mm-133 =I have the following eigenvalue problem:Let A and B be symmetric positive semi de'nite matrices.The underlying graphs of A and B are trees. (Actually the diagonalelementsare positive and the element A_ij= 0 (B_ij= 0) if vertices i and j arenotadjacent. A_ij= -1 (B_ij= -1) if the vertices i and j are adjacent)Let a, b be the smallest eigenvalues of the matrices A and B,respectively. a and b are simple eigenvalues and the correspondingeigenvectors x and y are positive (Perron-Frobenius Theorem).Ax=axBy=by,We know that 0 < a <= b, y^tAy=b and x^tBx>a.For which matrices A and B hold that a What I am looking for is an iterative algorithm, like : Chris> a[n+1]=f(a[n],...,b[n],...,c[n],...); Chris> b[n+1]=g(a[n],...,b[n],...,c[n],...); Chris> c[n+1]=h(a[n],...,b[n],...,c[n],...); Chris> ... Chris> such that Chris> F(a[k],b[k],c[k],...) for some (not large) k is EllipticE(z,k).Look in netlib for CarlsonOs elliptic integrals. There should be adescription of such an algorithm for Elliptic E.Ray =Correction : What I am looking for is an iterative algorithm, like :a[n+1]=f(a[n],...,b[n],...,c[n],...);b[n+1]=g(a[n],...,b[n], ...,c[n],...);c[n+1]=h(a[n],...,b[n],...,c[n],...);...such thatF(a[m],b[m],c[m],...) for some (not large) m is EllipticE(z,k).Chris =Given a matrix OAO, IOm looking to 'nd (numerically) a matrixOBO such that OA * B = 1O, where O1O is the appropriate unitmatrix.I had expected that one could treat the equation as a set o§inear equations OA * Bj = EjO. According to Numerical Recipes,if one has such a system of equations one can obtain the SVD ofthe matrix OAO: A = U * W * Vtwhere OVtO is the tranpose of OVO. Then the smallest solutionis: Bj = V * WO * Ut * Ej(where the entries in WO are either the inverse of thecorresponding entries in W, or zero if those entries are small).It seems clear to me (which doesnOt mean that itOs true) thatthis means that: B = V * WO * Utis a good choice. In practice, however, where OAO has morecolumns than rows (ie, is singular) IOm 'nding that OA * BO isnot the unit matrix, though itOs close on some rows and columns.This makes no sense to me. IOve con'rmed that the decompositionI get does, in fact, yield the matrix OAO.Is there a better way to go about this? Or should this work (inwhich case I should try another SVD algorithm or debug the oneIOve got)?-- . . . Except when they donOt, Because sometimes they wonOt. - Dr. Seuss-- Jason Cooper jcooper@acs.ucalgary.ca = >Given a matrix OAO, IOm looking to 'nd (numerically) a matrix >OBO such that OA * B = 1O, where O1O is the appropriate unit >matrix. > >I had expected that one could treat the equation as a set of >linear equations OA * Bj = EjO. According to Numerical Recipes, >if one has such a system of equations one can obtain the SVD of >the matrix OAO: > > A = U * W * Vt where OVtO is the tranpose of OVO. Then the smallest solution >is: > > Bj = V * WO * Ut * Ej > >(where the entries in WO are either the inverse of the >corresponding entries in W, or zero if those entries are small). >It seems clear to me (which doesnOt mean that itOs true) that >this means that: > > B = V * WO * Ut > >is a good choice. In practice, however, where OAO has more >columns than rows (ie, is singular) IOm 'nding that OA * BO is >not the unit matrix, though itOs close on some rows and columns. >This makes no sense to me. IOve con'rmed that the decomposition >I get does, in fact, yield the matrix OAO. > >Is there a better way to go about this? Or should this work (in >which case I should try another SVD algorithm or debug the one >IOve got)? -- > . . . Except when they donOt, > Because sometimes they wonOt. - Dr. Seuss > -- >Jason Cooper jcooper@acs.ucalgary.ca if A=U*W*Vt, then W has the same shape as A, hence in your case more columns than rows.then , if A is of full rank, W will consist of an invertiblediagonal matrix with a zero block appended to right.then B = V*W#*Ut where W# has the same shape as AO and consists of the inverted diagonal on the top and a zero block appended at the bottom. then by the rules obviously A*B=I where I has dimension=number of rwos of A.If A is not of full rank, you will get a projector (some diagonalelements in I change to zero) and in W# the reciprocal of zero is set to zero. (B is the Moore-Penrose-pseudoinverse). hence you were right.the problem with this approach is the decision which of the singularvalues (the nonzero elements of W) to consider as small. and if you set anything below eps to zero the deviation of the product from I can be large ifyour decision on the rank of A (implicit with this) was false.hence I assume the problem here and not in the software.hope this helpspeter =This is a question on the relationship between, the zero and non-zeroelements of the transition probability matrix and the correspondingtransient and absorbing states.Let T_a and T_b be the transition probability matrices for two 'nite statediscrete Markov chains A and B, respectively.We are given thata) i(T_a) = i(T_b), where i(X) is the indicator matrix of Xi.e., i(X) has 1.0 entries wherever X has non-zero entries and has 0.0entries wherever X has zero entries.b) T_a^n converges as n -> in'nity. Let the limiting matrix be Ta^{inf}Now, I think I can show that T_b^n also converges. However, I am trying to'nd out if i(Ta^inf) = i(Tb^inf).I have gotten this far:Since i(T_a) = i(T_b), we have i(T_a^2) = i(T_b^2) and so on ...I can show that for any 'nite k, i(T_a^k) = i(T_b^k),but does this property hold in the limit?In other words, if i(T_a) = i(T_b), then do A and B have the same transientand absorbing states?Kumar went looking for a short proof of FermatOs Last Theorem using basic> algebra following my physics training. I expected failures, false> starts, false positives, but mathematicians apparently arenOt> scientists, as they attack me for having them.Wrong! They refuted your arguments. You responded to discovery of your errors by attacking them.> Worse, theyOd use my own admissions of §awed approaches against me,> as if that proved I couldnOt succeed. And worst of all, when I did> succeed they kept up a campaign of personal attacks and smears.Succeed? At what? The only success you have had to date is in establishing that you are a crank and mostof the personal attacks and smears posted in this newsgroup were authored by you.> And itOs not just with the polynomial factorization as thereOs my> prime counting research as well.>> ItOs too much to be believable given what you can see with your own> eyes that none of my work is worth mention, when people can even see> the heat generated on several newsgroups. These people are expending> a LOT of energy, and you think itOs for nothing?The energy others are expending has been to keep the record straight. Your lies and distortions deservechallenges, and will continue to face them.> My work is available 24 hours a day, seven days a week at>> http://groups.msn.com/AmateurMath>> and if it were really so wrong, why do these people keep 'ghting?Obviously, as previously noted, to keep the record straight. DonOt you expect scientists andmathematicians to resist error? Take down that ridiculous treatise. It isnOt 't to wrap the garbage in.> What would happen if mathematicians even acknowledged the> *possibility* that IOm right? They know. Given people following and> keeping up with the details itOs clear IOm right.Clear to whom? You are wrong. Many peoply follow your posts for comic relief. Personally, I liken youto:1) Monty PythonOs Black Knight -- who, having been cut to pieces, insists ItOs only a §esh wound!2) An insect in a collectorOs jar, waving his ineffectual pinchers at the world,3) A stupidly grinning in§atable clown who keeps popping up whenever he is knocked down.> Scientists behave a certain way. Mathematicians are behaving another> way....and you are behaving like an insolent, arrogant and paranoid megalomaniac.> The proof is out there.>> James HarrisYeah, but not on your website. Let us know when you get back from the Twilight Zone with your latestX-File.--It takes a village to raise an idiot.--Democracy: The triumph of popularity over principle.--http://www.crbond.com =IOm dealing with a problem of nonlinear optimization with boundconstraints. I canOt use commercial software for this. So farL-BFGS-B from Netlib has produced satisfactory results. Now Ineed to know the value of the Hessian matrix at the end of thecomputation. L-BFGS-B doesnOt seem to have this information.(The L stands for Limited Memory, meaning that it stores onlyupdates to the Hessian.)Currently IOm looking at DRMNGB, also from Netlib. It seemsas if it stores the Cholesky decomposition of the Hessian.However, IOm having trouble 'guring out how to 'nd it andtransform it back into the full Hessian matrix. On my 'rstattempt the numerical values donOt seem right. Some thingswhich should be positive are not. As far as I can tell,it should be stored beginning at V(IV(42)), but itOs notclear in what order the terms are stored.Does anyone know about this? Or do you have another favoritesubroutine?-- * Patrick L. Nolan ** W. W. Hansen Experimental Physics Laboratory (HEPL) * * Stanford University * = >IOm dealing with a problem of nonlinear optimization with bound >constraints. I canOt use commercial software for this. So far >L-BFGS-B from Netlib has produced satisfactory results. Now I >need to know the value of the Hessian matrix at the end of the >computation. L-BFGS-B doesnOt seem to have this information. >(The L stands for Limited Memory, meaning that it stores only >updates to the Hessian.) > >Currently IOm looking at DRMNGB, also from Netlib. It seems > as if it stores the Cholesky decomposition of the Hessian. >However, IOm having trouble 'guring out how to 'nd it and >transform it back into the full Hessian matrix. On my 'rst >attempt the numerical values donOt seem right. Some things >which should be positive are not. As far as I can tell, >it should be stored beginning at V(IV(42)), but itOs not >clear in what order the terms are stored. > >Does anyone know about this? Or do you have another favorite >subroutine? > >-- >* Patrick L. Nolan * >* W. W. Hansen Experimental Physics Laboratory (HEPL) * >* Stanford University *drmngb is a bfgs based code and will hardly deliver a good approximation to the true Hessian, although as a minimizer it is o.k.why not this one: toms/636 keywords: estimating sparse hessian matrices, difference of gradients gams: G4f title: DSSM and FDHS for: estimating sparse Hessian matrices by: T.F. Coleman, B.S. Garbow, and J.J. More ref: ACM TOMS 11 (1985) 363-377 and 378 Score: 100%hope that helpspeter =are you saying you are feeding the optimizer at best 'rst derivativesand want to get out second derivatives? One way to do it is toevaluate your own with the 'nal solution vector. Any Hessian of theoptimizer will be an approximation at best ('nite difference,quasi-Newton). Only exact or automatic differentiation can give youthe Hessian with suf'cient acuracy.Hans Mittelmann ==> IOm dealing with a problem of nonlinear optimization with bound> constraints. I canOt use commercial software for this. So far> L-BFGS-B from Netlib has produced satisfactory results. Now I> need to know the value of the Hessian matrix at the end of the> computation. L-BFGS-B doesnOt seem to have this information.> (The L stands for Limited Memory, meaning that it stores only> updates to the Hessian.)> > Currently IOm looking at DRMNGB, also from Netlib. It seems> as if it stores the Cholesky decomposition of the Hessian.> However, IOm having trouble 'guring out how to 'nd it and> transform it back into the full Hessian matrix. On my 'rst> attempt the numerical values donOt seem right. Some things> which should be positive are not. As far as I can tell,> it should be stored beginning at V(IV(42)), but itOs not> clear in what order the terms are stored.> > Does anyone know about this? Or do you have another favorite> subroutine? =>Try the family:> (x/a)^n + (y/b)^n = 1>with n > 2 . Of course, n=2 is an ellipse centered at the origin.I forgot to mention one important thing. I need an ellipse-like curvethat is not symmetric in the way an ellipse (or the curve above) is. =>>Try the family: (x/a)^n + (y/b)^n = 1with n > 2 . Of course, n=2 is an ellipse centered at the origin.>>I forgot to mention one important thing. I need an ellipse-like curve>that is not symmetric in the way an ellipse (or the curve above) is.Like a longitudinal cut of an egg, or an etc,likehttp://www.mathematische-basteleien.de/eggcurves.htm= IOve done my part with my mathematical research by working hard toexplain it, arguing with people over details, writing a paper andsending it to math journals. And now I realize that part of theproblem is that the math is revolutionary.I use methods to factor polynomials into non-polynomial factors, whichis such a powerful approach that it leads to a short proof of FermatOsLast Theorem that is just a few pages.The math is mostly basic algebra.IOve given the information. IOve explained in detail butmathematicians can resist for a while, just as they resisted otherrevolutionary mathematics, but IOm hopeful that here at least *some*of you recognize that itOs not to the bene't of math society to actin that way.What does it take to consider the mathematics?Reviewing the math at my website, or considering the many posts whereI talk about the factorization (v^3+1)x^3 - 3vx + 1 = (a_1 x + 1)(a_2 x + 1)(a_3 x + 1)or variations on it.Now you can actually look at the factorization for v=1, and see thatas IOve stated for some time, ONLY TWO of the aOs can have a factor of2, and that factor is sqrt(2).The polynomial of course is 2x^3 - 3x + 1, and itOs easily factorable.My work gave a speci'c prediction. I can show you where thatprediction is the reality with a reducible polynomial. The methodsthemselves DO NOT CARE about reducibility, so they work when v givespolynomials not reducible over Q.many of you realize that if mathematicians refuse to do their part,then my task is dif'cult, even though IOm right.Factoring polynomials in a special way such that even the FLT equationitself can be factored, as weird as that may sound, is a great ideawhose time has come. Your task now is simply to accept the future,just as those before you accepted imaginary numbers, and thosebefore them accepted irrational numbers.You may be puzzled. You may not understand why thereOs so much drama. But then now maybe you can understand other math history a littlebetter. People like comfort. Thinking you know all you need to knowin certain areas is comforting. And then some guy comes along andstarts a revolution.I am that guy.James Harris =LOL.> LOL.Yeah, it may be funny to you, but what IOm talking about is a modernsituation where a mathematical argument given is simply ignored orlied about by mathematicians.2x^3 - 3x + 1 = (x-1)(2x^2 +2x -1) =(a_1 x + 1)(a_2 x + 1)(a_3 x + 1)My work has speci'c predictions, which are the mathematical reality.Mathematicians 'ghting work that is simple, demonstrable, and correctcanOt be good for the discipline.And itOs not funny.James Harris work that is simple, demonstrable, and correct> canOt be good for the discipline.>> And itOs not funny.>> James HarrisNo one is laughing at Mathematicians 'ghting work that is simple demonstrable, and correct. Thesubject of discussion is your error-ridden, false, oft-refuted, proof of FLT. They are laughing at*you*, idiot.--It takes a village to raise an idiot.--Democracy: The triumph of popularity over principle.--http://www.crbond.com Mathematicians 'ghting work that is simple, demonstrable, and correct> canOt be good for the discipline.>> And itOs not funny.>> James Harris> > No one is laughing at Mathematicians 'ghting work that is simple demonstrable, and correct. The> subject of discussion is your error-ridden, false, oft-refuted, proof of FLT. They are laughing at> *you*, idiot.Is that the math world? Is that what all of you see when you think ifmathematicians? People calling others idiot?IOm on whatOs supposed to be *your* turf. Basic algebra and astep-by-step argument which proves what IOm saying.Ultimately people attacking me are attacking mathematics itself.Now think about it. ThereOs my prime counting work, then thereOs mypaper, and also not least thereOs the actual short proof of FermatOsLast Theorem.If you believe people arguing with me thereOs only ONE small thingthat they 'ght.Think about it. Basic algebra proves my point. The argument isstraightforward and simple, but the consequences are huge.ItOs revolutionary for *social* reasons. And you can see society,your society, 'ghting as it has before. Learn the lessons from thebattles over irrational and imaginary, and try to let the math beyour guide--not your comfort zone.The work itself is conveniently available to you 24 hours a day, sevendays a week at http://groups.msn.com/AmateurMathand hey, you can even play with an applet (if you join the group).James Harris =documentstyle{report}begin{document}pagestyle{plain }raggedbottomdef half {frac{1}{2}}def hieruit {quad Longrightarrow quad}%subsection*{Should be the same but alas ...}%Consider the following integral:$$ int_{-half a}^{+half a} x^2 , dx / a = left[ frac{1}{3} x^3 right]_{-half a}^{+half a} / a = frac{2.a^3}{3.8.a} = frac{a^2}{12}$$Now consider:$$ lim_{sigma rightarrow infty} frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }$$We would expect that this more dif'cult integral nevertheless convergesto the same result as before: $a^2 / 12$ , because:$$ lim_{sigma rightarrow infty} e^{-half x^2 / sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[ x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} + sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } right)$$It should be correct to anticipate that $a<....................> So the two outcomes have the same order of magnitude, but they do not match. > WhatOs going wrong ? Please help !> %following .tex 'le, documentstyle[leqno]{report}newcommand{hsp}{hspace*{0.5cm}} newcommand{Erf}{{rm Erf}}begin{document}pagestyle{plain} raggedbottomdef half {frac{1}{2}}def hieruit {quad Longrightarrow quad}%subsection*{Should be the same but alas ...}%Consider the problem to determinebegin{equation}label{eu11} fbox{$displaystyle L(a):=lim_{sigma rightarrowinfty}I(a,sigma)hsp mbox{rm with} hsp I(a,sigma)= frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } $}end{equation}However:$$ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half a} x , e^{-half x^2 /sigma^2} , dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[ x.e^{-half x^2 / sigma^2} right]_{-halfa}^{+half a} + sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,dx$$ $$ hieruit hsp I(a,sigma)=frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }right)$$Let us de'ne $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,; h to0; .; $ Then problem (ref{eu11}) is equivalent withbegin{equation}label{eu12}fbox{$displaystyle L(a)=frac{a^2}{8}limlimits_{hto0}frac{1}{h^2}left(1-frac{he^{ -h^2}}{Erf(h)}right) hspmbox{rm where}hsp Erf(h):=intlimits_{0}^h e^{-t^2}; dt$}hsp .end{equation}But $displaystyle Erf(h)=he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $ where[{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k }cdotfrac{z^k}{k!}hsp,hsp (a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]Therefore [L(a)=frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=1}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}{h^ 2cdotsumlimits_{k=0}^infty frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=0}^inftyfrac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k }}{cdotsumlimits_{k=0}^inftyfrac{(1)_k}{(3/2)_k}h^{2k}}=frac{ a^2}{8}cdotfrac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12}; .]%end{document} => >....................> So the two outcomes have the same order of magnitude, but they do not match.> WhatOs going wrong ? Please help !> %> following .tex 'le,> documentstyle[leqno]{report}> newcommand{hsp}{hspace*{0.5cm}}> newcommand{Erf}{{rm Erf}}> begin{document}> pagestyle{plain} raggedbottom> def half {frac{1}{2}}> def hieruit {quad Longrightarrow quad}> %> subsection*{Should be the same but alas ...}> %> Consider the problem to determine> begin{equation}> label{eu11} fbox{$displaystyle L(a):=lim_{sigma rightarrow> infty}I(a,sigma)hsp mbox{rm with} hsp I(a,sigma)= frac{> int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx }> { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } $}> end{equation}> However:> $$> int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx => - sigma^2 int_{-half a}^{+half a} x , e^{-half x^2 /> sigma^2} ,> dleft(-half x^2 / sigma^2right) => $$ $$> - sigma^2 int_{-half a}^{+half a} x ,> dleft(e^{-half x^2 / sigma^2}right) => - sigma^2 left[ x.e^{-half x^2 / sigma^2} right]_{-half> a}^{+half a}> + sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} ,> dx> $$ $$> hieruit hsp I(a,sigma)=frac{> int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx }> { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } => sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} }> { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }> right)> $$> Let us de'ne $ h=frac{a}{2sigmasqrt{2}}; ,; h>0; ,; h to> 0; .; $ Then problem (ref{eu11}) is equivalent with> begin{equation}label{eu12}> fbox{$displaystyle L(a)=frac{a^2}{8}limlimits_{hto> 0}frac{1}{h^2}left(1-frac{he^{-h^2}}{Erf(h)}right) hsp> mbox{rm where}hsp Erf(h):=intlimits_{0}^h e^{-t^2}; dt> $}hsp .> end{equation}> But $displaystyle Erf(h)=> he^{-h^2}{}_1F_1left(1;frac{3}{2};h^2right) $ where> [{}_1F_1(a;b;z)=sumlimits_{k=0}^{infty}frac{(a)_k}{(b)_k} cdotfrac{z^k}{k!}hsp> ,hsp (a)_k=a(a+1)cdots(a+k-1)=frac{Gamma(a+k)}{Gamma(a)}]> Therefore> [L(a)=frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=1}^infty> frac{(1)_k}{(3/2)_k}h^{2k}}> {h^2cdotsumlimits_{k=0}^infty frac{(1)_k}{(3/2)_k}h^{2k}}=> frac{a^2}{8}limlimits_{hto 0}frac{sumlimits_{k=0}^infty> frac{(1)_{k+1}}{(3/2)_{k+1}}h^{2k}}> {cdotsumlimits_{k=0}^infty> frac{(1)_k}{(3/2)_k}h^{2k}}=frac{a^2}{8}cdot> frac{frac{(1)_1}{(3/2)_1}}{1}=frac{a^2}{12}; .> ]> %> end{document}I would like to read LaTex 'les. Is there some Windows software out therethat works immediately without having to run numerous fonts setup programscomputer. = [ deleted ]Here is your answer, as simple as I can make it:setlength{mathindent}{1.0cm}pagestyle{plain} raggedbottomdef half {frac{1}{2}}def hieruit {quad Longrightarrow quad}begin{document}%subsection*{Should be the same but alas ...}%Consider the following integral:$$ int_{-half a}^{+half a} x^2 , dx / a = left[ frac{1}{3} x^3 right]_{-half a}^{+half a} / a = frac{2.a^3}{3.8.a} = frac{a^2}{12}$$Now consider:$$ lim_{sigma rightarrow infty} frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx }$$We would expect that this more dif'cult integral neverthelessconverges to the same result as before: $a^2 / 12$ , because:$$ lim_{sigma rightarrow infty} e^{-half x^2 / sigma^2} = 1$$However:$$ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx = - sigma^2 int_{-half a}^{+half a} x , e^{-half x^2 / sigma^2} , dleft(-half x^2 / sigma^2right) =$$ $$ - sigma^2 int_{-half a}^{+half a} x , dleft(e^{-half x^2 / sigma^2}right) = - sigma^2 left[ x.e^{-half x^2 / sigma^2} right]_{-half a}^{+half a} + sigma^2 int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx$$ $$ hieruit frac{ int_{-half a}^{+half a} x^2 , e^{-half x^2 / sigma^2} , dx } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } = sigma^2 left( 1 - frac{ a.e^{-half (a/2)^2 / sigma^2} } { int_{-half a}^{+half a} e^{-half x^2 / sigma^2} , dx } right)$$It should be correct to anticipate that $a< > I would like to read LaTex 'les. Is there some Windows software out there> that works immediately without having to run numerous fonts setup programs> computer.Subscribe to the newsgroup comp.text.texand see what people there are asking about and recommending.You will probably want to install the MikTeX distribution, available from http://ftp.uci.agh.edu.pl/pub/tex/systems/win32/miktex/I strongly also recommend getting WinEdt, available from http://www.winedt.com/It is shareware, but worth the money ($40, or so). It is a LaTeX andTeX -aware editor that makes using these systems easy (I suggest LaTeXfor your own work).The book A Guide to LaTeX by Kopka & Daly is extremely helpful.Good luck.-- Julian V. NobleProfessor Emeritus of ^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo.