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From: John H Meyers on 18 Sep 2006 21:58 On Mon, 18 Sep 2006 18:15:58 -0500: > '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] It turns out that the RPL compiler issue [refuses to accept valid syntax 'LIMIT(expr1,expr2)'] is a previously reported bug of more than a year ago: http://bugs.hpcalc.org/long_list.cgi?buglist=136 Even though the ALG-mode compiler and EQW both accept the same syntax, that bug is marked "Resolution: WONTFIX" Is it really not possible to accept "any expression" (including just a name or a constant) in *both* arguments? [this would allow the expression var=value in second argument] The only work-around for editing a given single formula is to use EQW to edit the formula (which is a bit of a pain, no?); for a program or list containing such a formula, however, there is no work-around at all, as of ROM 2.09 (or as of however long WONTFIX lasts :) and thus you can not edit or compile such a program or list! Of lesser importance, but by way of explanation: > '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* This is only a natural consequence of LIMIT (lim) accepting '+\oo' as the second argument, and interpreting it as the value of the variable named in 'VX'; so long as 'VX' doesn't contain 'n' then the given result is returned (after 'n' STOVX you get the title of this thread instead :) [r->] [OFF]
From: Chris Smith on 18 Sep 2006 22:32 John H Meyers <jhmeyers(a)nomail.invalid> wrote: > > 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' Apparently, there were several sources of ambiguity. The post I responded to said "sigma from 1 to n of one over x plus x". I interpreted that as the sum from x=1 to infinity of 1/(x+x). You seem you have read the sum from ?=1 to infinity of (1/x)+x. It turns out that both diverge, but the first is a little less obvious, hence the uncertainly in my earlier post. (That also explains why I didn't understand acl's response at all.) -- Chris Smith
From: acl on 18 Sep 2006 22:36 John H Meyers wrote: > 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) > No disagreement. I was joking :) > 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] I didn't know that.
From: acl on 18 Sep 2006 22:41 Chris Smith wrote: > John H Meyers <jhmeyers(a)nomail.invalid> wrote: > > > 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' > > Apparently, there were several sources of ambiguity. The post I > responded to said "sigma from 1 to n of one over x plus x". I > interpreted that as the sum from x=1 to infinity of 1/(x+x). You seem > you have read the sum from ?=1 to infinity of (1/x)+x. It turns out > that both diverge, but the first is a little less obvious, hence the > uncertainly in my earlier post. (That also explains why I didn't > understand acl's response at all.) > Actually, you're right! I also read it as "... from x=1 to $\infty$ of x+1/x"... In which case, though, I think my post does make some sense (at least to the extend I usually do make sense, which doesn't say much, it must be admitted). But x<x+1/x (for positive real x), so the sum from x=1 to infty of x+1/x is, term by term, larger than a sum which obviously diverges. Cheers.
From: acl on 18 Sep 2006 22:44
acl wrote: > > Apparently, there were several sources of ambiguity. The post I > > responded to said "sigma from 1 to n of one over x plus x". I > > interpreted that as the sum from x=1 to infinity of 1/(x+x). You seem > > you have read the sum from ?=1 to infinity of (1/x)+x. It turns out > > that both diverge, but the first is a little less obvious, hence the > > uncertainly in my earlier post. (That also explains why I didn't > > understand acl's response at all.) > > > > Actually, you're right! I also read it as "... from x=1 to $\infty$ of > x+1/x"... In which case, though, I think my post does make some sense > (at least to the extend I usually do make sense, which doesn't say > much, it must be admitted). > > But x<x+1/x (for positive real x), so the sum from x=1 to infty of > x+1/x is, term by term, larger than a sum which obviously diverges. > Oh, I also switched notation between the previous posts and mine (the one replying to yours). I switched from sum(x+1/x,x=1..infinity) to sum(n+1/n,n=1..infinity). I suppose this does not enhance the clarity of my writings. |