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February 9

math lessons website

is there any website where I can do lessons in subtraction, addition, multiplication, and division in decimal, fraction, and other section of maths? —Preceding unsigned comment added by 74.14.117.32 (talk) 03:18, 9 February 2010 (UTC)

Subtraction, addition, multiplication and division in their basic sense do not constitute "math" (see this past section for more details) but that is probably besides the point. If you google "arithmetic games" or something of the sort, you should obtain numerous "fun training sites" in arithmetic (one "popular result" that pops up is this). But an excellent site for real mathematics is this, and in particular, this; this could be useful as well. PST 03:41, 9 February 2010 (UTC)

(posting out of order, because indentation would be wrong otherwise) Surely you jest that someone asking for arithmetic practice can make useful use of a site that "assumes you know the basics, i.e. high school algebra and geometry"? (It's certainly possible that the OP is a college student who just happens to want to brush up on arithmetic, but I find that very unlikely.) If you want to provide an appropriate/interesting "broader math" link, try the "recreational" link provided by MathReference. --Tardis (talk) 16:02, 9 February 2010 (UTC)

Whether those things constitute "math" depends on what you're doing with them. If you're working on understanding why multiplication is commutative and associative, why multiplication distributes over addition, why the division algorithm works the way it does, why LCMs are used in adding fractions, why a decimal expansion either terminates or repeats if and only if it's a quotient of integers, etc., that's definitely math. Michael Hardy (talk) 04:36, 9 February 2010 (UTC)

That is why I said "Subtraction, addition, multiplication and division in their basic sense do not constitute math..." (the key word here is "basic sense"). I definitely agree that one aspect of mathematics is what you do with particular tools, however basic the tools may be. But I think it is a safe guess that the OP's definition of "math" implicit in his/her post is "just the tools" (subtraction, multiplication etc.). I merely highlighted that such is not really mathematics. PST 05:04, 9 February 2010 (UTC)

What I was talking about was more basic. It's impossible for anyone to memorize algorithms, etc., until the parts I was talking about get done. Michael Hardy (talk) 22:10, 9 February 2010 (UTC)

Who cares? Just let it slide. Someone who asks for resources to help with addition of fractions doesn't need to be insulted by being told that their concept of "math" is wrong. For what it's worth, I happen to believe that the addition of fractions is part of math. —Bkell (talk) 15:23, 9 February 2010 (UTC)

You are of course entitled to believe that which you like but it is absurd to suggest that I have insulted the OP when you have not even responded to his/her question (and I have). PST 02:06, 10 February 2010 (UTC)

"Arithmetic is the oldest and most elementary branch of mathematics, used by almost everyone..." Arithmetic is a subset of mathematics in general. Nimur (talk) 15:32, 9 February 2010 (UTC)

And to refer someone who asked about basic numerical operations to a site on algebraic topology is just ridiculous - for goodness sake, PST, can you not just hold back, instead of insisting on jumping in every time? The phrase "personal fiefdom" comes to mind.→81.131.162.215 (talk) 16:16, 9 February 2010 (UTC)

At least I answered the question! Neither you nor Bkell have done that here. PST 02:06, 10 February 2010 (UTC)

It does seem that you have some very strong and not very mainstream ideas about what does or does not constitute mathematics, and are too eager to challenge any deviation from these. I think the consensus is that mathematics includes many things - basic topics and more advanced ones, theory and applications, methods of exploration and the results of this exploration, procedures that can be followed mindlessly and creative thinking, studying the known and researching the unknown. In your contempt for a typical student's notion that math is just a narrow subset of these, you are too defiant of the notion that math is just the complement subset.

More to the point, the kind of person that is thus misinformed is very unlikely to be impressed by your attempts to widen their horizons. The OP's response to the AT links is probably not "wow, I never thought there is so much more to math" but rather "what the *** does this guy want from me?". Only when an OP seems like the (rare) kind of person that can be "saved", you should gently nudge him in the right direction. -- Meni Rosenfeld (talk) 19:44, 9 February 2010 (UTC)

I definitely did not expect the OP to use my links as a motivation to begin study of algebraic topology; that would be an unlikely event if not an impossible one. The links "lightly suggested" (depending on how you look at it) that there are concepts in mathematics that have little to do with basic arithmetic. I agree that this is probably a failed attempt to convince the OP that there is more to mathematics than what he/she thinks. But unlike what is suggested by some posters directly above you, I do not feel that I have insulted the OP in any way. Some people have stronger views about particular matters than others, and while I agree that my views may not be shared by the general population, I did not state them impolitely in this instance. I definitely could have (and perhaps should have) given the OP a more tractable example of "non-arithmetic mathematics", but for some reason, I decided to go along a different path at the time of my posting. We all learn from our mistakes, so perhaps I will approach these questions differently next time. PST 02:22, 10 February 2010 (UTC)

Glad to hear it.±81.147.3.110 (talk) 17:23, 10 February 2010 (UTC)

Male/Female ratio problem

I once saw a problem that's something like follows:

"In East Asian families it's desirable to have a male heir in the family. The probabilities are 50% for either gender, but families will keep reproducing until they have 1 male child (and subsequently stop reproducing). What is the final male/female ratio due to this practice?"

and the answer is allegedly 1, as in it makes no difference at all. That is counter-intuitive but does kinda make sense after some thought, but I can't solidly prove or disprove it. Can someone confirm that it is the right answer and how does one come up with that answer? Thanks. --antilived^{T | C | G} 09:33, 9 February 2010 (UTC)

The number of girls in a complete family is distributed geometrically (with 0) with $p=1/2$, so the mean is 1. The number of boys in a complete family is also 1, hence the 1:1 ratio. -- Meni Rosenfeld (talk) 10:07, 9 February 2010 (UTC)

Actually, while the above may make it more intuitive, it's unneeded and doesn't answer the problem fully. The explanation is much simpler - every person born has an equal chance of being male or female, completely independently of how many girls or boys the parents had before. Thus the total number of males born is exactly (or, as accurately as the law of large numbers permits) equal to the number of females born. -- Meni Rosenfeld (talk) 10:18, 9 February 2010 (UTC)

(e/c) what you're talking about is a negative binomial distribution - the probability of getting m failures before getting n successes. in this case, n=1 (you are asking how many girls need to be born before a single boy is born) and the mean for the NBD is n*p/(1-p) = 1*.5/.5 = 1. in realistic terms, half the time a boy will be born first, but in other cases progressively large numbers of girls need to be born before a boy arrives, balancing out the preponderance of only-child boys.

interestingly, this seems to imply that if you change the problem so that you require 2 males in each family, then the ratio will turn out to be 2:1 in favor of women - insisting on more males in each family means you end up with far more women. odd... --Ludwigs2 10:21, 9 February 2010 (UTC)

No, it won't. The expected number of girls before the second boy is 2. Where did you get that number from? Algebraist 10:24, 9 February 2010 (UTC)

[ec] No, if $n=2$ then there are an average of 2 females in every complete family, but also 2 males, thus the ratio remains 1:1. -- Meni Rosenfeld (talk) 10:29, 9 February 2010 (UTC)

yeah, that's true. I was being bone-headed. happens to me far too frequently. --Ludwigs2 20:27, 9 February 2010 (UTC)

(edit conflict) That requires the fact that expectation is still linear when you apply it to sums of randomly many random variables, which isn't quite obvious. I think you need something like the optional stopping theorem. Algebraist 10:23, 9 February 2010 (UTC)

well, I think this was intended as a problem in abstract statistics, rather than a problem in real-world population dynamics. clearly in the real world it will be <1, since arbitrarily large family sizes are out of the question. --Ludwigs2 20:30, 9 February 2010 (UTC)

Er, why? If we accept the basic assumptions (not actually true in the real world, but not too far off) that male and female children are equally likely, and that different children have independent sexes, then I can't see how any real-world parenting strategy can get around optional stopping. Algebraist 23:11, 9 February 2010 (UTC)

Indeed. Considering the geometric distribution (or more generally, the negative binomial) is a distraction which leaves you confused about what happens with families that are not yet complete or never become complete. You only need to consider each birth independently (you might need some assumptions and proofs, but certainly not that each family is complete).

By the way, Human sex ratio is relevant for the real-world stuff. -- Meni Rosenfeld (talk) 09:20, 10 February 2010 (UTC)

Second integral in calculus

I'm trying to explain why the second integral of Integ(9x^2) is (3x^4)/4 + C. I just don't know the rule. Can you help me? Thanks! —Duncan^{What I Do} / _{What I Say} 20:44, 9 February 2010 (UTC)

What you said isn't true: $\int \int 9x^{2}\,{\text{d}}x\,{\text{d}}x\neq {\frac {3x^{4}}{4}}+C$, but it's close. You're just applying the power rule twice. --192.12.184.2 (talk) 20:57, 9 February 2010 (UTC)

Would it be 9x^4/12>? I'm a little unsure of how to use the power-rule on this... Thanks. —Duncan^{What I Do} / _{What I Say} 21:20, 9 February 2010 (UTC)

I think for the answer you need to replace C with

$Cx+D\,$

The first C comes from the first integration, which then gets integrated and another constant D arises for the second integration.--JohnBlackburne^words_deeds 21:27, 9 February 2010 (UTC)

Do you know how to do single integrals? If so, just integrate the function and then integrate the result. For example, the integral of 1 is $x+C$ and the integral of $x+C$ is ${\frac {x^{2}}{2}}+Cx+D$, so the second integral of 1 is ${\frac {x^{2}}{2}}+Cx+D$. -- Meni Rosenfeld (talk) 09:23, 10 February 2010 (UTC)

An antiderivate of an antiderivative of 9x² is (3x⁴)/4, because the derivative of 3x⁴)/4 is 3x³ and the derivative of that is 9x². Taemyr (talk) 12:55, 10 February 2010 (UTC)

February 10

Uniform minimum variance unbiased estimator

Hi there refdesk - I was wondering if I could get some advice on this problem, I fear it may be a bit above my current capabilities!

Let ${\hat {\theta }}$ be an unbiased estimator of $\theta \in \Theta \subseteq \mathbb {R}$ satisfying $\mathbb {E} _{\theta }({\hat {\theta }}^{2})<\infty$ for all $\theta \in \Theta$. We say ${\hat {\theta }}$ is a 'uniform minimum variance unbiased estimator' if ${\text{Var}}_{\theta }{\hat {\theta }}\leq {\text{Var}}_{\theta }{\tilde {\theta }}$ for all $\theta \in \Theta$ and any other unbiased estimator ${\tilde {\theta }}$. Prove a nec. and suff. condition for ${\hat {\theta }}$ to be a UMVU estimator is $\mathbb {E} _{\theta }({\hat {\theta }}U)=0$ for all $\theta \in \Theta$ and all estimators U with $\mathbb {E} _{\theta }(U)=0$ and $\mathbb {E} _{\theta }(U^{2})<\infty$ (i.e. '${\hat {\theta }}$ is uncorrelated with every unbiased estimator of 0').

May just be me but that seems exceedingly complicated, I don't think I can quite get my head around it I'm afraid! The more help/detail I can get, the better, so please do be generous :) Thankyou! Otherlobby17 (talk) 07:05, 10 February 2010 (UTC)

For $\Leftarrow$, use $U={\hat {\theta }}-{\tilde {\theta }}$ and the Cauchy–Schwarz inequality. For $\Rightarrow$, use ${\tilde {\theta }}={\hat {\theta }}+\lambda U$. -- Meni Rosenfeld (talk) 09:56, 10 February 2010 (UTC)

Wiki div operator

A div operator is mentioned at mediawiki. I expected it to be a modular division operator. However, {{#expr:3 div 2}} = 1.5. There is a {{div}} template that uses (a - (a mod n)) / n (humm). So now I'm curious. Does anyone know what the function of the div operator. Is it a simple division operator. I can't find any documentation on this. Thanks. –droll [chat] 08:41, 10 February 2010 (UTC)

I tried it for a few cases and it gave simple division every time, so it is highly unlikely that it is anything else. -- Meni Rosenfeld (talk) 09:25, 10 February 2010 (UTC)

Thanks. Things like this get to me. –droll [chat] 10:46, 10 February 2010 (UTC)

mw:Help:Extension:ParserFunctions##expr refers to meta:Help:Calculation for more details of the function of each operator. There it says about div: "Division, same as /, not integer division". PrimeHunter (talk) 00:38, 12 February 2010 (UTC)

Maximum cardinality of a distributive lattice of length n

I've been wondereing lately if there are distributive lattices of length n with more than 2ⁿ elements? If yes, are there distributive lattices of finite length with infinitely many elements?

In either case, proof or counterexample would be nice. An example for a distributive lattice with 2ⁿ elements is the Boolean lattice over some set with cardinality n. Yaan (talk) 13:17, 10 February 2010 (UTC)

The answer is no. First assume that the lattice is finite. If it has length 0, it clearly has size 1. If it has length n > 0, fix a chain a₀ < a₁ < ... < a_n. Since the chain cannot be expanded, 0 := a₀ must be the smallest element of the lattice, 1 := a_n the largest element, and a := a₁ an atom. Put $b=\bigvee \{c\mid a\nleq c\}=\bigvee \{c\mid c\land a=0\}$. By distributivity, $b\land a=0$, thus b < 1. It follows that the lattice is a union of two sublattices [0,b] and [a,1], each of which has length at most n − 1. By the induction hypothesis, the sublattices have cardinality at most 2ⁿ⁻¹, thus the whole lattice has cardinality at most 2ⁿ. This completes the proof for finite distributive lattices. If the lattice were infinite, pick arbitrary 2ⁿ + 1 elements, and take the sublattice generated by these elements. This is a finite distributive lattice with length at most n, but more than 2ⁿ elements, contradicting the previous part of the proof. —Emil J. 13:59, 10 February 2010 (UTC)

Alternatively, any distributive lattice of length n can be embedded in a Boolean algebra of length n (and therefore of size 2ⁿ), but I'm not sure how to prove that in an elementary way. —Emil J. 14:10, 10 February 2010 (UTC)

It follows immediately from Birkhoff's representation theorem. If a distributive lattice is represented by a poset, the length is exactly the number of elements in the poset, and each lattice element is represented by a lower set in the poset, so the number of lattice elements is at most the number of subsets of the poset, which is 2ⁿ. —David Eppstein (talk) 17:17, 10 February 2010 (UTC)

Thanks a lot everybody. Yaan (talk) 21:47, 15 February 2010 (UTC)

Limits of periodic functions at Infinity

Is there a distinguishing property of periodic functions limits? For example, a periodic function might have its limit undefined at ∞; but still has a defined boundary (i.e: [a,b]) at that limit.--Email4mobile (talk) 19:43, 10 February 2010 (UTC)

A periodic function has a limit at ∞ if and only if it is a constant function. A periodic function is bounded between a and b as x approaches ∞ if and only if it is bounded between a and b everywhere.

These seem like very easy questions. Did you mean to ask something else, or were you just confused about the definition of a limit approaching ∞? --COVIZAPIBETEFOKY (talk) 21:26, 10 February 2010 (UTC)

• It might be very easy to you, but new to me. I mean by common sense I knew that but I didn't know there is a rule. Thanks a lot.--Email4mobile (talk) 21:48, 10 February 2010 (UTC)

I didn't quote any rule. It might be a well-known theorem with a name, for all I know, but I'd never heard that question before in my life. I just figured it out logically; they're really not hard questions to work out. A periodic function can't have a limit to ∞ unless it's constant, because if it isn't constant, then it has at least two distinct values all the way out to ∞ (since it's periodic, two distinct values at x and x' imply two distinct values infinitely far out). Similarly, if the function is bounded over its period, then it's bounded over the entire real numbers, since the function simply repeats that one portion (and unbounded over any portion implies unbounded infinitely far out). --COVIZAPIBETEFOKY (talk) 01:01, 11 February 2010 (UTC)

People don't usually bother giving names to results with one-line proofs, so I doubt it has one. --Tango (talk) 01:09, 11 February 2010 (UTC)

What about the Frattini argument? Of course, this is just an arbitrary example and I agree that in general, arguments which follow directly from definitions do not warrant a name. PST 04:25, 11 February 2010 (UTC)

I said "usually". --Tango (talk) 12:16, 11 February 2010 (UTC)

The question may not be that trivial. It should be possible and interesting to extend the definition of limits to the interval arithmetic framework, so the interval-limit of a periodic function at ∞ is the set of values it can take. Mathematica does this - Sin[∞] returns Interval[{-1, 1}]. -- Meni Rosenfeld (talk) 07:34, 11 February 2010 (UTC)

Isn't that basically limsup and liminf? --Tango (talk) 12:16, 11 February 2010 (UTC)

Depends. If the function is discontinuous then its "interval-limit" can be different from $[liminf,limsup]$ (in which case it is not an interval at all). But anyway I'm not implying there's any new information in such a definition, only that it is a nice way to look at it. -- Meni Rosenfeld (talk) 12:50, 11 February 2010 (UTC)

9999 sticks and a triangle

There are given 9999 sticks with lengths of 1, 2, ..., 9998, 9999.

The players A and B alternately remove one of the sticks, player A begins. The game ends when only three sticks remain.

Player A wins if it can be formed from these sticks a "non-degenerate" triangle, otherwise B. Who can force a win? —Preceding unsigned comment added by 85.178.25.121 (talk) 23:03, 10 February 2010 (UTC)

This is a problem of the current Bundeswettbewerb Mathematik, the German mathematical Olympiad, so please don't discuss it till 1. March. Sorry to all that answered this question. --Schnark (talk) 09:11, 11 February 2010 (UTC)

Oh! Thanks for telling us. Eric. 131.215.159.171 (talk) 12:01, 11 February 2010 (UTC)

postulate of construction of angles

I have read its defination as ,"If one arm of an angle is on the edge of a half plane,then though the end of this arm one and only one ray can be drawn to the half plan to construct an angle of certain measure. " I could not understand it,please explain it. Is this defination correct.if not tell me correct defination.-mks-True path finder (talk) 23:45, 10 February 2010 (UTC)

The main confusion is the use of "half-plane". What they are saying is that you have a plane. You draw a line on it. That line cuts the plane in half. Now, you have a half-plane and a line (along the edge of it). If I give you an angle from 0 degrees to 180 degrees and ask you to draw another line on that half place at that angle to the original line, there is only one line you can draw. The limit of "one line you can draw" is based on another fishy word they used - "end". That original line must have an end point. So, the line you draw has to go through that end point at a specific angle. The entire point of all of this is that two lines form a distinct angle. You can't have an angle be 45 degrees and 50 degrees at the same time. -- kainaw™ 23:50, 10 February 2010 (UTC)

Thank u very much. —Preceding unsigned comment added by True path finder (talk • contribs) 00:02, 11 February 2010 (UTC)

February 11

Name that series

Does the series $\sum _{n=1}^{\infty }{\frac {n}{2^{n}}}=2$ have a name, such as the harmonic series or the power series does?58.147.58.179 (talk) 04:42, 11 February 2010 (UTC)

I think it's a power series; put x=1/2 and the a's to be the natural numbers Money is tight (talk) 08:40, 11 February 2010 (UTC)

So it is. I linked to the power series, but didn't look at that page because I was erroneously equating them with geometric series, when in fact geometric series are just one example of power series. Thanks. 58.147.58.179 (talk) 10:29, 11 February 2010 (UTC)

And the formula will work if you replace the number 2 on either side of the equation with any complex number ω, provided that the modulus |ω| < 1. You can prove it by finding the radius of convergence of the given power series. •• Fly by Night (talk) 17:28, 11 February 2010 (UTC)

You seem to be a bit confused. The power series $\sum _{n=1}^{\infty }nz^{n}$ is the Taylor series of ${\frac {z}{(z-1)^{2}}}$ about 0, and has radius of convergence 1. Thus, for any z with |z|<1, we have $\sum _{n=1}^{\infty }nz^{n}={\frac {z}{(z-1)^{2}}}$. The OP's series is the z=1/2 case of this. Algebraist 17:41, 11 February 2010 (UTC)

You're exactly right. I have your same exact expression saved in an earlier Matlab file. Not sure why I wrote what I did. Quite embarrassing really. Thank you very much for pointing out that error. •• Fly by Night (talk) 01:35, 12 February 2010 (UTC)

linear programming

I've been using linear programming to find solutions for minimizing ghe cost of making several products which is a comm application. I have noticed, however, that I cannot make just half the resulting number of products for half the minimum cost. How is this phenomenon explained? 71.100.8.16 (talk) 11:06, 11 February 2010 (UTC)

Fixed costs? Or perhaps you're using some kind of batch production that is less efficient for smaller batches? Perhaps there are minimum order quantities for your materials? It's hard to say unless you give more information. —Bkell (talk) 15:05, 11 February 2010 (UTC)

It could be Economies of scope, though that doesn't happen only with the linear case. Pallida  Mors 19:04, 11 February 2010 (UTC)

Symmetry groups

Given a group G, is it possible to construct a space for which G is the symmetry group? I thought at first that this was clearly untrue, since symmetry groups typically have functions as members, and so what does it mean to say that, say, Z mod 5 is the symmetry group of a space? But then I remembered that one of the goals of representation theory is to represent groups as matrix groups, so it didn't seem entirely unlikely. Is there a standard method of constructing the space, if true, and if untrue, does it work for some restricted class of groups we might consider?

The thought occurred to me today during a topology lecture, while discussing the construction of Eilenberg-MacLane spaces, and I wasn't sure where to even begin looking for an answer! (The reference desk seemed a good start...) Thanks for the help, Icthyos (talk) 17:33, 11 February 2010 (UTC)

It's sufficient to say that $\mathbb {Z} _{5}$ is isomorphic to the "real" symmetry group of the space; depending on how generally you mean to construe "space", the obvious example for that case is just 5 separate lines (or points, or planes, or...) with the group elements shuttling points cyclically among them. If you admit such examples, I'm sure any group can be used. --Tardis (talk) 18:23, 11 February 2010 (UTC)

Interpreting "space" very broadly, Cayley's theorem is the canonical result here. Algebraist 18:25, 11 February 2010 (UTC)

It is a theorem of Johannes De Groot that every group is the group of homeomorphisms of a compact Hausdorff space. As Algebraist hints, the idea of De Groot's proof is to use Cayley graphs, but then to replace the edges of the graphs by asymmetric subspaces. —David Eppstein (talk) 02:35, 12 February 2010 (UTC)

David Eppstein, are there more topological spaces than groups? If we map each space to its automorphism group, the theorem of Johannes De Groot shows this is a surjection. Pretty weird I'm sure we can define for each group some space and then show this is a surjection too. Money is tight (talk) 08:53, 12 February 2010 (UTC)

What do you mean by "more"? Both topological spaces and groups form a proper class. —David Eppstein (talk) 04:35, 13 February 2010 (UTC)

I know they form proper class, but that only means we can't talk about them in ZFC. But we can use NBG, where proper classes exist in the object language. In the article Axiom_of_limitation_of_size, it's an example of where we can talk about class functions; i.e. the function itself is a proper class of ordered pairs. Money is tight (talk) 06:25, 13 February 2010 (UTC)

That very axiom tells you that all proper classes have the same cardinality. In this specific example, I think that you don't even need the axiom to prove that both classes are equinumerous with V.—Emil J. 13:38, 15 February 2010 (UTC)

Thanks for the replies, this has given me much to ponder. (Woops, wasn't signed in) Icthyos (talk) 15:28, 14 February 2010 (UTC)

Summing a series

I found by experimentation that the sum of the series 1/2! + 2/3! + ... + n/(n+1)! appeared to be 1 - 1/(n+1)!, then proved this by induction. Can the sum be derived in any other way?—86.166.204.98 (talk) 20:14, 11 February 2010 (UTC)

$\sum _{k=1}^{n}{\frac {k}{(k+1)!}}=\sum _{k=1}^{n}\left({\frac {k+1}{(k+1)!}}-{\frac {1}{(k+1)!}}\right)=\sum _{k=1}^{n}\left({\frac {1}{k!}}-{\frac {1}{(k+1)!}}\right)=1-{\frac {1}{(n+1)!}}$. See also telescoping series. -- Meni Rosenfeld (talk) 21:21, 11 February 2010 (UTC)

(OP)Thanks. I saw how to do it soon after posting by taking the term ${\frac {1}{(n+1)!}}$ from RHS to LHS, whereby everything on the left collapsed to 1.—86.160.104.79 (talk) 13:44, 13 February 2010 (UTC)

Variability of taylor expansions in the complex plane under different branch-cuts

Hi all,

I've just completed the first part of the following problem:

Find the first two non-vanishing coefficients in the Taylor expansion about the origin of the following functions, assuming principal branches for (i), (ii) and (iii), making use of (where appropriate) series expansions for log(1+z), etc.

i)${\frac {z}{log(1+z)}}$

ii)${\sqrt {\cos(z)}}-1$

iii)$\log(1+e^{z})$

iv)$e^{e^{z}}$

Now I just figured they can't ask this question unless the functions are analytic about the origin, in which case the derivative should be the same in all directions at the origin, namely that along the real line, so then I just calculated the Taylor series as if they were real functions, getting:

i)$1+{\frac {z}{2}}$

ii)$-{\frac {z^{2}}{4}}-{\frac {z^{4}}{96}}$

iii)$\log(2)+{\frac {z}{2}}$

iv)$(1+z)e$

However, I didn't use any of the standard series expansions for these calculations, did I make things overly complicated for myself? Anyway, my problem is, I fear I've oversimplified things, because the next part of the question asks how each answer would differ if we didn't take the principal branch in each case? I also need to find the range of values for z for which each series converges. What have I done wrong, because I can't see how my answers would change if we took a different branch cut - are the answers even correct? Thanks very much for any explanation - and also, what would be the most sensible approach for finding the range of convergence for each series?

Many thanks in advance! 131.111.185.68 (talk) 20:20, 11 February 2010 (UTC)

The answers are correct. To hint at where the principal branch comes into play, consider the fact that if a non-principal branch was used, the answers to 1 and 2 could be $-{\frac {iz}{2\pi }}+{\frac {z^{2}}{4\pi ^{2}}}$ and $-2+{\frac {z^{2}}{4}}$. -- Meni Rosenfeld (talk) 21:28, 11 February 2010 (UTC)

euromillions lotto rollover

prize fund is £113 million cost of ticket £1.50 odds of winning 76 million to one. Is it worth purchasing the ticket? —Preceding unsigned comment added by 89.241.185.253 (talk) 21:34, 11 February 2010 (UTC)

Yes, think of all those important government projects you'll be funding !

Mathematically probably not. Although it sounds like a good bet, each time a big lottery gets in such a position it becomes very popular, and a lot of people place bets. So you're not betting on getting £113 million, you're betting on getting a fraction of that. And the bigger the pot the more chance it will be split, the more ways it will likely be split, as more people bet. Every once in a while someone will collect a big pot, but more often it will be won by more than one ticket I think.--JohnBlackburne^words_deeds 22:19, 11 February 2010 (UTC)

People win the lottery without purchasing a ticket. Therefore, the odds of winning with purchasing a ticket and without purchasing a ticket are nearly identical. -- kainaw™ 22:23, 11 February 2010 (UTC)

Just checked for a recent example and these siblings just won $100,000 this week without purchasing a lottery ticket. It was given to them as a birthday present. -- kainaw™ 22:31, 11 February 2010 (UTC)

The odds of winning the jackpot aren't useful, since that doesn't take into account the jackpot being split between multiple people or the smaller prizes. You need to know the total prize fund (the number you've quoted is actually the jackpot) and the total number of tickets bought in order to work out the expectation. A single rollover (which I think this is) is very unlikely to have a positive expectation. Multiple rollovers can, in theory. --Tango (talk) 22:52, 11 February 2010 (UTC)

The concept of the Kelly criterion is you can compute the optimal number of tickets to buy based on the size of your current bankroll and the winning probability and expectation. E.g., if the expected value is positive, then it makes sense to buy more than zero tickets, but since the probability of actually winning is small (the positive expectation comes from the size of the payout if you do win), you should not convert all your savings into lottery tickets. For almost everybody, the optimal number of tickets is much less than 1 ticket (i.e. running the numbers shows you should only bet a fraction of a cent rather than £1.50) and since that fractional bet is not possible, it's better to not bet at all. Buying a lottery ticket turns out to be a good investment (per Kelly) only if you are already a billionaire. 66.127.55.192 (talk) 23:19, 11 February 2010 (UTC)

A bit of math. From the news the jackpot was £85m a week ago. No-one won, but the jackpot gone up by £30m. But only 32% of winnings goes to the prize fund, so about £100m was added to all prize funds. And as only 50% of ticket sales go to winnings, about £200m was spent on tickets over the last week.

At £1.50 a ticket that's maybe 130m tickets bought to date. Still a day to go so probably a few tens of millions, if not more, will be sold this draw. The expected number of winners with 1 in 76 million is probably more than two, and the chance of there being only one winner is maybe less than 50%. And if there's only one winner they have not 1 in 76 million, but 1 in however many tickets sold chance: maybe 1 in 150 million. With the 50% that's a 1 in 300 million chance of scooping the whole jackpot.

I.e. the chance of winning a share of the jackpot is still 1 in 76 million, it's just very likely that it will be shared. Based on the above back of an envelope calculations it still seems a very poor bet (except for the lottery company).--JohnBlackburne^words_deeds 23:23, 11 February 2010 (UTC)

And don't forget about the taxes on the winnings. Rckrone (talk) 17:06, 12 February 2010 (UTC)

There aren't any. --Tango (talk) 23:43, 12 February 2010 (UTC)

And it turns out there were two winners, matching my rough calculations above (though the chance of it being exactly two was I think less than half). One winner was in the UK and won £56 million: at £1.50 a ticket and odds of 76 million to one a poor bet.--JohnBlackburne^words_deeds 23:49, 12 February 2010 (UTC)

The Kelly Criterion is irrelevant - the expectation for a lottery are almost always negative. The Kelly Criterion tells you how high the expectation needs to be for it to be worth you buying a ticket, but we know that will always be a positive number. The only way to have a positive expectation on a lottery (excluding multiple rollovers) is to be the one running the lottery. --Tango (talk) 13:43, 13 February 2010 (UTC)

February 12

Definition of Fundamental Domain from MathWorld--correct?

"Let be a group and be a topological G-set. Then a closed subset of is called a fundamental domain of in if is the union of conjugates of , i.e., and the intersection of any two conjugates has no interior. For example, a fundamental domain of the group of rotations by multiples of in is the upper half-plane and a fundamental domain of rotations by multiples of is the first quadrant . The concept of a fundamental domain is a generalization of a minimal group block, since while the intersection of fundamental domains has empty interior, the intersection of minimal blocks is the empty set."

• I hope this is right, since the last sentence is due to me. I would say if it is right , that would be nice, because it is concise and seems easy to understand and use. The current wikipedia article, like many wikipedia math articles (such as how affine space USED to be)go on and on without giving an immediate, precise definition. Maybe that's necessary in the nature of these things and the clear, concise definition on MathWorld is wrong? I am genuinely concerned that I have contributed a plausible, userfriendly but wrong explanation to MathWorld.Rich (talk) 03:05, 10 February 2010 (UTC) Pasted to Question Desk from Fundamental Domain(Talk) just now.Rich (talk) 00:14, 12 February 2010 (UTC)

Obviously you pasted the above and some symbols got omitted. Please try again. Michael Hardy (talk) 01:40, 12 February 2010 (UTC)

• sorry!98.207.84.24 (talk) 08:16, 12 February 2010 (UTC)

OK, I wondered which term was to be defined, and I was annoyed that your posting didn't tell us. But after getting a bit into it, I guessed that it's "fundamental domain". So I found that on MathWorld. I'm wondering about the word "conjugate". The way that word is used when speaking of conjugates of members of a group would make me expect to see gFg⁻¹ or the like instead of gF. But I'm not an expert in this area. Michael Hardy (talk) 01:47, 12 February 2010 (UTC)

• • Here's an improved transcription from Mathworld:

"Fundamental Domain

Let G be a group and S be a topological G-set. Then a closed subset F of S is called a fundamental domain of G in S if S is the union of conjugates of F , i.e., S=Union over all g in G of all gF, and the intersection of any two conjugates has no interior. For example, a fundamental domain of the group of rotations by multiples of 180 degrees in R^2 is the upper half-plane and a fundamental domain of rotations by multiples of 90 degrees is the first quadrant. The concept of a fundamental domain is a generalization of a minimal group block, since while the intersection of fundamental domains has empty interior, the intersection of minimal blocks is the empty set." (Transcribed from Eric Weisstein's MathWorld)Rich (talk) 23:24, 14 February 2010 (UTC)

The standard definition of a group action requires that the group act on a set from either the right or the left, so it is meaningful to interpret gFg⁻¹ as gg⁻¹F which is F for all g. Thus, it would not be very interesting to study the conjugates of F in that sense, and hence gF is referred to as the conjugate of F instead. (This seems the best explanation of the terminology "conjugation" in the given context; perhaps I am missing something, but I cannot think of another motivation for the terminology) Of course, there are other meaningful ways to merge conjugation with the theory of group actions; for instance, we can allow G to act on itself via conjugation.

With regards to the question, I am fairly certain that the last sentence is correct. In the most basic sense, for "A" to be a "generalization" of "B", every instance of B should be an instance of A. If we equip the set on which a group acts with the discrete topology (in fact, any topology finer than the smallest topology under which all minimal group blocks are closed will do), the instersection of the minimal group blocks is the empty set which of course has empty interior. The concept of a fundamental domain generalizes that of a minimal group block because the requirement on the intersection is less stringent; if the intersection were empty the requirement would be satisfied, but it would also be satisfied even if the intersection is more general (namely, has empty interior). Hope this helps. PST 03:45, 12 February 2010 (UTC)

A group block does not need to be closed in a topological sense so there are blocks that aren't fundamental domains. On the other hand the examples given in the MathWorld article are fundamental domains without being group blocks. Even if you assume a finite set with the discrete topology the concepts are still different. With a fundamental domain F=Fg implies g=1 but that's not the case with a block.--RDBury (talk) 23:37, 12 February 2010 (UTC)

What Minimal group blocks fail to be closed? Thanks, Rich (talk) 01:20, 13 February 2010 (UTC)

Let T be the smallest topology relative to which all the minimal group blocks in the G-set are closed. If we equip the G-set with certain topologies finer than T (such as T itself, or even the discrete topology), it can be ensured that all minimal group blocks are closed and that the G-set is in fact a topological G-set. Of course, fundamental domains are not necessarily minimal group blocks as RDBury points out. But fundamental domains are generalizations of minimal group blocks for transitive G-actions. Otherwise, they are not. For if we let G act trivially on a set S, the singleton sets in S are minimal group blocks but the orbit of any single point in S is trivial and thus no minimal group block can be a fundamental domain (no matter what topology we equip on the G-set).

But to clarify once more, it is indeed the case that, depending on the topology, all minimal group blocks are closed. Although the set on which the group acts initially does not have a topology, you can equip the set with a topology such that all minimal group blocks are closed (and of course such that the G-set is a topological G-set). (I made a silly error in my last post; unless we restrict our attention to transitive group actions, it is not necessarily the case that fundamental domains are generalizations of minimal group blocks). PST 05:10, 13 February 2010 (UTC)

Thanks Michael, RDBury, and PST for your helpful thoughts.98.207.84.24 (talk) 04:44, 14 February 2010 (UTC)

Integration of square root in the complex plane

Hi,

Just a quick one: I want to find $\int _{\gamma }z^{\frac {1}{2}}dz$ on the principal branch of the square root, where gamma is the circle |z|=1 and then the circle|z-1|=1, of radii 1. I've tried to parametrize in the obvious way, $z=e^{i\theta },\,1+e^{i\theta }$, respectively, and then I accidentally integrated for theta between 0 and 2pi - I think, retrospectively, this is wrong, since I got solutions out of -4/3 and 0: I have a feeling I should have gone from -pi to pi, but my work has been marked with my first answer correct and my second one wrong: if going from -pi to pi however, I wouldn't get -4/3 for the first integral, would I? So has my work just been marked correct wrongly, if that makes any sense?

In general, when integrating something with a branch cut, do we just choose our parameter range so that it 'goes around' the branch cut, e.g. -pi to pi doesn't cross the negative real axis, rather than 0 to 2pi which does? Thanks in advance, I don't need much detail in your answers, just yes/nos would be fine unless I've misunderstood something! 82.6.96.22 (talk) 05:47, 12 February 2010 (UTC)

You can choose the parameter range however you want, but be prepared for the possibility that the formula for the integrand might not be simple. For $0\leq \theta \leq 2\pi$ you have ${\sqrt {e^{i\theta }}}={\begin{cases}e^{i\theta /2}&0\leq \theta \leq \pi \\e^{i\theta /2-\pi }&\pi <\theta \leq 2\pi \end{cases}}$, while for $-\pi \leq \theta \leq \pi$, which goes around the branch cut, you can simply use ${\sqrt {e^{i\theta }}}=e^{i\theta /2}$.

I think the answer to the second question is indeed 0, and for the first it is actually ${\frac {-4i}{3}}$. -- Meni Rosenfeld (talk) 08:35, 12 February 2010 (UTC)

Stats resources/texts

Please recommend a stat resource/text that teaches you all the different manipulations involving expectation and variance. The stat text by Devore doesn't have comprehensive intro to these things. Eg. V(constant*random variable) = constant^2 * V(random variable) and many other manipulations involving summation sign and more than one random variable and constants. —Preceding unsigned comment added by 142.58.129.94 (talk) 19:58, 12 February 2010 (UTC)

This is mostly about probability rather than statistics, but it's online and covers the topics you mention. 66.127.55.192 (talk) 08:50, 13 February 2010 (UTC)

Integer closure under addition

I am currently somewhat confused about this. One article says that the integers are closed under addition and subtraction, whereas another one seems to give an example where they are not. What am I missing? Thanks in advance. It Is Me Here ^{t / c} 21:22, 12 February 2010 (UTC)

The integers are closed under finite sums; the second link refers to an infinite sum (not a convergent one either; its sum is 1/2 by a nonstandard definition). Here's another example of an infinite sum of integers whose value is not an integer:

1+1+1+1+1+1+1+1+1+1+...

The sum of that sequence is either 'undefined' or infinity, depending on context (it's not even a real number!). My point in bringing that up is that it's easy to see that integers aren't closed under infinite sums.

HTH, --COVIZAPIBETEFOKY (talk) 21:31, 12 February 2010 (UTC)

Closure under addition merely means that the sum of two integers is an integer. Addition is thought of in that context as a binary operation.

A consequence is that the sum of any finite number of integers is an integer. Michael Hardy (talk) 23:31, 12 February 2010 (UTC)

An example where subtraction isn't closed is for the non-negative integers 0, 1, 2, 3 etc. With those 3−2 gives 1 but 2−3 doesn't have a value so subtraction isn't closed. In topology you'll find an interesting combination, an infinite union of open sets gives an open set but only a finite number of intersections is guaranteed to give an open set. Dmcq (talk) 10:24, 13 February 2010 (UTC)

Thanks, all. It Is Me Here ^{t / c} 11:33, 13 February 2010 (UTC)

"subtraction isn't closed is for the non-negative integers" is a clumsy phrasing. "The set of all non-negative integers isn't closed under subtraction", or just "The non-negative integers are not closed under subtraction" is standard language. Michael Hardy (talk) 14:11, 14 February 2010 (UTC)

0.18, 9.45, and 0.38 estimation

Is Pluto's diameter 0.18 clser to 1/5 the size of earth of 1/6 the sizeof earth. Is this conventional to say Saturn is 9 times larger than earth or 10 times larger than earth. for Mercury is it conventional to say 1/3 the size of earth or 2.5 times smaller than earth?--69.233.255.251 (talk) 21:15, 12 February 2010 (UTC)

If you want to know whether 0.18 is closer to 1/5 or a 1/6, try a calculator. --Tango (talk) 23:49, 12 February 2010 (UTC)

Also keep in mind that "one time larger" means "twice as large," so "9 times larger" means "10 times as large," and "2.5 times smaller" doesn't make sense. —Bkell (talk) 00:49, 13 February 2010 (UTC)

That doesn't seem correct to me. I'm pretty sure in normal English "9 times larger" means the same thing as "9 times as large". 75.142.246.117 (talk) 04:19, 13 February 2010 (UTC)

After some googling it seems like the usage is mixed. I think I'm technically wrong about the meaning, but "times larger" is going to cause confusion. 75.142.246.117 (talk) 04:29, 13 February 2010 (UTC)

Yes, it's ambiguous. "900% larger" definitely means "10 times as large", but "9 times larger" is unclear and should be avoided. --Tango (talk) 13:46, 13 February 2010 (UTC)

Those who care about precision of language could say "larger by a factor of 9" to avoid the confusion if they insisted on using "larger". Colloquially, many imprecise people use "9 times larger" when they mean "9 times as large", and some of them can't even see the difference! The phrase "2.5 times smaller" doesn't make any logical sense, but it is used surprisingly often to mean "smaller by a factor of 2.5". Dbfirs 22:37, 13 February 2010 (UTC)

Remember that "size" in this case usually refers to the planetary volume, not its diameter. ~AH1^(TCU) 00:15, 14 February 2010 (UTC)

February 13

linear programming

When setting up constrains when using the simplex method can you use two equations for the same variable such as A=>B and A<=C in order to achieve an overall constrain that is a value between B and C? 71.100.8.16 (talk) 00:22, 13 February 2010 (UTC)

If B and C are also variables you'd set it up as B-A<=0, A-C<=0. Or introducing slack variables, B-A+X=0, A-C+Y=0.--RDBury (talk) 06:19, 13 February 2010 (UTC)

Lottery

Sorry if this is a stupid question, I'm not good at statistics:

Situation A:

1000 people each buy a lottery ticket for draw # 1

Situation B:

100 people each buy a lottery ticket for draw # 1; 100 other people each buy a lottery ticket for draw # 2; 100 other people each buy a lottery ticket for draw # 3; and so on for a total of 10 draws.

Is the probability that someone will win a prize equal in both situations? Poolofwater (talk) 15:54, 13 February 2010 (UTC)

That's going to depend on the rules of the lottery. How are winners selected and what prizes are there? --Tango (talk) 15:55, 13 February 2010 (UTC)

If the tickets are chosen at random yes. E.g. in theory someone in each of the 100 could win, so they win ten prizes. But also in theory the people in situation A could buy ten tickets with the same number that win. The chance might be vanishingly small in each case though.

If it's one person buying all the tickets then one way to increase your chance of winning a prize is but tickets with different numbers. E.g. if each draw has only 1000 numbers you can buy all 1000 in situation A and so guarantee a prize. In situation B you could still lose each draw. Of course real lotteries tend to have have much longer odds than 1 in 1000, so even if you bought your tickets at random they would likely all be different so the odds would be almost the same. And this is just your chance of winning something. Your expected winnings are the same whether you choose tickets at random or not.--JohnBlackburne^words_deeds 15:57, 13 February 2010 (UTC)

Thanks very much! Poolofwater (talk) 16:05, 13 February 2010 (UTC)

February 14

Integration

Can anyone tell me what ∫x dx({} refers to the "fractional part" function) will be in the limit of 2 to 20. Actually, it'll be better if you gave me the reasons rather than the actual steps. I mean, from 2 to 20, there are an awful number of numbers having non-zero fractional parts (apart from the integers, obviously). So how on earth can anyone integrate it? It can't be integrated piece-wise either. I tried to. Can anyone help?117.194.225.72 (talk) 12:10, 14 February 2010 (UTC)

Did you mean to put the curly brackets round x in the integration? Not that I recognize that notation anyway. It sounds like you are simply supposed to integrate the sawtooth wave function. Dmcq (talk) 12:32, 14 February 2010 (UTC)

(edit conflict) You want to integrate the fractional part of x from x = 2 to x = 20 ? Remember that ∫ f(x) dx from a to b is just the (signed) area enclosed between the graph of f(x) and the x-axis between x = a and x = b. This is quite straightforward to determine geometrically when f(x) is the fractional part of x. Note that this is the same as integrating piece-wise - between n and n+1 the fractional part of x is x-n, so you can integrate these pieces and then add them together. Gandalf61 (talk) 12:39, 14 February 2010 (UTC)

The fractional-part function is periodic, so if you integrate it from 0 to 1, you get the same thing as if you integrate it from 2 to 3, or from 3 to 4, or from 4 to 5, etc. There are 18 such intervals between 2 and 20, so the integral from 2 to 20 is 18 times that from 0 to 1. For x between 0 and 1, the fractional part of x is just x itself. So

$\int _{2}^{20}\{x\}\,dx=18\int _{0}^{1}x\,dx.$

Michael Hardy (talk) 14:09, 14 February 2010 (UTC)

What about the fractional part of numbers between 0 to 1 (having non-zero fractional part)? 0.556 ,for example? They should be taken into account too, right?In that case, integrating it should give an answer other than the value of x itself.... Correct me if I'm wrong. 117.194.231.5 (talk) 15:46, 14 February 2010 (UTC)

If x = 0.556, then the fractional part of x is also 0.556, i.e. the fractional part of x is x itself, just as stated above. Michael Hardy (talk) 22:40, 14 February 2010 (UTC)

Consider what the difference is between the fractional part of x and x itself between 0 and 1. Also, you should revisit your original impression that it can't be integrated piece-wise. Why did you say that? Finally, the suggestion above to consider its graph is a great idea. Have you sketched out a quick graph of the function and shaded in the area of integration yet? 58.147.58.28 (talk) 16:06, 14 February 2010 (UTC)

Thank you everyone. I just realised where I went wrong. I'm new to this stuff, so, while I knew that ∫f(x)dx is actually the area bounded by the function's graph and the x-axis, while trying to solve the above problem, I was constantly thinking of the summation of all the values of {x}. That would have made a different problem, something that went like: 0+0.001+0.002+0.003 and so on.. Thanks a bunch. Sketching the saw-tooth graph helped a lot. As simple as calculating the area of a triangle and multiplying it by 18, right? 117.194.227.81 (talk) 05:32, 15 February 2010 (UTC)

Correct. Michael Hardy (talk) 06:05, 15 February 2010 (UTC)

Sum of all combinations of n

What is the equation for the sum of all unique combinations of n objects. Eg, if n=2 (let's call the objects, A and B), the sum is 3 (combinations are A, B, and AB); for n=3, sum is 7 (combinations are A, B, C, AB, AC, BC, ABC). I know this seems similar to triangular numbers which is the sum of unique interactions between 2 objects among n numbers. Ephilei (talk) 19:06, 14 February 2010 (UTC)

Never mind. I found it. It's 2^n-1, one less than n's power set.Ephilei (talk) 19:23, 14 February 2010 (UTC)

Actually if you consider the empty subset too, they are exactly 2ⁿ. --84.221.209.11 (talk) 20:47, 14 February 2010 (UTC)

It is not very hard to prove either. We need to know about the binomial theorem and the binomial coefficients. Let us say that you have n objects. How many ways to pick none? There are nC0 (read as n choose none) ways of doing that. How many ways to choose one? There are nC1 (read as n choose one) ways of doing that. How many ways to choose two? There are nC2 ways of doing that. ... How many ways to choose n? There are nCn ways of doing that. So you need to make the following addition: nC0 + nC1 + nC2 + ... + nCn. This is given by applying the binomial theorem to (1+1)^n (which equals 2^n since 1+1=2). But you don't seem interested in the number of ways of choosing none, so we subtract nC0 = 1. That means there are 2^n-1 ways. •• Fly by Night (talk) 22:08, 14 February 2010 (UTC)

Quicker is to consider 3-digit binary numbers, with (say) A corresponding to a "1" in the RH position, B to a "1" in the central position and C to a "1" in the LH position. Then there are 8 possibilities 000, 001, 010, 011, 100, 101, 110 and 111, leaving 7 = 2^3 - 1 if the one containing neither A, B nor C is removed. Obviously extendable to any value for the number of objects—86.132.238.199 (talk) 23:06, 14 February 2010 (UTC)

Interesting! Thanks Ephilei (talk) 16:06, 15 February 2010 (UTC)

February 15

Definition of ${\sqrt {-1}}$

Why do we typically define ${\sqrt {-1}}$ to be $i$, instead of $\pm i$? Yakeyglee (talk) 03:11, 15 February 2010 (UTC)

Both $i$ and $-i$ are square roots of $-1$. Often the expression ${\sqrt {-1}}$ is avoided, because it's ambiguous; you're right when you say there isn't any particular reason it should be $i$ rather than $-i$. But when this expression is used, it is typically meant to stand for a single value, not two different values. The same goes for the expression ${\sqrt {4}}$, which always means just 2, not ±2, even though both 2 and −2 are square roots of 4. (The square root symbol ${\sqrt {\ }}$, when applied to a nonnegative real number, always indicates the principal square root, not just any square root. However, there is no compelling reason to say that $i$, rather than $-i$, should be the "principal" square root of −1.) —Bkell (talk) 03:26, 15 February 2010 (UTC)

To add to that, there's really nothing to distinguish the two square roots of -1. We arbitrarily pick one of them to be the principle square root ${\sqrt {-1}}=i.$ Then the other one must be $-i$. However, if we switched them, the complex numbers would look the same. Rckrone (talk) 05:08, 15 February 2010 (UTC)

Also notice that defining ${\sqrt {1}}$ to be $\pm 1$ isn't a great notation in computations, for then e.g. ${\sqrt {1}}+{\sqrt {4}}$ should mean either $\pm 1$ or $\pm 3$, and so on.--pma 18:09, 15 February 2010 (UTC)

What's the name for the argument of the logarithmic function?

The exponentiation a^x involves two numbers, each of which has its own name ("base" and "exponent") within the term a^x, so that the term a^x should be read: "Raising the base a, to the exponent x".

How about the logarithm log_ax ? Note that it involves two numbers as well, a being the base - as before; however, does x have - also here - its own name within the term log_ax ? In other words: how should the term log_ax be read? "Logarithm of the...(???) x, to the base a...? HOOTmag (talk) 20:50, 15 February 2010 (UTC)

Actually I'd say "a to the x" for the exponentiation. I may be wrong but I don't believe there is any special name for the argument of log. And I'd say something like "the log of x" or just "log x" where there's lots, or "log a of x" when the base needs to be stated. Dmcq (talk) 21:15, 15 February 2010 (UTC)

I was taught "log x, base a", for what that counts. - Jarry1250 ^{[Humorous? Discuss.]} 21:31, 15 February 2010 (UTC)

Cars and Fuel Consumption

Let's say I have an automobile that weighs w kilograms, and that I use u litres of fuel to drive d kilometres. Now assume that I make exactly the same journey, along the exactly the same roads, in exactly the same conditions; but that this time I carry c more litres of fuel then I did on the first journey. Let us assume (incorrectly) that the car consists of a single point with mass, and that there is no tyre friction or wind resistance; nor that the fuel changes temperature or that fuel consumption changes with velocity.

1. How much extra fuel, say e litres, would I have used by carrying this additional surplus of fuel?
2. How far less, say l kilometres will the same u litres of fuel carry me given that I am also carrying c more litres of fuel, and so weigh more.

Please forgive any problems with the way the question is posed; I'm sure you get my point. Basically: how much more fuel does it use to carry extra fuel in the tank? •• Fly by Night (talk) 22:31, 15 February 2010 (UTC)

It won't get you any further. If there is no friction or wind resistance then once in motion the car no longer requires any fuel and will never stop. The amount of fuel available will simply set the rate of acceleration and the speed at which the trip is completed - it will not effect the distance that the car can go. You need to assume a constant speed and a constant resistance for this question to make any sense. You will also need to give the weight of the car. The answer is not trivial - as the car burns fuel it will get lighter, you will need to use calculus.--Dacium (talk) 02:10, 16 February 2010 (UTC)

Round 0.45 or 0.49 or 0.50

For 5.55555 would be closer to 6? for 9.45 rounds to 9.5 bt I dunno if 45 cound round the 9 to a 10 or it pulls it back down to 9. In 8th grade math somebody told me .49 or lower I round back down .51 or above I round the number up. Would .50 bring the whole number up or keep it down? --69.229.36.56 (talk) 00:19, 16 February 2010 (UTC)

What? --COVIZAPIBETEFOKY (talk) 01:27, 16 February 2010 (UTC)