A33
==
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]:=fOut[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]=-20Andrzej> 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? -->
---------------------> 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 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 > 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 />
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> --> ---------------------> 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>
PKAndrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
=So? It's just as it should
be, isn't it?AndrzejAndrzej KozlowskiToyama International
UniversityJAPANhttp://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? -->
---------------------> 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 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 > 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 /
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> --> ---------------------> 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/
====
=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? -->
---------------------> 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 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> 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> --> ---------------------> 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: Ô 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? -->
---------------------> 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 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> 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 /
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> --> ---------------------> 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: Ô 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? -->
---------------------> Hayes> Mathematica Training and
Consulting> Leicester UK> www.haystack.demon.co.uk>
hay@haystack.demon.co.uk> Voice: +44 (0)116 271 4198message>
It seems clear to me that what 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> 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 /
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> --> ---------------------> 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<<Statistics`ContinuousDistributions`;>
<<Statistics`DataManipulation`;><<Graphics`Graphics`;><<
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 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 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 > 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
====
=This just means that Mathematica can't do it.
This looks difficult, nonlinear andall that.Kevin> Folks,>
What does it mean when I submit a differential equation, and
the same> equation appears as the output?> I have tried this
equation:> DSolve[y'[x] == (-x^2 + 2 y[x]^(-3))/(2 x y[x] - 
3
x^2), y[x], x]> Is it too complicated for the program?> Diana
Try using NDSolve and put in some BCs, it may not be
solvable> analytically, but should be solvable
numerically.>
====
=Helo,I watched Steve Jobs' keynote from
maxworld expo 02', where Theo Graypresent Mathematica, and
makes a joke - a funny integral of some sort - thatI din't
get it, because the video resolution was too poor.I found the
link on this page http://theodoregray.com/Can somebody, who
knows the idea, explain it quickly ... ?Borut
====
=Stan,I think
it is a shame that any engineering college would frown upon
usingMathematica. If used correctly, it can only speed
learning and give you alot more practice. It is also a shame
that students can't come to collegealready knowing the 
basics
of Mathematica so they could concentrate on thematerial to be
learned. The argument that the student is supposed to
belearning methods and not use Mathematica's advanced
automatic routines: onecan always skip the advanced routines
and program Mathematica to do thesteps of the method to be
learned.Anyway, on to your question. You could do something
like this...xsol = Solve[2x - 1 == 0, x][[1, 1]]answer = x /.
xsol // Nwhich returnsx -> 1/20.5answer0.5Look up ReplaceAll
and Rule in Help. What I like to do is save xsol as arule and
not store the x value in some variable answer, or in x itself.
Iwould also tend to stay with exact solutions as long as
possible. Then youcan use xsol and N something like
this...x^2 + Sin[x]% /. xsol% // Nwhich givesx^2 + Sin[x]1/4
+ Sin[1/2]0.729426Remember that % means the previous output
(in time) and when you use it forstatements within one cell
it is completely unambiguous. /. is the shortcutnotation for
ReplaceAll. So % /. xsol means to take the previous output,
andreplace all occurrences of x using the rule xsol (which is
x -> 1/2).David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/How
would I then use the result of Solve (0.5) and assign it to
anew variable, like answer for further calculations?I think
it has something to do with /. and -> rules of some sort,
butI can't get this simple concept to work!-Stan
====
=OK, here
is what it does for me....I am using Mathematica 4.2.1 with
Macintosh OS X 10.1.5....If i use the shoft key, and I select
a cell, then select another cellwhile skipping some cells, it
still selects all the cells between the 2that i have
selected....I then copy those and I paste them into a
newnotebook. Even though it selected some cells that i did
not wantselected, it does still paste the individual cells
into the newnotebook.However, if I use the command key
instead of the shirt key, I canselect individual cells to
copy with out it selecting the cellsinbetween, meaning i can
only selct the particular cells i want toselect, and they
still paste as individual cells.not sure if your using the
command key will enable you to paste intoseperate cells but
its worth a shot. try the Command key instead of theshirt
key...> I am having an extremely frustrating time with one
feature of > Mathematica 4.2 which I swear was not present in
earlier versions. It > find is quite dreadful and yet none
seems to have complained about it > so far so that I feel
that perhaps I missing something obvious. > Anyway, this is
the problem. I have two notebooks with a large number > of
cells each. I want to copy some cells from one notebook to
the > other. I am sure in earlier versions it used to be
possible to select > several contiguous cells by shift
clicking and then paste them into > another notebook. You
would then get several cells of the same type as > you
copied. However now (at least on Mac OS X) this seems
impossible. > What happens is that if you select a several
cells or a cell group and > past into another notebook you
get a messy expression contained in a > single cell from
which it does not appear possible to easily recover > the
original cells. I have tired using various Copy As but have
not > found anything that works. The only thing that works is
copying and > pasting individual cells, which is of course
very time consuming.> Is this also what happens on other
platforms? I am sure it was not the > case in earlier
versions, and if so was this an intentional change? It > so
it seems to me as bad an idea as I have come across in many
years. > Can anyone explain?Andrzej Kozlowski>
====
=I can
report I do have this problem with Mathematica 4.2 in Mac OS
X 10.2.2.I can select contiguous cells using shift and not
contiguous cells (using command) but copying them results in
one huge cell as you reported.I think is time for a Tech
Support reportCiao,aov> As a follow up on this issue:> I
reinstalled Mathematica 4.1 for Mac OS X and found that I
still have> this problem. On the other hand Hayes informed me
that he did not> have it on Mathematica 4.2 under Windows 98.
I then tried it myself,> with Mathematica 4.1 under Windows
98, and indeed I found I can copy> and paste groups of cells
and the cell structure is preserved. They do> not get merged
into a single cell with all the Mathematica markup> visible
as they do in my case under Mac OS X 10.2. I would like to
know> if other Mac OS X users have the same problem. If not,
it may be caused> by something weird in my installation,
although it seems unlikely as I> am not experiencing any
other problems. I have not yet tried reporting> this to
Technical Support yet since I am not sure if other Mac OS X>
users are also experiencing the same behaviour, but if it is
a bug it's> an awful one.> Andrzej> I am having an extremely
frustrating time with one feature of> Mathematica 4.2 which I
swear was not present in earlier versions. It> find is quite
dreadful and yet none seems to have complained about it> so
far so that I feel that perhaps I missing something obvious.>
Anyway, this is the problem. I have two notebooks with a large
number> of cells each. I want to copy some cells from one
notebook to the> other. I am sure in earlier versions it used
to be possible to select> several contiguous cells by shift
clicking and then paste them into> another notebook. You
would then get several cells of the same type as> you copied.
However now (at least on Mac OS X) this seems impossible.>
What happens is that if you select a several cells or a cell
group and> past into another notebook you get a messy
expression contained in a> single cell from which it does not
appear possible to easily recover> the original cells. I have
tired using various Copy As but have not> found anything that
works. The only thing that works is copying and> pasting
individual cells, which is of course very time consuming.> Is
this also what happens on other platforms? I am sure it was
not the> case in earlier versions, and if so was this an
intentional change? It> so it seems to me as bad an idea as I
have come across in many years.> Can anyone explain? Andrzej
Kozlowski>----Unix is user friendly - it's just picky about
it's friends.----Alfredo
Octavio,Patilla.comhttp://alfredo.octavio.net At: N
10¢È23.811', W 
066¢È58.953'----
====
=Your first problem I don't
understand. As for the second, this may give youa taste of
Mathematica's
powers:In[1]:=<<DiscreteMath`Combinatorica`In[2]:=KSubsets[{a
, b, c}, 2] /. {x_, y_} -> {x -> 0, y -> 0}Out[2]={{a -> 0, b
-> 0}, {a -> 0, c -> 0}, {b -> 0, c -> 0}}Tomas GarzaMexico
City----- Original Message -----> b = d and c = d = 1> 2)
from a list, say {a,b,c}, I would like to generate>
{{b->0,c->0},{a->0,c->0},{a->0,b->0}> I guess I have
understood the iterative way of doing that,> but what about a
more functionqal form? Francois --> Francois Lauze> The IT
University of Copenhagen, Glentevej 67> 2400 Kbh NV,
Denmark>
====
=Francois,If c = d = 1, then b = 1 also. So just
add the definition...f[f[a_, 1, 1], 1, e_] := f[f[a, b, 1/2],
b, 1/2]For your list problem, use
Combinatorica.Needs[DiscreteMath`Combinatorica`]?
KSubsetsKSubsets[l, k] gives all subsets of set l containing
exactly k elements, ordered lexicographically.KSubsets[{a, b,
c}, 2]Thread[# -> {0, 0}] & /@ %gives{{a,b},{a,c},{b,c}}{{a ->
0, b -> 0}, {a -> 0, c -> 0}, {b -> 0, c -> 0}}The last
statement works this way...{a,b} -> {0,0}Thread[%]{a, b} ->
{0, 0}{a -> 0, b -> 0}We then just map that function onto
each of the three sets.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/I guess
I have understood the iterative way of doing that,but what
about a more functionqal form?Francois--Francois LauzeThe IT
University of Copenhagen, Glentevej 672400 Kbh NV,
DenmarkReply-To: Mark Coleman
<mark@markscoleman.com>
====
=Greetings,single-image stereograms
in Mathematica (TMJ, Vol 5, #1). Is anyone aware of an update
to this work, or any additional Mathematica resources on
stereograms?-Mark
====
=MacOS 10.2. The cell structure is copied
and all is exactly the same.Ihave however on occasion lost my
mouse pointer when Mathematica is openand the display goes to
sleep and then wakes again. It doesn't happen withany other
program. I suspect its a combination of both MacOS 10.2
andMathematica because things such as mouse pointer probelms
also occur onRehardsYas> As a follow up on this issue:> I
reinstalled Mathematica 4.1 for Mac OS X and found that I
still have> this problem. On the other hand Hayes informed me
that he did not> have it on Mathematica 4.2 under Windows 98.
I then tried it myself,> with Mathematica 4.1 under Windows
98, and indeed I found I can copy> and paste groups of cells
and the cell structure is preserved. They do> not get merged
into a single cell with all the Mathematica markup> visible
as they do in my case under Mac OS X 10.2. I would like to
know> if other Mac OS X users have the same problem. If not,
it may be caused> by something weird in my installation,
although it seems unlikely as I> am not experiencing any
other problems. I have not yet tried reporting> this to
Technical Support yet since I am not sure if other Mac OS X>
users are also experiencing the same behaviour, but if it is
a bug it's> an awful one.> Andrzej> I am having an extremely
frustrating time with one feature of> Mathematica 4.2 which I
swear was not present in earlier versions. It> find is quite
dreadful and yet none seems to have complained about it> so
far so that I feel that perhaps I missing something obvious.>
Anyway, this is the problem. I have two notebooks with a large
number> of cells each. I want to copy some cells from one
notebook to the> other. I am sure in earlier versions it used
to be possible to select> several contiguous cells by shift
clicking and then paste them into> another notebook. You
would then get several cells of the same type as> you copied.
However now (at least on Mac OS X) this seems impossible.>
What happens is that if you select a several cells or a cell
group and> past into another notebook you get a messy
expression contained in a> single cell from which it does not
appear possible to easily recover> the original cells. I have
tired using various Copy As but have not> found anything that
works. The only thing that works is copying and> pasting
individual cells, which is of course very time consuming.> Is
this also what happens on other platforms? I am sure it was
not the> case in earlier versions, and if so was this an
intentional change? It> so it seems to me as bad an idea as I
have come across in many years.> Can anyone explain?> Andrzej
Kozlowski> 
====
=Are you really sure of this? The reason why I
am asking is because not only another user has confirmed the
problem but the Wolfram technical support has now
acknowledged this as a well known bug. The only possibility I
can see is that your version of 4.2 is actually newer than
mine. Could you check this? Mine is
4.2.0.0.Also:In[1]:=Out[1]=4.2 for Mac OS X (June 4,
2002)Andrzej> MacOS 10.2. The cell structure is copied and
all is exactly the same.I> have however on occasion lost my
mouse pointer when Mathematica is open> and the display goes
to sleep and then wakes again. It doesn't happen > with> any
other program. I suspect its a combination of both MacOS 10.2
and> Mathematica because things such as mouse pointer probelms
also occur on Rehards> Yas As a follow up on this issue:> I
reinstalled Mathematica 4.1 for Mac OS X and found that I
still have> this problem. On the other hand Hayes informed me
that he did > not> have it on Mathematica 4.2 under Windows
98. I then tried it myself,> with Mathematica 4.1 under
Windows 98, and indeed I found I can copy> and paste groups
of cells and the cell structure is preserved. They do> not
get merged into a single cell with all the Mathematica
markup> visible as they do in my case under Mac OS X 10.2. I
would like to > know> if other Mac OS X users have the same
problem. If not, it may be > caused> by something weird in my
installation, although it seems unlikely as I> am not
experiencing any other problems. I have not yet tried
reporting> this to Technical Support yet since I am not sure
if other Mac OS X> users are also experiencing the same
behaviour, but if it is a bug > it's> an awful one.>
Andrzej>> I am having an extremely frustrating time with one
feature of>> Mathematica 4.2 which I swear was not present in
earlier versions. It>> find is quite dreadful and yet none
seems to have complained about it>> so far so that I feel
that perhaps I missing something obvious.>> Anyway, this is
the problem. I have two notebooks with a large number>> of
cells each. I want to copy some cells from one notebook to
the>> other. I am sure in earlier versions it used to be
possible to select>> several contiguous cells by shift
clicking and then paste them into>> another notebook. You
would then get several cells of the same type >> as>> you
copied. However now (at least on Mac OS X) this seems
impossible.>> What happens is that if you select a several
cells or a cell group >> and>> past into another notebook you
get a messy expression contained in a>> single cell from which
it does not appear possible to easily recover>> the original
cells. I have tired using various Copy As but have not>>
found anything that works. The only thing that works is
copying and>> pasting individual cells, which is of course
very time consuming.>> Is this also what happens on other
platforms? I am sure it was not >> the>> case in earlier
versions, and if so was this an intentional change? >> It>>
so it seems to me as bad an idea as I have come across in
many years.>> Can anyone explain?>> Andrzej
Kozlowski>>Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/
====
=...What happens is that if
you select a several cells or a cell group andpast into
another notebook you get a messy expression contained in
asingle cell from which it does not appear possible to easily
recoverthe original cells...What a coincidence! Or maybe not
:-(I logged a telephone support call to WRI on this very
matter the day after Iupdated to OS10.2.2. Is that OS upgrade
involved? Hard to tell, because thesymptom appeared the day
AFTER. Did I go a whole day without copying cells?In my case,
the symptom extended to copying even one entire cell, to
thesame or a different notebook. It stopped me in my
tracks.At WRI's suggestion, I cleared the Front End cache. 
No
help.Then I uninstalled Mathematica 4.2 (on Mac, simply drag
the application tothe trash), restarted, and then reinstalled
Mathematica 4.2 and pasted myinit files back in. I was up to
full speed in 5 minutes (40 minutes afterthe first symptom.)
The symptom is gone and has not returned.As I told WRI, I
don't understand, but I am happy that that AWFUL copy bugis
gone!Tom Burton
====
=I am about to describe one of the
strangest experiences with a software bug that I have ever
had. It now looks that, partly thanks to the help of other
members of this list, I have managed to solve this most
irritating problem. But the true cause of it remains still
mysterious and so does the response I got from WRI's
technical support.First of all, while waiting for response
from the mathgroup, I decided to sent a bug report to WRI's
technical support. The answer came on the next day. Moreover,
it acknowledged this as a known bug and suggested a way round
it. The suggestion was to copy the notebook using Copy As
Complete Notebook menu item and then paste cells from the new
notebook. Unfortunately this solution proved imperfect. It is
true that it solved the visible markup problem but cells were
still being merged into a single cell and the only thing I
could do about it was to break them up again using the Divide
Cell menu item, hardly a convenient way to go about copying
cells. I more or less gave up (particularly once i received
the first response form a mathgroup user 
confirming that he
also had the problem) and tried to persuade myself that using
Mathematica in Classic was not all that bad when I got a
message from Yas Tesiram stating that he was using 4.2 on Mac
OS X 10.2 and did not have the problem at all. At this point I
was still pretty certain that there was indeed a bug, and that
it must have been fixed in a recent release that Yas managed 
to
get (I have to admit to suspecting WRI at this point about
deliberately keeping me in the dark about the existence of a
such a more recent release-- sorry about that). The
breakthrough came when I got a message from TOm Burton who
told me that he had this problem once but he cured it by
deleting and re-installing Mathematica. I immediately tried
his approach but it did not work. But now I was not prepared
to give up. I created a new user, logged out as myself and
logged in again as the new user. Guess what, the problem was
gone. I started frantically possible culprits in my ~/Library
and also for any applications that were set to run on start
up. I soon noticed a possible culprit, a little freeware
program called PTHPasteboard which creates
multiple-clipboards. I removed it and the problem was gone.
Everything seems to be fine again.Now the biggest mystery: why
did technical support so quickly acknowledged a bug, which may
not be one at all? Could this be something to do with the
frequency of complaints (eg.g on this list ) that they are
too slow to acknowledge bugs and to often refuse to accept
them as such? Actually, now I begin to feel that there may be
perhaps something to be said for this attitude, and certainly
it seesm preferable to its opposite: too great a readiness to
accept claims of bugs which may after all not be that at
all.Andrzej KozlowskiYokohama,
Japanhttp://www.mimuw.edu.pl/~akoz/http://
platon.c.u-tokyo.ac.jp/andrzej/> I am having an extremely
frustrating time with one feature of> Mathematica 4.2 which I
swear was not present in earlier versions. It> find is quite
dreadful and yet none seems to have complained about it> so
far so that I feel that perhaps I missing something obvious.>
Anyway, this is the problem. I have two notebooks with a large
number> of cells each. I want to copy some cells from one
notebook to the> other. I am sure in earlier versions it used
to be possible to select> several contiguous cells by shift
clicking and then paste them into> another notebook. You
would then get several cells of the same type as> you copied.
However now (at least on Mac OS X) this seems impossible.>
What happens is that if you select a several cells or a cell
group and> past into another notebook you get a messy
expression contained in a> single cell from which it does not
appear possible to easily recover> the original cells. I have
tired using various Copy As but have not> found anything that
works. The only thing that works is copying and> pasting
individual cells, which is of course very time consuming.> Is
this also what happens on other platforms? I am sure it was
not the> case in earlier versions, and if so was this an
intentional change? It> so it seems to me as bad an idea as I
have come across in many years.> Can anyone explain? Andrzej
Kozlowski
====
=Yes :PowerExpand[Sqrt[x^2]]But : It's dangerous
! You have to be sure that x is real positive !So you can
also do the following :Simplify[Sqrt[x^2],x>0]Bonne journ.8ee
!Florian Jaccardprofesseur de Math.8ematiquesEICN-HESLe
Locle-----Message d'origine-----Envoy.8e : jeu., 28. 
novembre
2002 20:08è : mathgroup@smc.vnet.netObjet : 
Simplification ?Is
there a way to force Mathematica to simplify the terms
likeSqrt[x^2]to give x
???__________________________________________________________
_Olivier DURUSSELEcole Polytechnique F.8ed.8erale de
LausanneSTI-IPR-LCSMME-EcublensCH-1015 LausanneT.8el.: +41 21
693 53 17
====
=............> Have an entry under Graphics
Primitives> Have an entry under Primitives> Have an entry
under Graphics Directives> Have an entry under
Directives............Dave,I almost always use the on line
Help BrowserSet the category to Master Index and these are
found.The first three are in the printed book's
index,----------------------- HayesMathematica Training and
ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198> >look at the online-help for Graphics[]
!>What is listed first ??>The following graphics primitives
can be used:>....>and>The following graphics directives can
be used:>What else can be done ? Should the text color red
?>and ßashing ?> JensHave an entry under Graphics Primitives>
Have an entry under Primitives> Have an entry under Graphics
Directives> Have an entry under Directives The object of an
index is to let people find stuff, who don't 
know theright>
way to ask. Dave Golber > So today I'm doing some graphics.
Oh: there are _options_ and>_directives_.> So I try to find
out what are all the graphics primatives?> When you look in
the index of the 1500 page book, or type Graphics>Primatives>
in the on-line Help, you don't find it. You have 
to look
underGraphics or> Graphics3D. In the book, it's in the
appendix.>
====
=Have an entry under Graphics Primitives> Have
an entry under Primitives> Have an entry under Graphics
Directives> Have an entry under DirectivesThe object of an
index is to let people find stuff, who don't 
know the right>
way to ask.Dave Golber> Well put . . . and relevant to this
group.Power tends to corrupt. Absolute power corrupts
absolutely. Lord Acton (1834-1902)Dependence on advertising
tends to corrupt. Total dependence on advertising corrupts
totally. (today's equivalent) 
====
=Alexander,My guess is that
your speed problems are due to the overhead of making so many
calls from Mathematica into Java. There is a more or less
constant overhead of making a call into Java that is due to
the internals of MathLink and J/Link. On a low-to-mid-range
PC, each call to Java costs about a millisecond (plus the
processing time for the actual Java code to run, which is
usually negligible). For many programs, this overhead is not
a concern, but your code is calling into Java at least twice
for every item read from the database, and if there are
100,000 items to read then you will have a slow program.There
is one simple optimization you can make that should cut the
time about in half. Your Switch statement tests against the
constants Types`CHAR, Types`DOUBLE, etc. These
harmless-looking expressions trigger a call into Java each
time they are evaluated. You can eliminate these unnecessary
calls by caching the values ahead of time: {$charType,
$doubleType, $dateType, ...etc. } = {Types`CHAR,
Types`DOUBLE, Types`DATE, ...etc.};Then the Switch becomes:
Switch[type, $charType, rs@getString[i], $doubleType,
rs@getDouble[i], ... etc. ]This one change will drop the
number of calls into Java from two to one for each item.If
you have lots of TIMESTAMP values in your data, you can also
speed up your handling of these types. Right now you make 6
calls into Java for each TIMESTAMP. You could probably drop
this to 2. In your DataSourceEvaluate Then in GetColumn you
would use it like this:If these optimizations are still not
enough, the only thing to do is to move at least some of the
loop (NestWhileList) down into Java code. Then you could call
into Java once for each result set instead of once for each
data item.Just to be clear, the reason we move the loop is
not because Java is faster than Mathematica but because we
want to minimize the number of times we cross the
Mathematica-Java barrier.A simple first try would be to write
Java code that gets the contents of a single row. Then you
could use essentially all of your current Mathematica code
unchanged. With this technique, you would only have one call
into Java for each row instead of each row x col. That might
be enough optimization. I like this approach because it
conforms with a general principle that I try to adhere to
when doing mixed-language programming: Write as much as
possible in the highest-level language. With J/Link, start
out writing everything in Mathematica, and only if the
performance isn't acceptable do you consider writing Java.
Then try to find the smallest possible piece of functionality
to move down into Java.Here is the RowGetter class. It is
little more than a direct translation of your GetColumn
function into Java. The getRow() method is a good example of
a method that sends its result back to Mathematica manually
instead of simply relying on J/Link's automatic behavior of
sending back the method's return value. Manual functions are
discussed in section 1.2.18 of the J/Link User
Guide.///////////////////// RowGetter class
//////////////////////import java.sql.*;import
com.wolfram.jlink.*;public class RowGetter { private int[]
types; public RowGetter(int[] types) { this.types = types; }
public void getRow(ResultSet rs) throws MathLinkException,
SQLException { ml.beginManual(); ml.putFunction(List,
types.length); for (int i = 0; i <= types.length; i++) {
switch (types[i]) { case Types.VARCHAR: case Types.CHAR:
ml.put(rs.getString(i)); break; case Types.DOUBLE:
ml.put(rs.getDouble(i)); break; case Types.TIME: case
Types.DATE: break; case Types.NUMERIC: case Types.INTEGER:
ml.put(rs.getInt(i)); break; case Types.TIMESTAMP: break;
default: // Do something to handle the fallthrough case:
ml.putSymbol($UnhandledJDBCType); } } }}///////////////////
End RowGetter class /////////////////////Then nothing about
your Mathematica code has to change except the NestWhileList:
rowGetter = JavaNew[RowGetter, types]; Rest[ NestWhileList[
rowGetter@getRow[rs]&, 1, rs@next[]& ] ]Todd GayleyWolfram
Research>I have a Mathematica program that uses
DatabaseAccess package to>retrieve large amount of data from
Sybase database. DatabaseAccess is>available for Windows only
since it uses odbc. Now, I wanted to run>access. At the end of
this message I'll insert the code that
emulates>DataSourceEvaluate and OpenDataSource from
DatabaseAccess but uses>jdbc. The implementation is simple
and straightforward. The problem is>the code works MUCH
slower than the equivalent odbc-based one on NT. I>retrieve A
LOT of data so this is really annoying. So I have
two>questions:1. Is it one of the Mathematica functions that
make it slow (the>conversion of the result of the sql query
into Mathematica table) or>is it some fundamental
java-related problem (JLink, jdbc etc)?2. How can I make it
run faster?Alexander>InstallJava[];>AddToClassPath[>
jconn2.jar,
jdbc2_0-stdext.jar];LoadJavaClass[java.sql.Types];>
LoadJavaClass[java.util.Calendar];Clear[GetColumn];>GetColumn
[i_, type_, rs_] := Switch[type,>
Types`CHAR,rs@getString[i],> Types`DOUBLE,rs@getDouble[i],>
Types`INTEGER,rs@getInt[i],> Types`VARCHAR,rs@getString[i],>
Types`TIMESTAMP,JavaBlock[> ToString[cal@get[Calendar`MONTH]
+ 1]<>/< ToString[cal@get[Calendar`DATE]]<>/<
ToString[cal@get[Calendar`YEAR]]]> ],>
Types`NUMERIC,rs@getInt[i]]ClearAll[DataSourceEvaluate];>
DataSourceEvaluate[conn_,sql_String] :=> JavaBlock[>
Module[{st = conn@createStatement[],rs,md, types,res},> res =
If [st@execute[sql], rs = st@getResultSet[];> md =
rs@getMetaData[];> types =
Array[md@getColumnType[#]&,{md@getColumnCount[]}];> Rest[>
NestWhileList[> Array[GetColumn[#,types[[#]], rs]
&,{Length@types}] &,> 1,> rs@next[] &]],> {}];> st@close[];>
res]]Clear[OpenDataSource];>OpenDataSource[database_String,
user_String,password_String] :=> Module[{driver =
JavaNew[com.sybase.jdbc2.jdbc.SybDriver],> dbUrl =
jdbc:sybase:Tds:<>$DBServer<>:<>$DBPort<>/<>database,> prop =
JavaNew[java.util.Properties]},> prop@setProperty[user,user];>
prop@setProperty[password,password];>
driver@connect[dbUrl,prop]];
====
=I'm a french student in
Computer Science, and crypto, and i want to knowif somebody
have the mathematica implementation of Rijndael, or somegood
url where i could find ressources about it.
Sincerely
====
=Andrzej:I run 4.1 under Mac OS 10.2.1. For what
it's worth, I copy a set of cells from a 4.1 notebook and
paste into the middle of another notebook. Everything seems
to go as it should.??
====
===========> As a follow up on this
issue:> I reinstalled Mathematica 4.1 for Mac OS X and found
that I still have> this problem. On the other hand Hayes
informed me that he did not> have it on Mathematica 4.2 under
Windows 98. I then tried it myself,> with Mathematica 4.1
under Windows 98, and indeed I found I can copy> and paste
groups of cells and the cell structure is preserved. They do>
not get merged into a single cell with all the Mathematica
markup> visible as they do in my case under Mac OS X 10.2. I
would like to know> if other Mac OS X users have the same
problem. If not, it may be caused> by something weird in my
installation, although it seems unlikely as I> am not
experiencing any other problems. I have not yet tried
reporting> this to Technical Support yet since I am not sure
if other Mac OS X> users are also experiencing the same
behaviour, but if it is a bug it's> an awful one.> Andrzej> 
I
am having an extremely frustrating time with one feature of>
Mathematica 4.2 which I swear was not present in earlier
versions. It> find is quite dreadful and yet none seems to
have complained about it> so far so that I feel that perhaps
I missing something obvious.> Anyway, this is the problem. I
have two notebooks with a large number> of cells each. I want
to copy some cells from one notebook to the> other. I am sure
in earlier versions it used to be possible to select> several
contiguous cells by shift clicking and then paste them into>
another notebook. You would then get several cells of the
same type as> you copied. However now (at least on Mac OS X)
this seems impossible.> What happens is that if you select a
several cells or a cell group and> past into another notebook
you get a messy expression contained in a> single cell from
which it does not appear possible to easily recover> the
original cells. I have tired using various Copy As but have
not> found anything that works. The only thing that works is
copying and> pasting individual cells, which is of course
very time consuming.> Is this also what happens on other
platforms? I am sure it was not the> case in earlier
versions, and if so was this an intentional change? It> so it
seems to me as bad an idea as I have come across in many
years.> Can anyone explain? Andrzej Kozlowski
====
=========
Hugh Walker Gnarly Oaks
====
=4.2.1 embeds the required
Mathematica fonts into the EPS graphic by *default*. So, you
don't have to fiddle with any options.But if you 
want to turn
font embedding off, the option name is IncludeSpecialFonts
under the ToPostScriptOptions category (there's another
IncludeSpecialFonts in the PrintingOptions category which
does not affect EPS, but is related to printing under
X)..Sincerely,John Fultzjfultz@wolfram.comUser Interface
GroupWolfram Research, Inc.>Do you refer here to the Option
Inspector? If so, what option name are>you referring to?>
...If you use version 4.2.1, you can configure font> export in
the preferences dialog (or give it as an> option)
====
=I'm
drawing various closed objects of arbitrary shape (thick
lenses, for then would like to Fill them with a specified
color. How to do the Fill?(I suppose I could do this as a
very complex polygon, replacing the arcs by many short line
segments and building up the whole object as a lengthy list
of points; but that seems like a pain.)Power tends to
corrupt. Absolute power corrupts absolutely. Lord Acton
(1834-1902)Dependence on advertising tends to corrupt. Total
dependence on advertising corrupts totally. (today's
equivalent) 
====
=I want to set my axes that one unit is five
centimeter. How can I do ?Sorry for my english, I'm french
:)
====
=I've been trying to run some modestly complicated
animations consisting of several ParametricPlot3D curves. I
can generate the sequence of images, but when I try to use
ShowAnimation, Mathematica starts using all the CPU, and if
there is sufficient number of frames, It starts swapping. What
that happens I either have to reboot the system, or wait
several minutes for Mathematica to let me have a few cycles
so I can kill it. I've seen what I can do with Java 3D, so I
know such animations can be run on this box.What is actually
happening when the animations are run? Even if I get them to
start, they never run smoothly. The always freze momentarily
about every 7 frames or so, and sometimes they will stall for
a fairly long period.I really think Mathematica is a wonderful
program for handling mathematical problems. It's ability to
graph mathematical functions is fantastic. I am interested in
studying the time evolution of physical systems, and believe
that 3D animation is the best way to understand the
mathematical expressions used to describe these systems. When
I get beyond the simplest systems, Mathematic seems unable to
support this aspect of my modeling.I don't have time to 
write
my own mapping between Java 3D and Mathematica, unless WRI
wants to fund the effort. Is there a good way to handle this
kind of situation using Mathematica?-- STHHatton's Law: 
There
is only One inviolable Law.
====
=I type the
following:ImplicitPlot[Sqrt[x-y]==0,{x,0,10}]and all ok,but
if i try to use ContourPlot in the following
mannerImplicitPlot[Sqrt[x-y]==0,{x,0,10},{y,0,10}]it
say:ContourGraphics::ctpnt: The contour is attempting to
traverse a cell inwhich some of the points have not evaluated
to numbers, and it will bedropped.I have not understud
why.Some one can help me?
====
=Andrzej, I have Mathematica 4.2
installed on my upgraded PowerTower Pro(G4800Mhz) with Mac OS
10.1.5 and do not have the problem you areexperiencing.Brian >
As a follow up on this issue:> I reinstalled Mathematica 4.1
for Mac OS X and found that I still have > this problem. On
the other hand Hayes informed me that he did not > have it on
Mathematica 4.2 under Windows 98. I then tried it myself, >
with Mathematica 4.1 under Windows 98, and indeed I found I
can copy > and paste groups of cells and the cell structure
is preserved. They do > not get merged into a single cell
with all the Mathematica markup > visible as they do in my
case under Mac OS X 10.2. I would like to know > if other Mac
OS X users have the same problem. If not, it may be caused >
by something weird in my installation, although it seems
unlikely as I > am not experiencing any other problems. I
have not yet tried reporting > this to Technical Support yet
since I am not sure if other Mac OS X > users are also
experiencing the same behaviour, but if it is a bug it's > 
an
awful one. I am having an extremely frustrating time with one
feature of > Mathematica 4.2 which I swear was not present in
earlier versions. It > find is quite dreadful and yet none
seems to have complained about it > so far so that I feel
that perhaps I missing something obvious. > Anyway, this is
the problem. I have two notebooks with a large number > of
cells each. I want to copy some cells from one notebook to
the > other. I am sure in earlier versions it used to be
possible to select > several contiguous cells by shift
clicking and then paste them into > another notebook. You
would then get several cells of the same type as > you
copied. However now (at least on Mac OS X) this seems
impossible. > What happens is that if you select a several
cells or a cell group and > past into another notebook you
get a messy expression contained in a > single cell from
which it does not appear possible to easily recover > the
original cells. I have tired using various Copy As but have
not > found anything that works. The only thing that works is
copying and > pasting individual cells, which is of course
very time consuming.> Is this also what happens on other
platforms? I am sure it was not the > case in earlier
versions, and if so was this an intentional change? It > so
it seems to me as bad an idea as I have come across in many
years. > Can anyone explain?> Andrzej Kozlowski>
====
=> As a
follow up on this issue:> I reinstalled Mathematica 4.1 for
Mac OS X and found that I still have > this problem. On the
other hand Hayes informed me that he did not > have it on
Mathematica 4.2 under Windows 98. I then tried it myself, >
with Mathematica 4.1 under Windows 98, and indeed I found I
can copy > and paste groups of cells and the cell structure
is preserved. [...]Andrzej[...]miss other features what are
very convinient in Windows:- Selecting arbitrary cells by
pressing the Ctrl-key- Clicking on a cell bracket after
pressing the Alt-key selects all cells of the same type in
the whole notebookRainer
====
=No problem using 4.1 and 10.2.2.
NB: Copying, switching applications,switching back to
Mathematica, then pasting does produce the problem you
describe.However, this problem has been around for awhile,
certainly it was thereusing 4.0 on OS 9.js-- Joshua A.
SolomonDepartment of Optometry and Visual ScienceCity
Universityhttp://www.staff.city.ac.uk/~solomon> As a follow
up on this issue:> I reinstalled Mathematica 4.1 for Mac OS X
and found that I still have> this problem. On the other hand
Hayes informed me that he did not> have it on Mathematica 4.2
under Windows 98. I then tried it myself,> with Mathematica
4.1 under Windows 98, and indeed I found I can copy> and
paste groups of cells and the cell structure is preserved.
They do> not get merged into a single cell with all the
Mathematica markup> visible as they do in my case under Mac
OS X 10.2. I would like to know> if other Mac OS X users have
the same problem. If not, it may be caused> by something weird
in my installation, although it seems unlikely as I> am not
experiencing any other problems. I have not yet tried
reporting> this to Technical Support yet since I am not sure
if other Mac OS X> users are also experiencing the same
behaviour, but if it is a bug it's> an awful one.> Andrzej> 
I
am having an extremely frustrating time with one feature of>
Mathematica 4.2 which I swear was not present in earlier
versions. It> find is quite dreadful and yet none seems to
have complained about it> so far so that I feel that perhaps
I missing something obvious.> Anyway, this is the problem. I
have two notebooks with a large number> of cells each. I want
to copy some cells from one notebook to the> other. I am sure
in earlier versions it used to be possible to select> several
contiguous cells by shift clicking and then paste them into>
another notebook. You would then get several cells of the
same type as> you copied. However now (at least on Mac OS X)
this seems impossible.> What happens is that if you select a
several cells or a cell group and> past into another notebook
you get a messy expression contained in a> single cell from
which it does not appear possible to easily recover> the
original cells. I have tired using various Copy As but have
not> found anything that works. The only thing that works is
copying and> pasting individual cells, which is of course
very time consuming.> Is this also what happens on other
platforms? I am sure it was not the> case in earlier
versions, and if so was this an intentional change? It> so it
seems to me as bad an idea as I have come across in many
years.> Can anyone explain?Andrzej Kozlowski
====
=There is no
hint of this trouble using4.2 for Mac OS X (August 22, 2002)
on Mac OS 10.2.2.If you start Mathematica with the Shift key
held down it will reset all the Options in the Preferences to
their default values and this might be worth
trying.------------------------------------------------------
---------> As a follow up on this issue:> I reinstalled
Mathematica 4.1 for Mac OS X and found that I still have >
this problem. On the other hand Hayes informed me that he did
> not have it on Mathematica 4.2 under Windows 98. I then
tried it > myself, with Mathematica 4.1 under Windows 98, and
indeed I found I > can copy and paste groups of cells and the
cell structure is > preserved. They do not get merged into a
single cell with all the > Mathematica markup visible as they
do in my case under Mac OS X 10.2. > I would like to know if
other Mac OS X users have the same problem. If > not, it may
be caused by something weird in my installation, although >
it seems unlikely as I am not experiencing any other
problems. I have > not yet tried reporting this to Technical
Support yet since I am not > sure if other Mac OS X users are
also experiencing the same behaviour, > but if it is a bug
it's an awful one.Christopher PurcellDRDC-Atlantic, 9 Grove
St., PO Box 1012,Dartmouth, NS Canada B2Y
3Z7chris.purcell@drdc-rddc.gc.ca
====
=(1) The function may have
no zeros.(2) The coefficients in your function have such 
wildly
differing magnitudes that you'll never get any reliable
numerical results unless you do a lot of rescaling.Hope this
helps...---Selwyn Hollis> i have to find all the roots in a
region of this equation for a sweep> of value in w :>
function[z_,w_]:=Module[{k0,sl},k0=w/(3*10^8);sl=(Sqrt[k0^2-z
^2])/(377*k0)-(I*Sqrt[e*k0^2-z^2])/(377*k0*Tan[o*Sqrt[e*k0^2-
z^2]])+I> w a (1 - b t w^2)/((1 - b t w^2)*(1 - g a w^2) - b
a w^2)];where a,b,g,t,o,e are costs:o=0.00406> e=1.001>
b=0.048*10^(-9)> t=0.4521*10^(-12)> g=0.396*10^(-9)>
a=0.176*10^(-12)I've tried with the following function 
posted
by Selwyn Hollis:shotgun[fn_, {z_Symbol, z1_?NumericQ,
z2_?NumericQ, dz_?NumericQ},> tol_?NumericQ,
fuzz_?NumericQ]:= Module[{gridpts,shot,solns}, (* Construct a
list of grid points. The mesh size is dz.*)> gridpts = Table[x
+ I*y, {x,Re[z1],Re[z2],dz},> {y,Im[z1],Im[z2],dz}]; (* It's 
a
bit faster if we compile Abs[fn]. *)> fc =
Compile[{{z,_Complex,2}}, Abs[fn]]; (* Keep only the grid
points where Abs[fn] < tol. *)> starters = Extract[gridpts,
Position[fc[gridpts], _?(#<tol&)]]; (* Map FindRoot through
the list of starting points.*)> shot =
Chop@FindRoot[fn,{z,#}]&/@starters; (* Sort by magnitude and
remove repetitions.> A repetition occurs when consecutive
solutions differ by > less than fuzz. *)> solns =
Sort[shot,(Abs[z/.#1] < Abs[z/.#2])&];> solns =
First/@Split[solns,(Abs[(z/.#1)-(z/.#2)] < fuzz)&]; (* Sort
again and remove repetitions. *)>
First/@Split[Sort[solns],(Abs[(z/.#1)-(z/.#2)] < fuzz)&] ]It
works fine with some function but not with the mine.> Does
anybody know how to change it, to make it works fine?>
Massimiliano CasalettiP.S. Sorry for my english> 
====
=>
function[z_,w_]:=Module[{k0,sl},k0=w/(3*10^8);sl=(Sqrt[k0^2-z
^2])/(377*k0)-(I*>
Sqrt[e*k0^2-z^2])/(377*k0*Tan[o*Sqrt[e*k0^2-z^2]])+I> w a (1
- b t w^2)/((1 - b t w^2)*(1 - g a w^2) - b a w^2)];Doing
some kind of fiber optics or wave propagation, one might guess
. . . .Power tends to corrupt. Absolute power corrupts
absolutely. Lord Acton (1834-1902)Dependence on advertising
tends to corrupt. Total dependence on advertising corrupts
totally. (today's equivalent) 
====
=The following works OK in
Mathematica 4.2 for Windows:input:With[{a = 1.1, b = 1.1},
NSolve[Exp[-Pi*a/x] + Exp[-Pi*b/x] == 1, x] ]output:{{x ->
4.9856}}Steve Luttrell> I am trying to solve the following
equation for given values of two> parameters a and b:
NSolve[Exp[-Pi*a/x] + Exp[-Pi*b/x] == 1, x] This is
straightforward for integer values of the parameters a and b>
(e.g. a=1, b=1). However, Mathematica is unable to compute
the> solution for non-integer values of a and b (e.g. a=1.1,
b=1.1; even> a=1.0, b=1.0 doesn't work). Instead, the error
message Solve::tdep:> The equations appear to involve the
variables to be solved for in an> essentially non-algebraic
way. is produced. Can anybody point out a way to overcome
this problem and to obtain a> solution? Eva>Reply-To:
kuska@informatik.uni-leipzig.de
====
=what ist withWith[{a =
1.1, b = 1.0}, FindRoot[Exp[-Pi*a/x] + Exp[-Pi*b/x] == 1, {x,
1}] ] JensI am trying to solve the following equation for
given values of two> parameters a and b:NSolve[Exp[-Pi*a/x] +
Exp[-Pi*b/x] == 1, x]This is straightforward for integer
values of the parameters a and b> (e.g. a=1, b=1). However,
Mathematica is unable to compute the> solution for
non-integer values of a and b (e.g. a=1.1, b=1.1; even>
a=1.0, b=1.0 doesn't work). Instead, the error message
Solve::tdep:> The equations appear to involve the variables
to be solved for in an> essentially non-algebraic way. is
produced.Can anybody point out a way to overcome this problem
and to obtain a> solution?Eva
====
=I do not have this problem
with my installation of Mathematica 4.2 on Mac OS X 10.2 as
long as I do not have the cursor in the target notebook
already in a cell-- the cursor should be a horizontal line
indicating insert cell mode. It also works for non-contiguous
cells.Alex> As a follow up on this issue:> I reinstalled
Mathematica 4.1 for Mac OS X and found that I still have>
this problem. On the other hand Hayes informed me that he did
not> have it on Mathematica 4.2 under Windows 98. I then tried
it myself,> with Mathematica 4.1 under Windows 98, and indeed
I found I can copy> and paste groups of cells and the cell
structure is preserved. They do> not get merged into a single
cell with all the Mathematica markup> visible as they do in my
case under Mac OS X 10.2. I would like to know> if other Mac
OS X users have the same problem. If not, it may be caused>
by something weird in my installation, although it seems
unlikely as I> am not experiencing any other problems. I have
not yet tried reporting> this to Technical Support yet since I
am not sure if other Mac OS X> users are also experiencing the
same behaviour, but if it is a bug it's> an awful one.>
Andrzej> I am having an extremely frustrating time with one
feature of> Mathematica 4.2 which I swear was not present in
earlier versions. It> find is quite dreadful and yet none
seems to have complained about it> so far so that I feel that
perhaps I missing something obvious.> Anyway, this is the
problem. I have two notebooks with a large number> of cells
each. I want to copy some cells from one notebook to the>
other. I am sure in earlier versions it used to be possible
to select> several contiguous cells by shift clicking and
then paste them into> another notebook. You would then get
several cells of the same type as> you copied. However now
(at least on Mac OS X) this seems impossible.> What happens
is that if you select a several cells or a cell group and>
past into another notebook you get a messy expression
contained in a> single cell from which it does not appear
possible to easily recover> the original cells. I have tired
using various Copy As but have not> found anything that
works. The only thing that works is copying and> pasting
individual cells, which is of course very time consuming.> Is
this also what happens on other platforms? I am sure it was
not the> case in earlier versions, and if so was this an
intentional change? It> so it seems to me as bad an idea as I
have come across in many years.> Can anyone explain? Andrzej
Kozlowski
====
=[I fear this is a FAQ but I've dutifully
searched the current 4.2 docs andthis formus archive and
could find only recommendations for Mathematica 
3.0]I'm trying
to copy/export graphics from Mathematica to other apps such as
MS Word orPowerPoint. Graphs print great directly from
Mathematica (correctly rotates they-axis text even though it
doesn't appear correctly on the screen). Butattempts to
copy/paste into other applications (using PICT, PICT
withembedded postscript, and all other options) or
Export[file.GIF,%] etc. andthen importing that 
file to other
apps usually produces something ugly andthe y-axis text never
prints rotated from these other apps.I'm addicted to 
producing
publication-quality graphics with Mathematica, but I needto be
able to include them in those documents produced by other
apps. Iknow the workaround is simply to use Mathematica as my
word processor for technicaldocuments, but I can't convert 
all
my colleagues with whom I exchangeworking drafts.Any
suggestions appreciated, including RTFM if you will give me a
hint whereto look. gary
====
=Vince,I've often found that using
PlotRange to extend the printed range just a bitbeyond the
limits I want makes missing tick marks appear on either
end.Gary> When I plot a graph with the following FrameTicks
specification:FrameTicks - {> Automatic,> {> {-[Pi], -[Pi],
{0.01, 0}, {AbsoluteThickness[1.5]}}> },> Automatic,> {>
{-[Pi], -[Pi], {0.01, 0}, {AbsoluteThickness[1.5]}}> }> }the
tick to the right of the -Pi label on the left axis does not
appear.> The only way I can make it appear in Mathematica is
with:FrameTicks - {> Automatic,> {> {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}},> {-[Pi], , {0.01, 0},
{AbsoluteThickness[1.5]}}> },> Automatic,> {> {-[Pi], -[Pi],
{0.01, 0}, {AbsoluteThickness[1.5]}}> }> }> where I have
repeated the frametick line but omitting the -Pi label.I'm
using Mathematica 4.1.1.0 on Windows XP.> Vince
Boros
====
=Vince,The tick appears for me with Mathematica 4.2
under Windows 98----------------------- HayesMathematica
Training and ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198> When I plot a graph with the following
FrameTicks specification: FrameTicks - {> Automatic,> {>
{-[Pi], -[Pi], {0.01, 0}, {AbsoluteThickness[1.5]}}> },>
Automatic,> {> {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}}> }> } the tick to the right of the
-Pi label on the left axis does not appear.> The only way I
can make it appear in Mathematica is with: FrameTicks - {>
Automatic,> {> {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}},> {-[Pi], , {0.01, 0},
{AbsoluteThickness[1.5]}}> },> Automatic,> {> {-[Pi], -[Pi],
{0.01, 0}, {AbsoluteThickness[1.5]}}> }> }> where I have
repeated the frametick line but omitting the -Pi label. I'm
using Mathematica 4.1.1.0 on Windows XP.> Vince Boros
====
=In
reply to my question, it turned out that in the menu item
Edit -> Preferences -> Formatting Options -> Font Options, I
had Background set to GrayLevel[1]. I changed Background to
None, and the problem disappeared. So what had been going
wrong is that I had been writing black text on a white
background on top of the graph, instead of black replied
offering suggestions.Vince> When I plot a graph with the
following FrameTicks specification:FrameTicks - {> Automatic,>
{> {-[Pi], -[Pi], {0.01, 0}, {AbsoluteThickness[1.5]}}> },>
Automatic,> {> {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}}> }> }the tick to the right of the
-Pi label on the left axis does not appear. > The only way I
can make it appear in Mathematica is with:FrameTicks - {>
Automatic,> {> {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}},> {-[Pi], , {0.01, 0},
{AbsoluteThickness[1.5]}}> },> Automatic,> {> {-[Pi], -[Pi],
{0.01, 0}, {AbsoluteThickness[1.5]}}> }> }> where I have
repeated the frametick line but omitting the -Pi label.I'm
using Mathematica 4.1.1.0 on Windows XP.> Vince Boros>

====
=Vince,It would help if you gave us an actual Plot
statement that exhibited theproblem. I suspect that your
PlotRange is {-Pi, Pi} and then Mathematicadrops the lower
tick mark. (Don't ask me why - it's a 
feature.) If you
wouldextend the vertical PlotRange slightly, it would
probably solve the problem.In the second statement you got
rid of the tick label yourself and so youdidn't notice that
Mathematica had already dropped it.David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/
{-[Pi], -[Pi], {0.01, 0}, {AbsoluteThickness[1.5]}} } }the
tick to the right of the -Pi label on the left axis does not
appear. The only way I can make it appear in Mathematica is
with:FrameTicks -> { Automatic, { {-[Pi], -[Pi], {0.01, 0},
{AbsoluteThickness[1.5]}}, {-[Pi], , {0.01, 0},
{AbsoluteThickness[1.5]}} }, Automatic, { {-[Pi], -[Pi],
{0.01, 0}, {AbsoluteThickness[1.5]}} } }where I have repeated
the frametick line but omitting the -Pi label.I'm using
Mathematica 4.1.1.0 on Windows XP.Vince Boros
====
=I am a new
Mathematica user with a numerical integration problem. My
2DNintegration gives the warning slwcon (numerical
integration convergestoo slow). I am integrating over a thin
strip through a 2DInterpolating Function. The Interpolating
function is zero for mostvalues. All stips pass through some
non-zero region. When the strip ismostly in the non-zero
region there is no problem. Howvever, when thestrip passes
through only a small part of the non-zero region, thewarning
occurs and the integration is slow. The integration
parametersI have so far found to work best are:
WorkingPrecision -> 9,AccuracyGoal ->Infinity, Precision Goal
-> 2, Method -> Gausskronron.If I increase the PrecisionGoal
both the number of warnings and the timeincrease.Ann
Fitchard
====
=Yes, I have the same problem with 4.2 under OS X
(10.2.2). And it *is awful* indeed. What's worse (because
it's much harder to detect and correct) is the arbitrary
introduction of spaces and linefeeds to quoted strings which
are cut and pasted. For my use of Mathematica, this has cost
me untold hours. No combination of copy-as and/or paste-as
has affected this problem.I sure hope to hear that this is
going to be fixed soon!Michael WilliamsBlacksburg> As a follow
up on this issue:> I reinstalled Mathematica 4.1 for Mac OS X
and found that I still have> this problem. On the other hand
Hayes informed me that he did not> have it on Mathematica 4.2
under Windows 98. I then tried it myself,> with Mathematica
4.1 under Windows 98, and indeed I found I can copy> and
paste groups of cells and the cell structure is preserved.
They do> not get merged into a single cell with all the
Mathematica markup> visible as they do in my case under Mac
OS X 10.2. I would like to know> if other Mac OS X users have
the same problem. If not, it may be caused> by something weird
in my installation, although it seems unlikely as I> am not
experiencing any other problems. I have not yet tried
reporting> this to Technical Support yet since I am not sure
if other Mac OS X> users are also experiencing the same
behaviour, but if it is a bug it's> an awful one.> Andrzej> 
I
am having an extremely frustrating time with one feature of>
Mathematica 4.2 which I swear was not present in earlier
versions. It> find is quite dreadful and yet none seems to
have complained about it> so far so that I feel that perhaps
I missing something obvious.> Anyway, this is the problem. I
have two notebooks with a large number> of cells each. I want
to copy some cells from one notebook to the> other. I am sure
in earlier versions it used to be possible to select> several
contiguous cells by shift clicking and then paste them into>
another notebook. You would then get several cells of the
same type as> you copied. However now (at least on Mac OS X)
this seems impossible.> What happens is that if you select a
several cells or a cell group and> past into another notebook
you get a messy expression contained in a> single cell from
which it does not appear possible to easily recover> the
original cells. I have tired using various Copy As but have
not> found anything that works. The only thing that works is
copying and> pasting individual cells, which is of course
very time consuming.> Is this also what happens on other
platforms? I am sure it was not the> case in earlier
versions, and if so was this an intentional change? It> so it
seems to me as bad an idea as I have come across in many
years.> Can anyone explain? Andrzej KozlowskiIn-reply-To:
<200211281908.OAA24008@smc.vnet.net>
====
=RP> In[4]:=
Sum[Binomial[0,k],{k,0,Infinity}]RP> Out[4]= 0RP> I disagree.
Should be 1.I confirm that you identified a bug 
the long-liver,
I have sent your dataalong with the reference to you to
support@wolfram.com . Indeed, In[1] := Sum[Binomial[0, k],
{k, 0, Infinity}] Out[1] = 0 In[2] := NSum[Binomial[0, k], {k,
0, Infinity}] Out[2] = 1.This bug exists in the following
versions of Mathematica 4.2 for Microsoft Windows (June 5,
2002) 4.2 for Microsoft Windows (February 28, 2002) 4.1 for
Microsoft Windows (November 2, 2000) 4.0 for Microsoft
Windows (April 21, 1999) Microsoft Windows 3.0 (April 25,
1997)RP> Perhaps my desired result for In[3] is expecting too
much.I hardly think so. For example, another system returns 1
for the counterpartof your input.RP> In[3]:=
Sum[Binomial[0,k],{k,0,n}]RP> Out[3]= 0RP> I disagree. Should
be If[n == 0, 1, 0].Again, you are right because this sum is
equal to 1 + 0 + 0 +... 0 = 1.Please note that In[4] :=
Binomial[0, k] Out[4] = Sin[k*Pi]/(k*Pi)while it should be
If[k==0, 1, Sin[k*Pi]/(k*Pi)] .RP> In[5]:=
Sum[Binomial[n,k],{k,0,Infinity}]RP> nRP> Out[5]= 2RP> I
agree. But inconsistent with Out[4].RP> After all, x/x
reduces to 1, not If[x == 0, Indeterminate, 1].RP> But Out[5]
demonstrates that Mathematica knows the binomial theorem.RP>
It ought to be able to come up with the correct result for
In[4].Yes, it seems to be too obvious to discuss further.Best
wishes,Vladimir BondarenkoMathematical and Production
DirectorSymbolic Testing Group http://www.CAS-testing.org/
GEMM Project (95% ready)Mail : 76 Zalesskaya Str, Simferopol,
Crimea, Ukraine
====
=> In[3]:= Sum[Binomial[0,k],{k,0,n}]
Out[3]= 0 I disagree. Should be If[n == 0, 1, 0].Of course, I
meant If[n >= 0, 1, 0].Rob PrattDepartment of Operations
Researchhttp://www.unc.edu/~rpratt/In-reply-To:
<200211281907.OAA23937@smc.vnet.net>
====
=S> here is my
problem:S> Lets say you solve an equation using SolveS>
Solve[2x - 1 == 0, x] // N{{x -> 0.5}}S> How would I then use
the result of Solve (0.5) and assign it to aS> new variable,
like answer for further calculations?Just use [[...]] (or its
prefix equivalent, Part) to select a part of theoutput you 
want
to use for your further calculations. In[1] := answer =
(Solve[2x - 1 == 0, x] // N)[[1, 1, 2]]; In[2] := answer^2
Out[2] = 0.25Best wishes,Vladimir BondarenkoMathematical and
Production DirectorSymbolic Testing Group
http://www.CAS-testing.org/ GEMM Project (95% ready) Mail :
76 Zalesskaya Str, Simferopol, Crimea, Ukraine
====
=Yes,
sometimes guys in engineering faculties have odd fixations.
But I thinkyou're quite right in sticking to Mathematica.
Anyway, the result of a Solvestatement is a list of rules,
each of them saying something like a solutionis -> something.
In your case, there is only one solution, so the list hasonly
one element, and the solution is the first (and only) element
of thelist. So, if you writeIn[1]:=a = N[Solve[2*x - 1 == 0,
x]]Out[1]={{x->0.5}}then the first element of a
isIn[9]:=a[[1]]Out[9]={x->0.5}Now you want to use this
particular value of x. There are several ways to goabout it.
One, and perhaps the more instructive, is to look at the
FullFormof
a[[1]]:In[2]:=FullForm[{x->0.5}]Out[2]//FullForm=List[Rule[x,
0.5]]which again says that you're dealing with a list. The
first (and only)element of this list is a rule, and the second
part of this rule is thenumerical value you're looking for:
In[3]:=a[[1,1,2]]Out[3]=0.5Another way of getting at it is
using a ReplaceAll statement. If you writeIn[4]:=x /. {x ->
0.5}this can be read as x, where x takes the value 0.5,
which, when executed,will yield 0.5. Then you may write
directlyIn[5]:=N[Solve[2*x - 1 == 0,
x]][[1,1,2]]Out[5]=0.5This is valid in general. If your
equation has several solutions, then thefirst index will lead
you to the one you wish to use. Suppose you wanted touse this
solution as input to some other problem, e.g., to be used as
theargument of a function, like, say Sin[x]. Then you may
writeIn[6]:=Sin[x]/.a[[1]]Out[6]=0.479426orIn[7]:=Sin[x/.a[[1
]]]Out[7]=0.479426i.e. you may use the rule on the argument
or on the function itself (in thissimple example).Tomas
GarzaMexico City----- Original Message -----> Lets say you
solve an equation using Solve Solve[2x - 1 == 0, x] // N {{x
-> 0.5}} How would I then use the result of Solve (0.5) and
assign it to a> new variable, like answer for further
calculations? I think it has something to do with /. and ->
rules of some sort, but> I can't get this simple concept to
work!> -Stan
====
=ans = Solve[2x-1==0,x][[1]]//Nanswer =
x/.ansNote that the [[1]] construct effectively turns ans
into {x->0.5}, withoutit the above answer would be {0.5} not
0.5.Kevin> Even though it's frowned upon by the engineering
college, i use> mathematica a lot of the stuff that would
take pages of code in> another system. I have most basic
functions and operations down, but I am> lacking one basic
thing I need to do to be able to program more> problems
successfully: here is my problem: Lets say you solve an
equation using Solve Solve[2x - 1 == 0, x] // N {{x -> 0.5}}
How would I then use the result of Solve (0.5) and assign it
to a> new variable, like answer for further calculations? I
think it has something to do with /. and -> rules of some
sort, but> I can't get this simple concept to work!>
-Stan>Reply-To: kuska@informatik.uni-leipzig.de
====
=sound
complicated ;-)answer = First[x /. Solve[2x - 1 == 0, x] //
N]First[] because you will use only the first of the
solutions. JensEven though it's frowned upon by the
engineering college, i use> mathematica a lot of the stuff
that would take pages of code in> another system. I have most
basic functions and operations down, but I am> lacking one
basic thing I need to do to be able to program more> problems
successfully:here is my problem:Lets say you solve an equation
using SolveSolve[2x - 1 == 0, x] // N{{x -> 0.5}}How would I
then use the result of Solve (0.5) and assign it to a> new
variable, like answer for further calculations?I think it has
something to do with /. and -> rules of some sort, but> I
can't get this simple concept to work!> -Stan
====
=Stan,We have
fullsolution = Solve[{x + y == 2, x^2 + y^2 == 10}, {x, y}]
{{x -> -1, y -> 3}, {x -> 3, y -> -1}}The full solution
contains two answers {x->-1,y->3} and {x->3,y->-1}This lets
us state the full solution in a way that relates to the
variablesand also lets us do things like a x + b y /.
solution {-a + 3 b, 3 a - b}Which gives the results of
substituting using each of the two answers.With your example
fullsolution = N[Solve[2 x - 1 == 0, x]] {{x -> 0.5}}We can
get answer = x /. fullsolution[[1]] 0.5 answer
0.5----------------------- HayesMathematica Training and
ConsultingLeicester
UKwww.haystack.demon.co.ukhay@haystack.demon.co.ukVoice: +44
(0)116 271 4198> Even though it's frowned upon by the
engineering college, i use> mathematica a lot of the stuff
that would take pages of code in> another system. I have most
basic functions and operations down, but I am> lacking one
basic thing I need to do to be able to program more> problems
successfully: here is my problem: Lets say you solve an
equation using Solve Solve[2x - 1 == 0, x] // N {{x -> 0.5}}
How would I then use the result of Solve (0.5) and assign it
to a> new variable, like answer for further calculations? I
think it has something to do with /. and -> rules of some
sort, but> I can't get this simple concept to work!>
-Stan>
====
=You can use FindRoot after plotting the graph of
your equation.For examplePlot[Exp[-Pi*0.5/x] +
Exp[-Pi*3.0/x], {x, 0, 10}]FindRoot[Exp[-Pi*0.5/x] +
Exp[-Pi*3.0/x] == 1, {x, 2}]> NSolve[Exp[-Pi*a/x] +
Exp[-Pi*b/x] == 1, x]>However, Mathematica is unable to
compute the> solution for non-integer values of a and b (e.g.
a=1.1, b=1.1; even> a=1.0, b=1.0 doesn't work). Instead, the
error message Solve::tdep:> The equations appear to involve
the variables to be solved for in an> essentially
non-algebraic way. is
produced.**************milkcartIn-reply-To:
<200211281908.OAA23991@smc.vnet.net>
====
=EH> I am trying to
solve the following equation for given values of twoEH>
parameters a and b:EH> NSolve[Exp[-Pi*a/x] + Exp[-Pi*b/x] ==
1, x]EH> This is straightforward for integer values of the
parameters a and bEH> (e.g. a=1, b=1). However, Mathematica
is unable to compute theEH> solution for non-integer values
of a and b (e.g. a=1.1, b=1.1; evenEH> a=1.0, b=1.0 doesn't
work). Instead, the error message Solve::tdep:EH> The
equations appear to involve the variables to be solved for in
anEH> essentially non-algebraic way. is produced.EH> Can
anybody point out a way to overcome this problem and to
obtain aEH> solution?Your may wish to use FindRoot.a = 1.1; b
= 1.1;FindRoot[Exp[-Pi*a] + Exp[-Pi*b/x] == 1, {x, 1/10^100,
10}]{x -> 107.748}a = Random[Real, 1, 1000];b = Random[Real,
1, 1000];FindRoot[Exp[-Pi*a] + Exp[-Pi*b/x] == 1, {x,
1/10^100, 10}]{x -> 0.741772}Best wishes,Vladimir
BondarenkoMathematical and Production DirectorSymbolic
Testing Group http://www.CAS-testing.org/ GEMM Project (95%
ready)Mail : 76 Zalesskaya Str, Simferopol, Crimea,
Ukraine
====
=Indeed, NSolve gives a list of numerical
approximations to the roots of apolynomial equation (cf.
OnLine Help). Use instead FindRoot, and of courseyou'll have
to supply numerical values for a and b, as well as
initialvalues. This shouldn't be difficult if 
you begin by
plotting your functionto see where to set initial values for
the root finding process. Forexample, if you 
defineIn[1]:=g[x_,
a_, b_] := Exp[(-Pi)*(a/x)] + Exp[(-Pi)*(b/x)] - 1and plot
this function, say for a = 2.5 and b = 3.0, you'll 
find that
theroot lies somewhere between 11 and 13:In[2]:=Plot[g[x,
2.5, 3.], {x, 1, 15}, PlotRange -> {{8, 15}, {-1, 1}}];Then
use FindRoot:In[3]:=FindRoot[g[x, 2.5, 3.] == 0, {x, {11,
13}}]Out[3]={x -> 12.428204959984097}Tomas GarzaMexico
City----- Original Message -----> a=1.0, b=1.0 doesn't 
work).
Instead, the error message Solve::tdep:> The equations appear
to involve the variables to be solved for in an> essentially
non-algebraic way. is produced. Can anybody point out a way
to overcome this problem and to obtain a> solution?
Eva
====
=Eva,Instead of NDSolve you could try something like
this...f[a_, b_] := Exp[-Pi*a/x] + Exp[-Pi*b/x]
Plot[f[1.1,2.5]-1,{x,0,10}];FindRoot[f[1.1, 2.5] == 1, {x,
6}]{x -> 7.71189}David
Parkdjmp@earthlink.nethttp://home.earthlink.net/~djmp/ Can
anybody point out a way to overcome this problem and to
obtain asolution?Eva
====
=Does somebody know an algorithm to
simplify the product Sqrt[a]*Sqrt[b] ?With many
thanksGernot
====
=If i go:x [CTRL]+_i get index-mode.Power-mode
is obtained byx [CTRL]+^However, i don't know how to enter
those twodirectly above eachother. Is it possible?Also -
[Integral] gives me a nice sign but whatdo i type to get an
integral with limits? LaTeXsytax is supposed to work but i
didn't
succeeded...--V.8anligenKonrad-------------------
====
=Using
Mathematica 4.2 under Jaguar Mac OS X 10.2, I am unable to
print anything on my LaserWriter 8500 both with AppleTalk or
TCP/IP connection. MS Word prints without problem. What
happens?-- Daniel AMIGUET Avenue du L.8eman 59 CH 1005
Lausanne +41 21/728 0730Reply-To: ng
<georgakopoulosNOSPAM@LETSSTOPGETTINGSPAMmindspring.com>
====
=
Store your graphic in a variable, then add to it the
necessary graphicsprimitives using a Show[ ]
command.graph1=...your graphic...maxx=your maximum
xShow[{graph1, Graphics[{
Line[{{0,-.1},{maxx,-.1}}],Line[{{0,-.2},{maxx,-.2}}]}]}];
Luck,Nicholas For reasons too involved to go into, I need to
produce a plot with 4> horizontal scales running in parallel
along the bottom edge. I would> appreciate
recipes/suggestions for doing this.> kj>
====
=honestly say that
this seems like an unbelievably powerful combination. There
are, shall we say, issues. Nonetheless, once I get used to
what best goes into Java Code, and what best goes into
Mathematica, I believe this will redefine my idea of what
Mathematica graphics are. I'm not sure how much direct
mapping I can do between Mathematica's native graphics
functions and Java 3D. I don't know much about extracting
data from Mathematica's graphics. If I can get a raw numeric
representation of Mathematicas graphics, it should be fairly
easy to feed this to Java3D. This is cool! -- STHHatton's
Law: There is only One inviolable Law.
====
=Dear MathGroupI
have the task of plotting dipole magnetic field lines in
3Dspherical coordinate system. Before I actually start with
my work, I was trying learnwith the following simple
code:ex=-y[t];ey=x[t];x0=1.0;y0=1.0;t1=6;lines = FieldLine[
{x[t], ex, x0}, {y[t], ey, y0}, {t,
t1}];c=Show[Graphics[lines, AspectRatio -> Automatic, Frame
-> True]];a=Show[AddArrow[lines, 0.5]];In these command
lines, c plots contour plot. But a gives thefollowing
message:Show::gcomb: An error was encountered in combining
the graphics objects in Show[{Arrow[{0.570594, 1.29399},
<<3>, HeadLength -> 4], <<4>}].If any knows the solution to
this problem please let me know. I use Gangadhara--
_____________________________________________________________
______________Bangalore - 560034,
Indiahttp://www.iiap.ernet.in/personnel/ganga/ganga.html_____
_____________________________________________________________
________
====
=I have two lists (in general of different length)
that have thefollowing structure Random[Integer, {1, 2000}]},
{100}]; Random[Integer, {1, 2000}]}, {200}];I am then
interested in obtaining all the strings in s1 that matchwith
those in s2. At the present time I use the following agorithm
tofind the matchesmyList = Flatten[Outer[List, s1, s2, 1],
1];Select[myList,(First[First[#]] == First[Last[#]]) &]This
works fine, but when s1 and s2 are large ( e.g. 3000 or
moreelements) then Outer seems inefficient. My question: does
anyone havea suggestion that would be more efficient than my
kludge approach.Note I have tried Intersection, which finds
all the matches, i.e.myList2 = Intersection[s1,s2, SameTest
-> (#1[[1]] == #2[[1]] &)];But I have not been successful in
selecting a