2 - b^6 - 121*b^4 - 2) + > (55/10)*b^8 + a/(2*b)>> In[9]:=> g>> Out[9]=> -(54767/66192)>> In[10]:=> N[%]>> Out[10]=> -0.8273960599468214> PK> ==== Andrzej, Bobby, PeterIt looks as if using SetAccuracy succeeds here because the inexact numbersthat occur have finite binary representations. If we change them slightly toavoid this then we have to use Rationalize:1) Using SetAccuracy Clear[a,b,f] f=SetAccuracy[333.74*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+5.4* b^8+a/(2*b), Infinity]; a=77617.1; b=33096.1; a=SetAccuracy[a,Infinity];b=SetAccuracy[b,Infinity ]; f - 15640321149084868351974949239896188679725401538739519428131155 149493891236234 52500771916869370459119776018798804630436149786919912931962574 3010292363124675/ 10867106143970760551000357827554793888198143135975649579607989 867743572824016 06539536129829321813712324363677397376040962) Rewriting as fractions a=776171/10; b=330961/10; f=33374/100*b^6+a^2*(11*a^2*b^2-b^6-121*b^4-2)+54/10*b^8+a/(2 *b) -(5954133808997234115690303589909929091649391296257/ 41370125000000)3) Using Rationalize Clear[a,b,f]f=Rationalize[333.74*b^6+a^2*(11*a^2*b^2-b^6-121 *b^4-2)+5.4*b^8+a/(2*b),0]; a=77617.1; b=33096.1; a=Rationalize[a,0];b=Rationalize[b,0]; f -(5954133808997234115690303589909929091649391296257/ 41370125000000)I use Rationalize[. , 0] besause of results like Rationalize[3.1415959] 3.1416 Rationalize[3.1415959,0] 31415959/10000000--Allan---------------------Allan HayesMathematica Training and ConsultingLeicester UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice : +44 (0)116 271 4198> Well, first of of all, your using SetAccuracy a