matt grime

Johnny, when I said, in my argument that you think is wrong, when k=1, that n! >k^{n-k} for all n sufficiently large, why do you ignore the fact that n! *is* strictly greater than 1 for all n sufficiently large? n greater than 2 in this case. I did not state it was true for all n; it isn't; it is true for all n greater than some number dependent on k, but that is immaterial. Perhaps the > should have been a => sign, I honestly can't remember exactly what I intended to type but it really is not important. Once more you miss the point of the argument and focus on unnecessary aspects, just like you missed the point about absolute convergence of power series. It is not important that the power series for (1+x)^t may converge for some t when x is 1, but that it diverges for |x|>1

I think I have the "johnny l'hopital conjecture" Your "derivative" of a sequence a_n is simply a_n -a_{n-1}, isn't it? You could use that do differentiate log(n!) directly by the way and the answer is log(n), which, if used in the above generalized l'hopital, would give a limit of log(n)/1 which diverges. Conjecture: Suppose that a_n and b_n are sequences and that a_n and b_n diverge. Then the ratio a_n/b_n tends to the same thing as (a_n-a_{n-1})/(b_n-b_{n-1}) when at least of the sequences a_n - a_{n-1} or b_n - b_{n-1} converges (and both do not converge to zero).

Can I suggest you first prove that n!^{1/n} diverges, since it is quite straight foward. It may even convince you of somethings.

I say it diverges because it does diverge as is elementary to show,and indeed can be shown to be true becase the series expansion of e^x converges. Also you take log(1)+log(2)+..log(n) and diff wrt n and stated that the first terms vanish because there is no n, which is false, since I can rewrite the sum as log(n) +log(n-1)+log(n-2)+...log(n-(n-2)) + log(n-(n-1)) in which every term has an n in it. Plus, there is, of course, the fact that you've not proved your "generalized" l'hopital is true, and this example shows it may be false. Not that it's clear you can differentiate log(n!) properly.

Applying it to the original question you're going to get that n!^{1/n} converges, which is a shame since it is easy to prove that it diverges, as I did in that original thread - a proof you ignored. (why do you "presume" your answers?)

We do not need negative numbers to describe space. We managed for many thousands of years with mathematics but without needing negative numbers whilst being perfectly able to navigate our way round. We do not *need* any mathematics. It is merely a useful tool. This whole "needing" physical interpretations held mathematics back for centuries. We are suffciently clever to be able to describe negative numbers as pairs of naturals, rationals as pairs of integers and complex numbers as pairs of reals, avoiding much ontological baggage in the process.

Having read your edited and very lng initial post it seems all you want is the difference f(n+1)-f(n) which is simply log((n+1)/n) in the case you care about. Look up difference equations. What use it is is something unclear.

Ok, shall we get it absolutely correct? The radius of convergence is 1, so it converges absolutely for all |x|<1. It will diverge for all |x|>1, and may or may not converge when |x|=1 dependent on the alpha. Eg it diverges when x=1 and alpha is -1.

Only valid for |x| less than 1 or alpha an integer, before you go too far down whatever line of enquiry this is.

As the statement of l'hoptal's rule does not allow for the kinds of functions you are talking about I am perfectly correct to state it is not valid to use it. You can attempt to define a different rule with different hypotheses that generalizes L'hopital, but it will not be the same theorem.

Not sure I agree with that. the ratio of consective terms is [math]\frac{\log [( n+1)!](n)}{\log[n!] (n+1)}[/math] which is [math] (1- \frac{1}{n+1})(1+ \frac{\log(n+1)}{\log(n!)}[/math] which converges to 1 I believe so the ratio test doesn't show anything in this case.

L'hopital's rule applies to the ratio f/g when f and g are smooth functions that have a taylor/laurent series in the neighbouhood in which we are intersted in. So of course l'hopital is invalid for this situation by the very hypothesis of when it applies. If you are asking is there an analogue for functions that do not possess laurent series then that is a different theorem, which does not exist almost surely.

For anyone wanting to answer this you should know Johnny wished to apply l'hopital's rule to somthing for which l'hoptals rule is not valid, in particular this involved differentiating the series log(1)+log(2)+...+log(n) with respect to n where n is naturally a natural number.

Rather easy induction proof by the look of it, wouldn't you say?

Well, my point was that we'd already had that information.

That'd be a specialized example gamma function mentioned in the first reply, in case anyone was wondering.

Stirling's approximation doesn't even give n! for n integer; it gives an asymptotic approximation.

The error was you never bothered to work out the dot product part and include it, you only worked out the cross prodcut part. The fact that there is no dot product part in the method you used exactly accounts for the reason why it fails to appear in your answer. It really is no simpler, nor more complicated than that. Quote: "What kind of multiplication allows you to write: ii=-1 ? Not the ordinary multiplication of two real numbers that we learn in grade school." "What kind"???? Never mind, Johnny. You do understand that I can declare the composition of any two elements of any ring to be anything I care as long as they satisfy the rules. What has "grade school" maths got to do with this? You seem happy enough with the complex numbers as a+ib. They're just symbols to manipulate.

That's a bit messy and hard to follow, but it can be proved directly and easily. Let k be any positive integer, then n! > k^{n-k} for all n usfficiently large. Proof: Assume n>K, then the RHS is k multiplied by itself k-n times, but the LHS is n! = (k)!*(k+1)(k+2)....(k+(n-k)) > k!k^{n-k} >k^{n-k} as required Thus n!^{1/n} > k^{(n-k)/n} but the rhs tends to k as n tends to infinity thus n!^{1/n}> k-1 for n sufficiently large. But k was artbitrary thus n!^{1/n} must tend to infinity. Actually that is rather unnecessary since we know that the sum of k^n/n! is exp{k} it follows that k^n/n! tends to zero, and inparticular k^n/n! <1 for all n sufficiently large, but there's no harm in having two ways of proving something. As for your proof you are abusing the equals sign a lot. Yes [math] \frac{1}{n}\sum_{r=1}^{n}\log r[/math] does diverge, but I don't see how you can use l'hopital, which deals with real valued differentiable functions to conclude that.

Firstly you must sepcify distributivity on both sides; i think there are examples to show this, try googling for them. Secondly yes 0 is unique as is the identity in *any* group. That has been shown to you, explained to you at least three times now and still remains a trivial exercise, and the commutativity of the group is not important since identities are declared to be two sided. You can either assume + is commutative and show the unique left identity is also a right identity, or you can simply say R,+ is an abelian group. Your choice. There is redundancy in the group axioms when we restrict to it being abelian, that is all, and thus there are several equivalent ways to give the group axioms, just as there several equivalent ways to axiomatixe the real numbers, and just as we can show there to be lots of redundancy in other axiomatizations of other things.

Seeing as i^2=-1=j^2=k^2 then treating them as "imaginary units" would seem better than treating them as vector, though why you can't simply call them quaternions is a mystery. We rarely point out something is a vector anyway, since the fact it is in a vector space implies that already. You assumed i, j and k had length 1 when you called them unit vectors. That is what a unit vector is for heaven's sake: a vector of length 1.

I don't understand your question. Mathematics models some parts of "reality", that's all. Reality is imperfect - you can't actually measure many things absolutely, can you? Who says space is even infinitely divisible, which is what you're assuming as well. Reality has a tendency not to follow the rules exactly, since mathmatics is just an idealized model. Your query is all relative: what's a decimal place? Measuring in cm's to 1 dp is the same as measureing in mm to no dp.

You're confusing the mathematical world with the real one. Try drawing two lines exactly of length 1 unit, whatever that may be, at exactly 90 degrees to one another and then chekcing that the resulting "third distance" really is sqrt(2)

Back to rings. The purpose of ring theory is simply to offer a self contained theory of certain objects that occur in algebra. There are many things we consider interesting and we have chosen to codify certain parts of them, ie pick out a set of rules common to all of them. We can then study what results follow simply from these axioms. Then we can add some more and study, say, commutative rings, which are important in geometry, or rings that are also vector spaces, ie algebras. Or perhaps we care about rings that have no zero divisors, which are called integral domains. As far as I'm concerned a Ring is simply the simplest algebraic object that is interesting in generality. More precisely it is the most basic case when the axioms it satisfies start to be interesting and also not too strong. Less restricting axioms are too general, they don't allow us to state and prove very many general results, and stronger axioms become a little too restrictive. Of course, a group is arguably a simpler object, but I think they are in some sense too algebraic, that is the study of groups alone is purely algebraic, whereas rings are of more general interest to, say, number theorists. Ok a ring is a goup under addition, but it is frequently an infinite one.

The space of operators on some vector space, eg the smooth functions on some compact subset of R^n is a ring, it has a spectral theory. It isnt' very complicated, but if you're still trying to understand quaternions, which are simply a 4 dimensional real division algebra, then talking about the spectral theory of operators is not going to be helpful. However, we can only second guess what it is you want to talk about, so why not formulate some questions?