mm-110
.
Please help to solve these two problems. One member
have posted this problem to a math group on msn. I have no
idea to solve thisProblems are going like this:1. DigitsShow
that for any natural n, at least one of two numbers, n or
n+1,can berepresented in the following form: k + S(k) for a
certain k, whereS(k) isthe sum of all digits in k. For
instance, 21 = 15 + (5+1)2. Party!There is a group of people
at a party. Show that you can introduce some of them to each
other so that after the introduction, no more than
two people in the group would have the same number of friends
(initial conguration doesnt work because they all initially
have 0 friends).Ajhar
=I am posting a code which is written
in c++ which canfind the bonacci seriesup to any n number
.The code is like this://starting of the
program#inclue<iostream.h>void main(){int
number,f1=0,f2=1,temp=0;cout<<Enter a number to which you
wants to show the bonacci series: ;cin>>number;//the main
logic is going like thiscout<<Fibonacci series starts
here;cout<<n<<f2;for(int
i=0;i<=number;i++){temp=f1+f2;cout<<n<<temp;f1=f2;f2=temp;}}/
/end of programyou can run this program on turbo c++ or vc++
compilerAjharhttp://roxvary.netrms.com
Im trying to
write a program in VBA to solve a series of non-linear>
equations. Unfortunately, I hardly know anything about
numerical> analysis. So if you hava a program or idea how to
solve it please> share with me.> The problem is the
following:> C(1)=sum(j=1 to n){alpha(j,1)beta(j)prod(i=1 to
k)[c(i)^alpha(j,i)]}> ...> C(k)=sum(j=1 to
n){alpha(j,k)beta(j)prod(i=1 to k)[c(i)^alpha(j,i)])> where >
prod(i= 1 to
k)c(i)^alpha(j,i)=c(1)^alpha(j,1)*c(2)^alpha(j,2)...> I
need to solve these equations for c(i). The alpha matrix is
such> that the rst kxk part is a unity matrix.> Anybody, any
suggestions?> greetings:> ZsoltI can do programming but i
doest quit understand your problem will you explain your
problem?Ajhar
Im somewhat versed in math but no
expert by far. Ive been browsing the> internet, news-groups
and all, but still I have not found a good answer to> my
question.> Im looking for a Java (J2ME) implementation (or
something from which i can> create this implementation) for
the approximations (only using additions,> subtractions,
multiplications, divisions and mayb a square-root function
and> logarithm) for one or both of these two functions:>
exp(x) e^x (or pow(2,y), ... raising the power of 2 on>
computers can be faster)> and/or> pow(x,y) x^y> where both
x and(!) y can be real numbers (not necessarily integers).>
The approximations need to be relatively fast (J2ME,
MIDP1.0... limited> device capabilities, no oating-point
support) but quite accurate.> Ive already have good
implementations for the log(x) (using the Maclaurin> series
for 1/(ln[x+1])) and squareroot(x) (using Newtons
Iteration).> I need the approximations where y is between 0
and 1 and where y is a large> number:> (e^x =
e^(largenum+fraction) = e^largenum * e^fraction)> PS: The
x and y are real numbers represented in J2ME by a class
representing> some form of xed-point values.
approximate exp(x)
=In the following Q=ln(2)=
.69314718055994530942... and symbol [.] is devoted for
integral part. Also x=[x]+{x}where {x} is fractionary part of
x. First observe that
2^{a+N}e^{Z(a)}where N=integer:=[x/Q] , F:={x/Q} ,
Z(a)=(F-a)*Q ,a being an arbitrary real
number.
=Please verify
(1): take logarithm in both sides.Suppose x >0 , let q be a
positive integer and put k=[q*F] . Ifk/q =< F={x/Q} < (k+1)/q
, select parameter a to be (2k+1)/(2q) . Then |Z(a)| =< Q/8 <
1/10 . Therefore using (1)(2)
e^x=approx.=C(k)*2^N*G(Z(A))where G(z) will be a ,,good
approximation of e^z for z small, e.g. for |z|<1/10 .==In the
following I give such approximation G(z). Dene(*) A(z)=z^6 +
840*z^4 + 75600*z^2 +665280 B(z)= 42*z^4 + 10080*z^2 + 332640
H(z):= A(z)+ z*B(z). Then put
= (3) G(z) =
H(z)/H(-z) .
= It may be shown that |(e^z
- G(z))/e^z| =< 10^{-26} for |z|< Q/8 .In conclusion , tofind
an approximation DEX(X) of e^X=exp(X) you can proceed as
follows ( q=4 ) : STEP 1: Test the variable X , and determine
T=|X| . Suppose that you work with real numbers in intervals
(16^{-64},16^{63}). This means that you can approximate e^X
only for X in [-175,175] .Dene (for instance) DEX(T)=
.0996...E-76 if T =< -175 , DEX(T)=1 when T=0 and DEX(T)=
.10035...E+77 when T>= 175 .STEP 2: Find N=[T/Q] , F={T/Q}
.STEP 3: Find P=2^N , K=[4F] , A=(2K+1)/8 , Z=(F-A)Q .STEP 4:
Select constant C , C:= 2^A , as a function of possible values
of K ,K in{0,1,2,3} .STEP 5: Taking into account (3), consider
approximation e^Z=approx = H(Z)/H(-Z) .STEP 6: Finding nal
approximation, namely e^T=approx=C*P*G(Z)and DEX(X)=
1/(C*P*G(Z)) if Z is in (-175,0) , or DEX(X) = C*P*G(z) when
X belongs to interval (0,175) . Note: Its possible to prove
that the roots z_k of H(z) satisfy 2 =< |z_k| =< 42
,therefore H(-z) =/=0 for |z| small ,e.g |z|< 2 . Other
better functions that (*) are available.If you want tofind
betterapproximations for e^z , please inform me. I appreciate
that (1) was important.
=Those who cant design,
teach.Those who cant teach, design.And the rest are just
clueless.Dan :-)
And I keep stressing that I am talking
about the U.S., of which youve> admitted some lack of
knowledge.I said that I am not too familiar with US based
research grants. Ialso said that I have no idea what the Fox
news fan club is supposedto be. That is is where I admitted
some lack of knowledge about theUSA, as you say.However I
also said that: I know very well what publish or perish
is(unjustied patronizing tone again, I note), William. I
have somescientic publications, William, and we have pretty
much the sameacademic system here in Israel, as in the USA
(including gettingIn fact Ive spent some time in the USA
academic system, in FortCollins C0. I am also quite familiar
with the UK academic system, inaddition to the Israeli, of
course. So, when I talk about the peoplefrom the academia, I
know, it means quite a lot of people, from threedifferent
countries, but having similar academic systems. That
doesntmake my statements absolute, as you say. But, IMHO,
they aresignicant enough in countering statements
like:...but they often cant do any signicant research
either. (Dependsreply to my: A Prof with a tenure that
doesnt have much contractNote your rather strong words often
and any. As to what looks likean insurance policy (or a sort
of disclaimer) of Depends on...,well, it doesnt make much
sense that the academic system and academicfreedom concepts
should be signicantly different in, say chemistrydept from
that of math, or biology. Even in the great USA. Well, ifsome
mysterious reports they send back from the front lines...
givea different picture, I guess, I am expected to accept
them as factsabout the USA. Well, I might not do that,
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hA6KGK001939;
=Im desperately
looking for an algorithm to put a reduciblenonnegative matrix
in Frobenius canonical form by a symmetricpermutation of rows
& columns. In graph-theoretic terms, this amountsto
relabeling the graph nodes so that the absorbing nodes (or
groups onodes) receive the highest labels. I will really
appreciate it ifanyone out there can lend me a hand.
Im desperately looking for an algorithm to put a reducible>
nonnegative matrix in Frobenius canonical form by a
symmetric> permutation of rows & columns. In graph-theoretic
terms, this amounts> to relabeling the graph nodes so that
the absorbing nodes (or groups o> nodes) receive the highest
labels. I will really appreciate it if> anyone out there can
lend me a hand.Look at Dulmage-Mendelsohn decomposition.For
example in Matlab, help dmpermArnold Neumaier
I am a
student implementing a simple mode solver myself,> I need
help.>> The one dimensional Helmholtz equation>
d^2U(x)/dx^2+k^2(N(x)^2-n_eff^2)U(x)=0>> can be solved using
Finite Difference method:czh, lets have an example of this
BVP (boundary value problem).k^2 = 1.234^2N(x) = (x +
0.002345)n_eff^2 = 2.453x_0 = 0x_40 = 10n = 39y_0 = 0y_40 =
-1.435h = 0.250Solution: y_1 to y_39Use plausible data or
modify and solve these equations using anappropriate
program.>>
d^2U(x)/dx^2+k^2(N(x)^2-n_eff^2)U(x)=0((y_2-2*y_1+y_0)/h^2)+
1.234^2*(((1*h)+0.002345)^2-2.453)*(y_1)=0((y_3-2*y_2+y_1)/h^
2)+1.234^2*(((2*h)+0.002345)^2-2.453)*(y_2)=0((y_4-2*y_3+y_2)
/h^2)+1.234^2*(((3*h)+0.002345)^2-2.453)*(y_3)=0((y_5-2*y_4+y
_3)/h^2)+1.234^2*(((4*h)+0.002345)^2-2.453)*(y_4)=0((y_6-2*y_
5+y_4)/h^2)+1.234^2*(((5*h)+0.002345)^2-2.453)*(y_5)=0((y_7-2
*y_6+y_5)/h^2)+1.234^2*(((6*h)+0.002345)^2-2.453)*(y_6)=0((y_
8-2*y_7+y_6)/h^2)+1.234^2*(((7*h)+0.002345)^2-2.453)*(y_7)=0(
(y_9-2*y_8+y_7)/h^2)+1.234^2*(((8*h)+0.002345)^2-2.453)*(y_8)
=0((y_10-2*y_9+y_8)/h^2)+1.234^2*(((9*h)+0.002345)^2-2.453)*(
y_9)=0((y_11-2*y_10+y_9)/h^2)+1.234^2*(((10*h)+0.002345)^2-
2.453)*(y_10)=0((y_12-2*y_11+y_10)/h^2)+1.234^2*(((11*h)+
0.002345)^2-2.453)*(y_11)=0((y_13-2*y_12+y_11)/h^2)+1.234^2*(
((12*h)+0.002345)^2-2.453)*(y_12)=0((y_14-2*y_13+y_12)/h^2)+
1.234^2*(((13*h)+0.002345)^2-2.453)*(y_13)=0((y_15-2*y_14+y_
13)/h^2)+1.234^2*(((14*h)+0.002345)^2-2.453)*(y_14)=0((y_16-2
*y_15+y_14)/h^2)+1.234^2*(((15*h)+0.002345)^2-2.453)*(y_15)=0
((y_17-2*y_16+y_15)/h^2)+1.234^2*(((16*h)+0.002345)^2-2.453)*
(y_16)=0((y_18-2*y_17+y_16)/h^2)+1.234^2*(((17*h)+0.002345)^2
-2.453)*(y_17)=0((y_19-2*y_18+y_17)/h^2)+1.234^2*(((18*h)+
0.002345)^2-2.453)*(y_18)=0((y_20-2*y_19+y_18)/h^2)+1.234^2*(
((19*h)+0.002345)^2-2.453)*(y_19)=0((y_21-2*y_20+y_19)/h^2)+
1.234^2*(((20*h)+0.002345)^2-2.453)*(y_20)=0((y_22-2*y_21+y_
20)/h^2)+1.234^2*(((21*h)+0.002345)^2-2.453)*(y_21)=0((y_23-2
*y_22+y_21)/h^2)+1.234^2*(((22*h)+0.002345)^2-2.453)*(y_22)=0
((y_24-2*y_23+y_22)/h^2)+1.234^2*(((23*h)+0.002345)^2-2.453)*
(y_23)=0((y_25-2*y_24+y_23)/h^2)+1.234^2*(((24*h)+0.002345)^2
-2.453)*(y_24)=0((y_26-2*y_25+y_24)/h^2)+1.234^2*(((25*h)+
0.002345)^2-2.453)*(y_25)=0((y_27-2*y_26+y_25)/h^2)+1.234^2*(
((26*h)+0.002345)^2-2.453)*(y_26)=0((y_28-2*y_27+y_26)/h^2)+
1.234^2*(((27*h)+0.002345)^2-2.453)*(y_27)=0((y_29-2*y_28+y_
27)/h^2)+1.234^2*(((28*h)+0.002345)^2-2.453)*(y_28)=0((y_30-2
*y_29+y_28)/h^2)+1.234^2*(((29*h)+0.002345)^2-2.453)*(y_29)=0
((y_31-2*y_30+y_29)/h^2)+1.234^2*(((30*h)+0.002345)^2-2.453)*
(y_30)=0((y_32-2*y_31+y_30)/h^2)+1.234^2*(((31*h)+0.002345)^2
-2.453)*(y_31)=0((y_33-2*y_32+y_31)/h^2)+1.234^2*(((32*h)+
0.002345)^2-2.453)*(y_32)=0((y_34-2*y_33+y_32)/h^2)+1.234^2*(
((33*h)+0.002345)^2-2.453)*(y_33)=0((y_35-2*y_34+y_33)/h^2)+
1.234^2*(((34*h)+0.002345)^2-2.453)*(y_34)=0((y_36-2*y_35+y_
34)/h^2)+1.234^2*(((35*h)+0.002345)^2-2.453)*(y_35)=0((y_37-2
*y_36+y_35)/h^2)+1.234^2*(((36*h)+0.002345)^2-2.453)*(y_36)=0
((y_38-2*y_37+y_36)/h^2)+1.234^2*(((37*h)+0.002345)^2-2.453)*
(y_37)=0((y_39-2*y_38+y_37)/h^2)+1.234^2*(((38*h)+0.002345)^2
-2.453)*(y_38)=0((y_40-2*y_39+y_38)/h^2)+1.234^2*(((39*h)+
0.002345)^2-2.453)*(y_39)=0These equations are computer
generated.--Website temporarily closed
Can someone
point me to information on the web about fast factorial>>
calculation? This needs to be for exact value. I can handle
the part>about>> it being too large to represent, but would
like tofind a faster method>than>> just multiplying every
number.>> Adam,>> Try this one
http://www.luschny.de/math/index.htm.>> Could tell what kind
of application are you working for?>>Its labview. NI is
sporting a contest to code the fastest factorial>program. I
have a few ideas on how to calculate numbers larger
than>representable in normal computer formatting, but wanted
to see what>algorithms might I use that would be quicker than
2x3x4... Just for funs>tho.>>An interesting property is that
for n = 2m,n! = 2^m (m!)^2Another interesting point I
developed some time ago,
=fr&lr=&ie=UTF-8&selm=956tbb%24rg8%40deadzone.rsn.hp.com&rnum
=3Michel
> Can someone point me to information on
the web about fast factorial>> calculation? This needs to
be for exact value. I can handle thepart>about>> it
being too large to represent, but would like tofind a
fastermethod>than>> just multiplying every number.> Adam,>> Try this one
http://www.luschny.de/math/index.htm.>> Could tell what
kind of application are you working for?>>Its labview.
NI is sporting a contest to code the fastest factorialprogram. I have a few ideas on how to calculate numbers
larger than>representable in normal computer formatting,
but wanted to see what>algorithms might I use that would be
quicker than 2x3x4... Just for funs>tho.>> An
interesting property is that for n = 2m,> n! = 2^m (m!)^2I
used to think of myself as somewhat of a math wiz but Im not
getting thisequation. Cant seem to make it balance for sample
numbers. In fact lookingat {n,m}={4,2}, I dont see how it
could work at all. 4! contains a factorof 3 which I dont see
the right side coming up with no matter how I look atit. Can
you clue me in as to what I am missing?
=I suspect for this
homework assignment that the original poster was supposed to
stumble across the Gamma function and, more importantly for
large numbers, the LogGamma function.$.02 -Ron Shepard
I
suspect for this homework assignment that the original poster
was> supposed to stumble across the Gamma function and, more
importantly> for large numbers, the LogGamma function.No, it
really isnt a homework assignment. It is a labview
programmingcontest (for fun). The point is to code it to do
up to 10000 factorialquickly. Do you think that use of the
loggamma function would be a moreefcient algorithm? Does it
produce exact answers?
=Adam Russell schrieb> The point is
to code it to do up to 10000 factorial> quickly.If you want
to compare your code with my Javaimplementation, go
here:<http://www.luschny.de/math/factorial/
FastFactorialFunctions.htm>and then click on benchmark.Java
Factorial Benchmark - Timings(in seconds)N! where N =
16000/32000/64000/128000/256000/512000/1024000PrimeSwing 0,2/
0,8/ 3,8/ 17,6/ 97,3/ 524,9/ 2480,1So for 10000! you can
expect 0.1 seconds with Java,computing on a PC, which is two
years old. Fast enough?> Do you think that use of the
loggamma function would be a more> efcient algorithm? Yes,
but...> Does it produce exact answers?No. Only an
approximation.
=Michel OLAGNON schrieb> An interesting
property is that for n = 2m,> n! = 2^m (m!)^2Sure? I tried
Maple:OLAGNON := proc(m) n := 2*m; m!^2*2^m
end;seq(OLAGNON(i),i=0..7);1, 2, 16, 288, 9216, 460800,
33177600, 3251404800seq(i!,i=0..7);1, 1, 2, 6, 24, 120, 720,
5040Ups. But I know what you mean. If you read carefullythe
function given above, you willfind:46 Integer
recFactorial(int n)47 {48 if (n < 2) return 1;49 return
(recFactorial(n/2)^2) * swing(n);50 }This is a correct, more
general and recursive formulationof the observation you
obviously refer to.Gruss Peter
>46 Integer
recFactorial(int n)>47 {>48 if (n < 2) return 1;>49 return
(recFactorial(n/2)^2) * swing(n);>50 }>>This is a correct,
more general and recursive formulation>of the observation you
obviously refer to.Very interesting for n=4. What does the
swing do?-- Surendar Jeyadev jeyadev@wrc.xerox.bounceback.com
n)>47 {>48 if (n < 2) return 1;>49 return
(recFactorial(n/2)^2) * swing(n);>50 }> Very interesting
for n=4.Not more and not less interesting then for anyother
n. y := recFactorial(n/2)^2< y := recFactorial(n/2); y :=
y^2;> What does the swing do?How much of the thread do you
read, before you write?
Michel OLAGNON schrieb> An
interesting property is that for n = 2m,>> n! = 2^m
(m!)^2>>Sure? I tried Maple:>OLAGNON := proc(m) n := 2*m;
m!^2*2^m end;>>seq(OLAGNON(i),i=0..7);>1, 2, 16, 288, 9216,
460800, 33177600, 3251404800>>seq(i!,i=0..7);>1, 1, 2, 6, 24,
120, 720, 5040>>Ups. But I know what you mean. If you read
carefully>the function given above, you willfind:>>46 Integer
recFactorial(int n)>47 {>48 if (n < 2) return 1;>49 return
(recFactorial(n/2)^2) * swing(n);>50 }>>This is a correct,
more general and recursive formulation>of the observation you
obviously refer to.>Yes, indeed, I had not meant to solve the
whole thing, justto give a hint for even n, and let the OP
nd out how touse it for any n.Michel
=Hallo Liste,ich
hoffe es ist ok, wenn ich nicht in English schreibe?Der
Simple Algorithmus f.9fr station.8are Str.9amungen besteht
aus folgendenTeilen:1) L.9asen der Impulsgleichung, u_n+1 =
f(u_n) (n.8achsterIterationsschritt u ist eine Funktion vom
Vorg.8anger)2) Massenquelle berechnen aus der
Konti-Gleichung3) Druckkorrektur4)
Geschwindigkeitskorrekturen anpassen5) siehe 1)Ich habe nun
folgendes Problem:Ich m.9achte zum L.9asen der
Impulsgleichung ausschlielich die nichtlinearen Terme als
Iterationsvorg.8anger benutzen.z.B.: u_n+1= (u_n) + ...Dies
hat nur leider (bei mir) zur Folge, da
dieDruckkorrekturgleichungen nicht mehr linear anb.8angig
voneinander sind.Mir wurde erz.8ahlt, da diese Variante zum
L.9asen der Impulsgleichungsehr wohl konvergieren w.9frde.Was
hat man allerdings nun dabei zu beachten, insbesondere bei
derDruckkorrekturgleichung, damit dieses Verfahren auch das
macht, was essoll?Ist diese Art der Diskretisierung falsch
der Impulgleichung falsch?GruKai
=I have searched for C/C++
source code for calculating thechisquare_inv function, but
with no success. Does anyone know where tosearch?
I
have searched for C/C++ source code for calculating the>
chisquare_inv function, but with no success. Does anyone know
where to> search?> - Function File: chisquare_inv (X, N)>
For each element of X, compute the quantile (the inverse of
the> CDF) at X of the chisquare distribution with N degrees
of> freedom.> take a lool at the GSL-library and its inverse
of theChi-square
functionhttp://sources.redhat.com/gsl/ref/gsl-ref_19.html#
SEC301Hope that helps.Axel
=I have a function of the
form:t(p,q,c) = {Xp + Yq + Zc if t < Tmax {Tmax otherwiseand
a relationq = t^{-1}(p,c)where t^{-1}(p,c) is the inverse of
t(p,q,c) in qApparently t^{-1}(p,c) can be calculated
numericallyusing the Newton-Raphson method but I cant seehow
it is applied. Any hints or references
appreciated.rgdsrob
=However, all these techniques assume
the matrix is NON-singular, has therebeen any generalization
to singular matrices, but still without utilizingthe
transpose of the matrix in the calculations?i.e. the only
operation required is to supply a routine that generates
A*xfor a certain x.Alien+> I have been looking for techniques
based on GMRES or Transpose-Freemethods> like CGS/QMR that can
be utilized for singular matrices. I understand all> these
techniques require nonsingular matrices. Any one knows if
these> techniques have been extended to singular and/or very
ill-conditioned> systems?> Alien+>
However, all
these techniques assume the matrix is NON-singular, has
there> been any generalization to singular matrices, but
still without utilizing> the transpose of the matrix in the
calculations?> i.e. the only operation required is to supply
a routine that generates A*x> for a certain x.> Alien+>
I have been looking for techniques based on GMRES or
Transpose-Free> methods> like CGS/QMR that can be utilized
for singular matrices. I understand all> these techniques
require nonsingular matrices. Any one knows if these>
techniques have been extended to singular and/or very
ill-conditioned> systems?Applying iterative methods to
singular or ill-conditioned systems usually has a
regularizing effect, irrespective of the method; nothing
special needs to be done. One stops when the residual is
slightly above the expected noise level in theright hand side
(or with more sophisticated methods like L-curves)Arnold
Neumaier
> I would want to point out that the main
advantage for utilizing QMR/CGS,>> etc. is that they avoid
the utilization of the Transpose,> Not QMR. For Gmres or
BiCGstab (much better than CGS) youre right.There is a
transpose-free QMR (TFQMR) available in qmrpack on
Netlib.
There is a transpose-free QMR (TFQMR) available
in qmrpack on Netlib.Which is not as the name suggests an
implementation of QMR without usingtransposes, but rather CGS
with residual smoothing (Walker and Zhou,Weiss) applied to
it.V.-- homepage: cs utk edu tilde lastname
Could
somebody up there help me out by plugging>the following into
Mathematica or equivalent and>>(I m sure there is a result
because the Mathworld Integrator>gives me one for the
indenite case - but its a bit
messy)>>Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2 +
2*Cos[x]]],{x,0,Pi}]>I know this is more of a sci.math
question, but overthere i m not>getting a lot of
response...Yes, well in sci.math.num-analysis you should
expect to get informationabout to evaluate the integral
numerically (for any xed D ), notsymbolically. For that, you
should turn to sci.math.symbolic.To work symbolically it may
be easier to get rid of the trig:use the half-angle
substitutions cos(x) = (1-t^2)/(1+t^2) andsin(x) = 2t/(1+t^2)
(and use d sin(x) = cos(x) dx to concludedx = 2dt/(1+t^2) ) to
write this as the integral over (0, infty) of
2(1-t^2)/(1+t^2)^2 * log( 2t^2/(1+t^2) + sqrt( D^2 +
4/(1+t^2) ) )You can further eliminate the logarithms by
doing integration by parts.This then leaves you with a purely
algebraic integrand.For almost all D, the equation D^2 +
4/(1+t^2) = u^2 describes aRiemann surface of genus 1, so at
best you should expect the antiderivativeto involve elliptic
functions. Thats likely to be a mess as you say.(Though
since you are only interested in a denite integral, you
maybe able to compute it with residues; I didnt try.) In any
event,are you sure thats going to be useful? Might it not,
perhaps, besimpler to call the integral F(D) and then deduce
whateverinformation you need about the function F straight
from its denitionof as an integral?dave
=alex
<heldring@voltor.upc.es> schrieb im Newsbeitrag> Could
somebody up there help me out by plugging> the following into
Mathematica or equivalent and>> (I m sure there is a result
because the Mathworld Integrator> gives me one for the
indenite case - but its a bit messy)>>
Integrate[Cos[x]*Log[-Cos[x] + 1 + Sqrt[D^2 + 2²
+2*Cos[x]]],{x,0,Pi}]Alex, numerially solved using Gaussian
Quadrature Integration forD = 1.23456f = Cos(x) * Log(-Cos(x)
+ 1 + Sqrt(D ^ 2 + 2 + 2 * Cos(x)))Integral =
0.499159841831736--