"Operator not implemented (SERIES)"
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From: acl on 18 Sep 2006 11:45 Chris Smith wrote: > acl <achilleaslazarides(a)yahoo.co.uk> wrote: >>Hi Ben, >>Out of curiosity, what do you expect this limit to be? >>Cheers. > > Good question. The ratio test is indeterminate (the limit is one), but > the series doesn't appear to be converging to anything as n gets up to > the 1 million range. I'd guess the correct answer is +inf. I am not > great at this stuff, but I don't see a way to establish that as fact, > rather than supposition. (I'm not Ben; just jumping in.) > I think it's pretty obvious that the whole sum does diverge. Even if one doesn't know that sum(1/n,n=1..inf) diverges, clearly sum(n,n=1..inf) does. So sum(n+1/n,n=1..inf) will be made up of two parts, one which diverges and one which may or may not. So.... PS: Strange things can sometimes happen when thinking about finite operations, such as sum(n+1/n,n=1..M), and extrapolating to the case M=inf. These strange things do not occur here.
From: John H Meyers on 18 Sep 2006 13:22 On Sun, 17 Sep 2006 06:26:47 -0500, Steen Schmidt wrote: > I'd venture a guess and say that the operator > that is not implemented in the SERIES function > (which is apparently used by lim) is Psi. An "infinite" sum is a limit, although "lim" was not specifically used; what surprised the OP is that 1/X can be summed to infinity, as can X (each yielding '+\oo'), but that the sum of 1/X and X can not -- the error message also seems incomplete without some way to know what "operator" it's talking about. However, anyone ought to be able to glance at this problem and announce the result without touching a button; it's only the loss of confidence which amplifies the significance of something which looks simple but doesn't compute. Best wishes from http://www.mum.edu and http://www.maharishischooliowa.org
From: John H Meyers on 18 Sep 2006 16:43 On Mon, 18 Sep 2006 10:06:04 -0500: >> Out of curiosity, what do you expect this limit to be? > > Good question. The ratio test is indeterminate (the limit is one), but > the series doesn't appear to be converging to anything as n gets up to > the 1 million range. I'd guess the correct answer is +inf. The first thread ("CAS oddity") on this topic omitted an important detail, which is -- what was the summation variable? If we're summing X+1/X from X=0 to infinity, the result is obviously '+\oo' -- but if we're summing from n=0 to infinity (as first stated in this alternate thread), then the answer depends on X, does it not? At any rate, I was originally assuming that the summation variable was the same as the variable in the expression, in which case either separate term (X or 1/X) can be summed by the CAS, but the the expression 'X+1/X' can not -- ROM version 2.09 with CAS version 4.20060602 gives the "Operator not implemented (SERIES)" error. --
From: acl on 18 Sep 2006 17:23 John H Meyers wrote: > On Mon, 18 Sep 2006 10:06:04 -0500: > >>>Out of curiosity, what do you expect this limit to be? >>Good question. The ratio test is indeterminate (the limit is one), but >>the series doesn't appear to be converging to anything as n gets up to >>the 1 million range. I'd guess the correct answer is +inf. > > The first thread ("CAS oddity") on this topic omitted an important detail, > which is -- what was the summation variable? > > If we're summing X+1/X from X=0 to infinity, the result is obviously > '+\oo' -- but if we're summing from n=0 to infinity (as first stated > in this alternate thread), then the answer depends on X, does it not? How? For any real positive X, (X+1/X)>0, and the sum is basically then constant*Sum(1,n=0..infinity); which is infinite. Except if you mean X<0 whence the constant is negative and the result -\infty. So if the summation is over n, the result is a function of X, f(X), such that f(X)=-\infty if X<0 and f(X)=+\infty if X>0. This can be extended to complex X (and we get directed infinities). In fact, it can even be extended to X being a tensor (in which case we can define 1/X to be the inverse of X etc). There is practically no limit to how pointless an exercise in pedantry this can be made into. But, somehow, I don't think this was what was being asked.
From: John H Meyers on 18 Sep 2006 19:15
On Mon, 18 Sep 2006 16:23:47 -0500, acl wrote: > There is practically no limit to how pointless an exercise in pedantry > this can be made into. But, somehow, I don't think this was what was > being asked. I didn't think so either, until I reread the original post of this thread, which says: "Anyone know what causes this message? I get when trying to find the limit as n approaches infinity of sigma from 1 to n of one over x plus x." He did literally use two different variables there (n vs. x) Using ROM 2.09 (VER 4.20060602): 'lim(\GS(m=1,n,1/X+X),n=+\oo)' ==> [Syntax error at last 'n'] This illustrates a previously mentioned issue, which is that there is NO valid RPL 'expression' syntax for LIMIT(lim) which *names* the variable being varied [the RPL compiler is deficient here] Let's try Steen's interpretation: | As I understand it (after reading your question a dozen times) | you are trying to evaluate 'lim(SIGMA(X=1,n,1/X+X),n=inf)' 'lim(\GS(X=1,n,1/X+X),n=+\oo)' ==> Syntax error at last 'n' 'lim(\GS(X=1,n,1/X+X),+\oo)' ==> '(2*Psi(n+1)-(2*Psi(1)-(n^2+n)))/2' Well, how can a *limit* as n -> +\oo depend on a particular value of 'n'? The delivered answer is actually '\GS(X=1,n,1/X+X)' and not its *limit* Once again, the RPL compiler (or is it during execution?) is deficient; it has in effect *ignored* the "lim" function, and has returned only its first argument! As of ROM 2.09 (VER 4.20060602), algebraic expressions for LIMIT(lim) can be used properly *only* in ALGebraic mode (or with `back quotes`): `lim(\GS(X=1,n,1/X+X),n=\oo)` ==> '?' "Operator not implemented (SERIES)" How about `lim((2*Psi(n+1)-(2*Psi(1)-(n^2+n)))/2,n=+\oo)` Yep, it's '?' "Operator not implemented (SERIES)" Another approach: '\GS(n=1,\oo,1/X+X)' in Exact mode evaluates to '\oo'; in Approximate mode it starts trying to actually add it up :) [r->] [OFF] |