Wikipedia:Reference desk/Archives/Mathematics/2010 February 9

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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)