Wikipedia:Reference desk/Archives/Mathematics/2009 March 9

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March 9

sum of n terms of a series

Hello all. If $a_{1},a_{2}\cdots a_{n}$ are in AP and $a_{i}>0$ for each i, then what will be the sum of the first n terms of the series: ${\frac {1}{{\sqrt {a_{1}}}+{\sqrt {a_{2}}}}}+{\frac {1}{{\sqrt {a_{2}}}+{\sqrt {a_{3}}}}}\cdots$? Essentially I am looking for a method to find expressions for similar problems also.--Shahab (talk) 06:33, 9 March 2009 (UTC)

In your particular case, if you first rationalize, you get the same denominators, and you are left with alternate sums of square roots of the $a_{k}$ with nice cancellations. But in general, a simple closed formula like this one, is not available for partial sums of a series; look at Euler summation formula. --pma (talk) 08:28, 9 March 2009 (UTC)

Thanks. Your link is what I had been searching for since a long time.--Shahab (talk) 08:55, 9 March 2009 (UTC)

Fractional derivatives of exp(-1/x^2)

Just expanding on the question asked above, are all the fractional derivatives of $e^{-1/x^{2}}$ also equal to zero? Zunaid 09:25, 9 March 2009 (UTC)

Of course you mean: equal to zero at x=0. Yes, they are. They define the fractional integrals of f, or $\textstyle D^{-\alpha }f=J^{\alpha }f$, for $\scriptstyle \alpha >0$, as the convolution

$D^{-\alpha }f(x)=\gamma _{\alpha }*(f\chi _{\mathbb {R} _{+}})\,(x)=\int _{0}^{x}\gamma _{\alpha }(x-t)f(t)dt$,

where $\textstyle \gamma _{\alpha }(x):={\frac {x_{+}^{\alpha -1}}{\Gamma (\alpha )}}$ (check your link). Then in general, $\textstyle D^{s}f$ is defined via the identity $\textstyle D^{s}f=D^{m}D^{-\alpha }f$, with any positive integer $\textstyle m>s$ and $\textstyle \alpha :=m-s>0$. For the function $\textstyle f(x)=e^{-1/x^{2}}$ (and vanishing at 0), all $\textstyle D^{s}f$ vanish at x=0. Indeed, in this case $f\chi _{\mathbb {R} _{+}}$ is a smooth function, and $D^{m}(f\chi _{\mathbb {R} _{+}})=(D^{m}f)\chi _{\mathbb {R} _{+}}$, so we can switch derivation and convolution in the above definition of $\textstyle D^{s}f$:

$D^{s}f=D^{m}D^{-\alpha }f=D^{m}\left(\gamma _{\alpha }*(f\chi _{\mathbb {R} _{+}})\right)=\gamma _{\alpha }*D^{m}(f\chi _{\mathbb {R} _{+}})=D^{-\alpha }D^{m}f$,

showing that all fractional derivatives $\textstyle D^{s}f(x)$ vanish at 0, as all the $\textstyle D^{m}f(x)$ do. PS: I took the liberty of changing the header. --pma (talk) 16:35, 9 March 2009 (UTC)

Communicating meaning with distant space aliens - no pictures allowed

[This discussion began at the Language Reference Desk. -- Wavelength (talk) 14:56, 9 March 2009 (UTC)]

Imagine that the two-way communication of signals between us and some space-aliens orbiting a distant star has been established. They are blind and immobile and cannot use pictures or diagrams of any kind. There is no pre-established code or alphabet. While I can imagine that eventually the meaning of mathematical or logical symbols might eventually be established (for example tranmitting many messages such as "..+..=...." would give meaning to + and =), would it be possible to eventually build up enough meaning from a zero base so that in time they would understand what was meant by the message "Last thursday my Uncle Bill went to the supermarket"? Helen Keller springs to mind. 89.240.206.60 (talk) 02:01, 8 March 2009 (UTC)

I don't see how it's possible to go from 2+2=4 to any non-math concept. Remember, it was impossible to decipher hieroglyphics without help from the Rosetta Stone, even though they were written by human beings, and this would be n times worse (n >> 1). Clarityfiend (talk) 05:22, 8 March 2009 (UTC)

Earth has blind, immobile animals called barnacles, and some humans have done research on how to talk with animals (http://www.howtotalkwithanimals.com/), but I have never heard of anyone attempting to communicate with a terrestrial barnacle. Instead of contemplating communication with alien barnacle-like creatures, why not ponder how we humans can communicate better with each other? -- Wavelength (talk) 06:45, 8 March 2009 (UTC)

LINCOS was a whole elaborate language (developed at length in a book) based more or less around that premise (though I think there were some abstract mathematical images included)... AnonMoos (talk) 07:00, 8 March 2009 (UTC)

H. Beam Piper's much-reprinted story Omnilinual has terrans cracking the Martian language by finding a periodic table. Unfortunately, the idea in the story simply doesn't work: the English names for common elements only make sense in the context of the history of science, not modern science (eg oxygen = 'acid-maker' and hydrogen = 'water-maker]; these are Graeco-Latin rather than English, but German for example translates the roots and still perpetuates the errors), so why assume that the Martian names would be meaningful? --ColinFine (talk) 18:51, 8 March 2009 (UTC)

For what it's worth, I coincidentally ran into the following article today ---> Pioneer plaque ... in which NASA scientists are, in fact, trying to communicate with distant space aliens ... albeit with the use of pictures. (Joseph A. Spadaro (talk) 22:20, 8 March 2009 (UTC))

The essential bottleneck to get through may be that of naming geometric shapes, such as a triangle. A triangle could then be used to build up other shapes. The triangle could be named after being identified by its mathematical properties. If however they have no sense of the spatial, then you are stuffed. 89.243.72.122 (talk) 23:56, 8 March 2009 (UTC)

How can an organism distinguish between a random collection of perceptible stimuli and a purposeful collection of perceptible stimuli produced by intelligent design? How can it distinguish between a message and a non‑message?

-- Wavelength (talk) 02:09, 9 March 2009 (UTC)

Humans or even sheepdogs or bees seem to have no problems with doing that. And if we humans recieved a signal from a distant star in the form of the Fibonacci series or any other simple mathematical series, then that would indicate that the sender was an intelligent being. 89.242.94.128 (talk) 11:37, 9 March 2009 (UTC)

The series should not be too simple, as then we could not be sure it was not generated by some nonsentient physical process. The Fibonacci sequence in particular is a very bad example, as it is known to appear in nature without any involvement of intelligence, see Fibonacci number#Fibonacci numbers in nature. — Emil J. 13:42, 9 March 2009 (UTC)

This fellow's research into a generalization of information theory that assumes no prior common language might be of interest, for a formal take on a specific variation of the question, which he calls "Universal Semantic Communication". The general strategy is to frame it as goal-oriented communication, which allows us to conclude that we've successfully communicated something when we can achieve some goal as a result of the communication faster than we would've been able to do without it. --Delirium (talk) 02:55, 9 March 2009 (UTC)

It might be worth posting this question on the mathematics desk. I am sure that they would have ways of encoding mathematics that they would think recognisable (and going from simple operations to advanced formula). They might even have some insights in how to jump out of Mathematics. -- Q Chris (talk) 13:49, 9 March 2009 (UTC)

[The copied text ends here. -- Wavelength (talk) 22:46, 9 March 2009 (UTC)]

Computers, like these space aliens discussed, are "blind and immobile and cannot use pictures or diagrams of any kind." Yet you communicate with them and benefit from that. Bo Jacoby (talk) 10:38, 10 March 2009 (UTC).

Commuting family of matrices

Hi, I am working on a representation theory problem but I need some linear algebra to do it. I have talked to the professor and he has given me much of the problem but I just still do not understand simply because I do not think we ever covered what he is saying in linear algebra. Here's a theorem I do have

If ${\mathcal {F}}\subseteq M_{n}$ is a commuting family, then there is a vector $x\in \mathbb {C} ^{n}$ that is an eigenvector of every $A\in {\mathcal {F}}$.

My professor is talking about something that is more powerful. He says, I think, if a commuting family leaves any subspace invariant, then that commuting family has a common eigenvector in that invariant subspace. Is this true?

Thanks for any help StatisticsMan (talk) 15:41, 9 March 2009 (UTC)

They are the same thing. If they have a common invariant subspace, then they act as operators on that subspace. Just think of the elements of F as functions rather than lists of numbers and it should be clear; you just restrict the functions to that invariant subspace. JackSchmidt (talk) 15:59, 9 March 2009 (UTC)

Okay, thanks. StatisticsMan (talk) 19:05, 9 March 2009 (UTC)

—Preceding unsigned comment added by 71.80.236.201 (talk) 17:44, 9 March 2009 (UTC)

If by "profile" you mean an article about you, then the answer is at Wikipedia:Notability (people). There are a lot of specialized subguidelines of that, like Wikipedia:Notability (academics) which would apply to mathematicians (I assume that would be the relation to mathematics here). If you have more questions about Wikipedia policy, Wikipedia:Village pump (policy) is a more suitable place. —JAO • T • C 17:50, 9 March 2009 (UTC)

Question on Perfect Squares

For which n does the following hold:

$1^{2}\pm 2^{2}\pm 3^{2}\pm \ldots \pm n^{2}=0$?

where it is possible to choose either + or - in any of the cases.

There are a lot of solutions for number 1 to 100, but how do I generalise this to the n case? I appreciate the contribution made already, albeit I typed the question into the incorrect section on Wikipedia.

--84.70.242.151 (talk) 17:48, 9 March 2009 (UTC)

I hope you don't mind I took the liberty of formatting the equation so it's clearer what the question is. I don't have much of an answer though. —JAO • T • C 17:55, 9 March 2009 (UTC)

Let's say that it is necessary that either n=0 mod 4 or n=-1 mod 4, otherwise the sum is odd. As you mentioned, here PrimeHunter shows that from n=7 to 100 it is also sufficient, while there are no solutions for 0<n<7. In general, observe that an algebraic sum of 8 consecutive squares with signs exactly (+ - - + - + + -) always vanishes. Therefore you can build solutions for any n=8m, n=7+8m, n=11+8m, n=12+8m, (with $\scriptstyle m\geq 0$) just taking the PrimeHunter's solutions resp for n=0, n=7, n=11, n=12, and attaching to them a null algebraic sum of the subsequent 8m consecutive squares, with signs choosen with the periodicity (+ - - + - + + -). This does all n>6 with n=0 mod 4 or n=-1 mod 4. A more challenging question would be, find an asymptotics for the number of solutions (it is exponential, at least $c2^{n/8}$, for you can replace any group of 8 signs with the opposite, if you wish) --pma (talk) 18:56, 9 March 2009 (UTC)

Ok, that does make sense. Thanks for the input. The 8 consecutive squares result is a new one to me...can use that again sometime no doubt! I wouln't have thought to use congruence here. I would have thought a series approch would have been the way to go about it. Thanks though --84.70.242.151 (talk) 23:47, 9 March 2009 (UTC)

Notice that the sequence (+ - - + - + + -) is produced starting by + and iterating the operation of adding an inverted copy. Once more produces (+ - - + - + + - - + + - + - - +), that gives you a null algebraic sum of 16 consecutive cubes, &c, just in case. --pma (talk) 01:30, 10 March 2009 (UTC)

It is the constant term in

$\prod _{i=1}^{n}\left(x^{-i^{2}}+x^{i^{2}}\right).$

Maybe the asymptotics can be obtained by the saddle-point method or something like that. An equivalent formulation is that it is

$2^{n}Prob(X_{1}+X_{2}+\cdots +X_{n}=0),$

where $X_{1},X_{2},\ldots ,X_{n}$ are independent random variables with $X_{i}$ having mass 1/2 at $-i^{2}$ and mass 1/2 at $i^{2}$. It is plausible that some local central limit theorem exists which gives the answer, but maybe not. McKay (talk) 04:49, 10 March 2009 (UTC)

Oh, very nice... so maybe we can put $\textstyle x=e^{it}$ in your product and write the equality between trigonometric polynomials:

$2^{n-1}\prod _{k=1}^{n}\cos(k^{2}t)={\frac {a_{0}(n)}{2}}+\sum _{k=1}^{\infty }a_{k}(n)\cos(kt)$

where $\textstyle a_{k}(n)$ is the number of representations of k as an algebraic sum of the form $\pm 1\pm 2^{2}\pm ...\pm n^{2}$; in particular we can write, for the number of solutions of the OP's problem

$a_{0}(n)={\frac {2^{n}}{\pi }}\int _{0}^{\pi }\prod _{k=1}^{n}\cos(k^{2}t)\,dt$,

and start the asymptotics...--pma (talk) 14:47, 10 March 2009 (UTC)

Following McKay's hint, and estimating the integral by something like the saddle-point method, I obtained:

$a_{0}(n)=2^{n}{\sqrt {\frac {10}{\pi n^{5}}}}\left(1+o(1)\right)\quad as\,\,n\to \infty ;\,\,n=-1\,or\,0\mod 4$,

which is not bad. For instance, for $\textstyle n=24$ we get $\textstyle a_{0}(24)=10838$ while $2^{24}{\sqrt {\frac {10}{\pi 24^{5}}}}=10607..$ --pma (talk) 08:41, 12 March 2009 (UTC)

I can follow you all the way up to that last step, but then I'm lost. How did you get that? Black Carrot (talk) 05:23, 13 March 2009 (UTC)

(deinde) It's a computation; a nice exercise of advanced calculus indeed. I list here the main facts, and will put other details at request.

1. First observe that the integrand $\scriptstyle p_{n}(t):=\prod _{k=1}^{n}\cos(k^{2}t)$ verifies either $\scriptstyle p_{n}({\frac {\pi }{2}}+t)=-p_{n}({\frac {\pi }{2}}-t)$ for all t, or $\scriptstyle p_{n}({\frac {\pi }{2}}+t)=p_{n}({\frac {\pi }{2}}-t)$ for all t, according whether the number of odd indices in the product is odd or even, that is, whether n=1 or 2 (mod 4), respectively n=0 or 3 (mod 4). In the former case the integral on $[0,\pi ]$ is 0 (as we already know); in the latter we can write the integral as twice $\scriptstyle \int _{0}^{\pi /2}p_{n}(t)dt$.

2. The picture of $\scriptstyle p_{n}(t)$ on $\scriptstyle [0,{\frac {\pi }{2}}]$ is this: it is 1 at zero; decreases speedily to the first zero $\scriptstyle {\frac {\pi }{2n^{2}}}$ and oscillates with very small amplitude all along the interval $\scriptstyle [{\frac {\pi }{2n^{2}}},\,{\frac {\pi }{2}}]$ . As a matter of fact, the contribution to the integral on the latter interval (although longer) is less and less relevant: after estimation

$\int _{\frac {\pi }{2n^{2}}}^{\frac {\pi }{2}}p_{n}(t)dt=o\left(\int _{0}^{\frac {\pi }{2}}p_{n}(t)dt\right),\quad (as\,n\to \infty )$,

and we are left with the estimation of the interval on $\scriptstyle [0,{\frac {\pi }{2n^{2}}}]$.

3. Note that on $\textstyle [0,{\frac {\pi }{2n^{2}}})$ the function $\textstyle p_{n}(t)$ is positive so we can write it as $\textstyle p_{n}(t)=\exp \left(\sum _{1\leq k\leq n}\log \cos(k^{2}t)\right)$. A linear change of variables, namely $\textstyle dt=n^{-5/2}ds$ gives:

$\int _{0}^{\frac {\pi }{2n^{2}}}p_{n}(t)dt=n^{-5/2}\int _{0}^{\infty }q_{n}(s)ds$,

where $\textstyle q_{n}(s)$ is a sequence of functions with support in $\textstyle [0,{\frac {\pi {\sqrt {n}}}{2}}]$, converging punctually to $\textstyle \exp(-s^{2}/10)$ (very easy to check; try it), and, what is important, dominated in $L^{1}$ (by a function of the form $\scriptstyle \exp(-Ms^{2})$). Therefore by the dominated convergence theorem

$\int _{0}^{\infty }q_{n}(s)ds=\int _{0}^{\infty }e^{-{\frac {s^{2}}{10}}}ds\,(1+o(1))={\frac {\sqrt {10\pi }}{2}}\,(1+o(1))$.

This is an efficient standard way for estimating integrals: dominated convergence theorem after rescaling. In general, the change of variables to desingularize a sequence of integrals may be difficult to find, but in this and similar cases, $\textstyle dt=n^{\alpha }ds$ does it, with a proper choice of $\textstyle \alpha$ found by the Taylor expansion at 0. As a last remark, notice that the original question may be stated for r-th powers as well (r a positive integer); the same mod 4 subdivision in cases hold, and an analogous integral formula and asymptotics can be done analogously (with 2r+1 and 4r+2 in general instead of 5 and 10). --pma (talk) 12:07, 15 March 2009 (UTC)

An alternative to human mathematical concepts

Could a system of advanced mathematics develop without abstract concepts like vectors, imaginary numbers, or matricies, or are they somehow necessary? Could an alien species understand just as much science as humans do without, or with other, abstract ideas? As an example, humans use differential equations and complex numbers to solve simple harmonic motion. Is it possible to express simple harmonic motion using other ways that aren't used simply because they were discovered late in history? --99.237.96.33 (talk) 20:31, 9 March 2009 (UTC)

You just have to look at Newton's Principia to see for instance that a geometric way of expressing things could be used and an alien could possibly do that without a single line of equations. My own feeling is that people think about things in very different ways but still come to the same conclusions normally, so I've no difficulty thinking alien math might be quite unrecognizable and need quite a bit of reinterpretation - math is as much or more about how to go about it as the actual results. Dmcq (talk) 21:16, 9 March 2009 (UTC)

The basic concepts will probably need to be the same, but the way they are thought of could be completely different. For example, they will probably have complex numbers (they occur quite naturally when solving technological problems), but even with Earth maths we have two completely different ways of thinking about them (i=sqrt(-1) and i="90deg anticlockwise rotation") - an ETI could do things very differently. The answers, however, ought to be the same - maths is pretty absolute (we don't know if they would assume the continuum hypothesis or not, but we can be pretty sure they will have something close to Euclid's postulates, and if you have the same basic axioms, you'll draw the same basic conclusions, regardless of how you formulate them). --Tango (talk) 23:09, 9 March 2009 (UTC)

I can imagine an alien race that really has no advanced math or theoretical science, but simply learns things by trial and error. Take bridge design, for example. They could just keep trying different designs and choose the one that seems to hold up the best, without ever calculating the forces involved. I suspect that this is how the early bridges were built here on Earth, but an alien civilization might have continued this practice far longer than us. StuRat (talk) 00:33, 10 March 2009 (UTC)

Would that be like the discovery and continuation of roasting pigs by burning down huts? -hydnjo (talk) 01:32, 10 March 2009 (UTC)

That's not quite the same, no. There it's a complete coincidence (serendipity). In my bridge building example, there is some logic at play, just not explicit mathematical calculations. Perhaps first they place a log across a stream, but then they notice that for long spans the bridge sags down into the water in the middle. Perhaps they try using a log with an upward curve to it, to avoid this, and note that it works well. But then, for even longer spans, this arch bridge fails, so they end up tying cables to support the bridge in the middle, making a suspension bridge. StuRat (talk) 14:54, 15 March 2009 (UTC)

The standard assumption is that any civilisation that has developed the technology necessary to receive a radio transmission from the stars will have a similar technological basis to us, including the mathematics behind it. I think it's a good assumption because if it doesn't hold we probably can't meaningfully communicate with them anyway - you need some common ground to start the process. --Tango (talk) 00:41, 10 March 2009 (UTC)

Topological Vector Spaces

Hi, I am working on a several part problem and I am stuck on this part. Specifically, if you have the book, it's Problem 10.30 part c from Royden. We have a base B at $\theta$ for a translation invariant topology, so the set of all basis elements for the whole set are of the form x + B. In this specific part, we are trying to show

   If multiplication by scalars is continuous (at $\langle 0,\theta \rangle$) from $\mathbb {R} \times X$ to X,
   then If $U\in B$ and $x\in X$, there is an $\alpha \in \mathbb {R}$ such that $x\in \alpha U$.

I, with friends, have been working on this part for a while and we do not know what to do. I am assuming it's pretty simple but we just can't see it. Hopefully I have provided enough information for you to be able to understand. Can any one help me out with a hint? Thanks StatisticsMan (talk) 23:09, 9 March 2009 (UTC)

(I assume that you mean that θ is the origin of X). Are you sure about the hypotheses? Consider an infinite dimensional real vector space X, and the topology generated by the finite codimensional affine subspaces of X (that is, open sets are union of these). This is a translation invariant topology; actually it makes (X,+) a topological group; moreover, the multiplication by scalars is continuous at (0,θ) in the pair. But it does not make X a TVS; indeed the property you stated does not hold (just take U an hyperplane and x a vector not in U). In fact now I see that an even simpler counterexample is the discrete topology on any X of positive dimension. That property follows immediately if you ask instead, that for any x in X the multiplication by scalar with x is continuous at 0 from R to X. --pma (talk) 00:29, 10 March 2009 (UTC)

Sorry, I am assuming it is a topological vector space I think... the problem is to prove a proposition which is almost an if and only if, and the problem is confusing as to what you can assume at each point. At any point we can assume it is a vector space, but at this point I think we can assume it is a topological vector space. Is this your problem with what I was saying? Certainly, if this is needed, I am pretty sure we can just assume it. Thanks StatisticsMan (talk) 00:53, 10 March 2009 (UTC)

Well, if X is a TVS then the multiplication by scalars is continuos at any $(t,x)\in \mathbb {R} \times X$. But you just need separate continuity in the scalar variable, as I said: assume that the map $t\mapsto tx$ is continuous at t=0. Then for any U in B there is a nbd [-a,a] of 0 in $\mathbb {R}$ that is sent into U by this map. In particular consider ax....--pma (talk) 01:06, 10 March 2009 (UTC)

I get that if $t\mapsto tx$ is continuous at $t=0$, then for any open set O containing theta, there must exist an open set U in R containing 0 such that f[U] subset O. U contains an open interval around 0, that contains one that is centered at 0, so [-a, a] as you said. Since [-a, a] \in U and f[U] subset O, we know ax in O. Now, of course, x in 1/a O, which is what we wanted to show. But, I guess I do not understand if I can assume that function is continuous? Are you saying in a topological vector space that it is not given that is true?

In a TVS, it is continuous by definition; but of course in the "conversely" part of the main theorem you can't assume it, for "X is TVS" is the thesis. So, what you are saying proves that, if X is a TVS, then 4 holds for any U nbd of θ. Conversely, if you have a vector spaces with a family B satisfying properties 1 through 5, and you give it the topology generated by set of B and their translates, properties 4 and 5 tell you that, for any x in X, the map taking a real t into tx is continuous at t=0 (Do you see why? If x=θ there is nothing to prove; otherwise, the α given by 4 is not 0, and (1/α)x is in U; by 5 the whole interval [-1/|α|, 1/|α|] times x is contained in the U of B given by 4). This is not all you need to prove, it is just the separate continuity of the multiplication in the first variable at 0. See below as to the full continuity of the multiplication by scalars. --pma (talk) 09:56, 10 March 2009 (UTC)

Let me just write out the entire Proposition I am trying to prove and the two parts I have already proven (which I was trying to avoid :)... ):

Let X be a topological vector space. Then we can find a base B at theta that satisfies the following:

1. If U, V in B, then there is a W in B with $W\subset U\cap V$.

2. If U in B and x in U, there is a V in B such that $x+V\subset U$.

3. If U in B, there is a V in B such that $V+V\subset U$.

4. If U in B, and x in X, there is an $\alpha \in \mathbb {R}$ such that $x\in \alpha U$.

5. If U in B and $0<|\alpha |\leq 1$, then $\alpha U\subset U$ and $\alpha U\in B$.

Conversely, given a collection B of subsets containing theta and satisfying the above conditions, there is a topology for X making X a topological vector space and having B as a base at theta. The topology will be Hausdorff if and only if

6. $\cap \{U\in B\}=\{\theta \}$.

And, the first three parts of the problem are (remember, I am on the third part):

Prove Proposition 14:

(a) A collection B of subsets containing theta is a base at theta for a translation invariant topology if and only if 1 and 2 hold.

(b) Addition is continuous from $X\times X$ to X if and only if 3.

(c) If multiplication by scalars is continuous (at $\langle 0,\theta \rangle$) from $\mathbb {R} \times X$ to X, then 4.

So, does any of this let us prove that this function you constructed is continuous. I appreciate that you are helping. I'm just lost. StatisticsMan (talk) 04:17, 10 March 2009 (UTC)

Ok, the only possibly uncorrect point is c): it is not true taken alone, as I showed to you; but it's not a problem; you can replace it with the continuity at 0 in R of the map $t\mapsto tx$. Anyway, I think I understand where the mess came from: proving the sufficiency and the necessity part of a big proposition at once (it happens to me too, sometimes). In order to clarify things I suggest that you keep disjoint the two main parts of the principal theorem and attack separately each of them (then, after, you may have a look at the exact meaning of each axiom or combination of axioms). So, first assume that X is a TVS, that is, the vector space operations $\cdot :\mathbb {R} \times X\to X$ and $+:X\times X\to X$ are continuous; in this case you have to build a base B of nbds of θ satisfying 1...5, using the continuity of + and $\cdot \,$. Notice that the set of axioms is not standard: you may find another equivalent base with slightly different properties. In fact you may first build a base B', partially satisfying the axioms, then find an equivalent base B improving B'. In any case the idea is to exploit the continuity assumptions in order to get a nbd base as rich of properties as possible. Then do part 2: now X is just a vector space, but you have the family B satisfying 1...5: you give X the topology generated by sets in B and their translates, and you need to prove that it makes continuous the vector space operations. The continuity of the sum comes from 3 as you wrote; once you have it, to prove the continuity of the multiplication by scalars, it is useful to write $ax-a_{0}x_{0}=(a-a_{0})(x-x_{0})+(a-a_{0})x_{0}+a_{0}(x-x_{0})$, that reduces the task to just proving continuity in the pair (a,x) at (0,θ); continuity in the first variable a at 0 for all fixed $x_{0}$; continuity in x at θ, for all fixed $a_{0}$. Well, it's still a bit generic, I hope it helps however. Which part makes problems to you: I) "X TVS $\implies$ there exists a B satisfying 1..5", or II) "X with a B satisfying 1..5 $\implies$ X is TVS with nbd base B" ? --pma (talk) 08:21, 10 March 2009 (UTC)

Ah, maybe I see what happened. Philologically, there has been a contraction: I guess that originally point (c) stated: "If multiplication by scalars is continuous (at $(0,\theta )$) from $\mathbb {R} \times X$ to X, then 5"; and there were a further point (d) stating: "If multiplication by scalars is separately continuous in the real variable at 0, for all x, then 4".

Last remark: to prove the continuity of $x\mapsto a_{0}x$ at θ, for all fixed $a_{0}$, you can write $a_{0}=\pm n+\alpha$, with n a positive integer and $|\alpha |<1$, and then use the (already proven) continuity of the sum and the axiom 5. Is it OK? Now you should have everything, although scattered here and there... --pma (talk) 10:21, 10 March 2009 (UTC)

The problem is we do not know if 5 is true yet. So, not sure if it's allowable to use it. StatisticsMan (talk) 15:17, 10 March 2009 (UTC)

This depends on what implication you are proving. If you are proving (I), you have to find a base B of nbd's (not all nbds, just a base) with the properties 1..5; if you are proving (II), it is true by hypothesis... --84.221.208.38 (talk) 17:54, 10 March 2009 (UTC)

Unindenting because it's harder to read now. I talked to my professor and it turns out to not be too hard. I understand it now. For part c, we either are given it's continuous everywhere, so it is at (0, x) for any x, or we're given only (0, theta) and by translation invariance, it is also at (0, x). So, for any U in B, there exists an open set O in R times X such that (0, x) in ) and times(O) subset U. In particular, there exists a subset of O of the form (-epsilon, epsilon) times (x + V) where V in B, and this is true because (0, x) is in there. Therefore, epsion/2 x \in U so x in 2/epsilon U so we're done. Thanks for your help though. It was helpful to think about it in other ways. StatisticsMan (talk) 23:52, 10 March 2009 (UTC)

Yes, it is very easy. Just do things with a bit of order. When you start a proof, just ask yourself if it is clear to you what are the hypotheses, and what is the thesis. The same when you communicate with other people: make it clear what you want to prove, under what hypotheses. Anyway, I'm glad you got it :)--pma (talk) 09:51, 11 March 2009 (UTC)