Wikipedia:Reference desk/Archives/Mathematics/2008 January 3

= January 3 =

Differential equation
I'm trying to solve this:

dy/dx = 4x + 4x/sqrt(16-x^2)

I figured,

y = ∫(4x + 4x/sqrt(16-x^2))dx y = 2x^2 + ∫4x(16-x^2)^(-1/2)dx y = 2x^2 + ∫(64x-4x^3)^(-1/2)dx

let u = 64x-4x^3 <-- But I don't know how to isolate x from here so that I can express x in terms of u. du = (64-12x^2)dx dx = du/(-12x^2+64)

So I'm stuck here:

y = 2x^2 + ∫u^(-1/2)(du/(-12x^2+64))dx

How do I isolate that x so that I can substitute for it in terms of u and finish the integration?

Thanks!

--anon —Preceding unsigned comment added by 70.23.86.90 (talk) 04:46, 3 January 2008 (UTC)
 * Check again... you can't distribute the $$4x$$ in the numerator into the radical that's in the denominator. Let $$u=16-x^2$$... then $$du = -2xdx$$, so you have $$-2du$$ in the numerator. Hence, the second part is just $$-2\int{u^{-1/2}du}$$... so $$-4\sqrt{u} = -4\sqrt{16-x^2}$$. (+ the first part of the antidifferentiation + C, of course) -- Kinu t /c  05:28, 3 January 2008 (UTC)
 * Check again... you can't distribute the $$4x$$ in the numerator into the radical that's in the denominator. Let $$u=16-x^2$$... then $$du = -2xdx$$, so you have $$-2du$$ in the numerator. Hence, the second part is just $$-2\int{u^{-1/2}du}$$... so $$-4\sqrt{u} = -4\sqrt{16-x^2}$$. (+ the first part of the antidifferentiation + C, of course) -- Kinu t /c  05:28, 3 January 2008 (UTC)


 * (edit conflict) You have an error. 4x is not under the radical sign. So distribution of $$4 x \left(16-x^2\right)$$ is wrong.
 * So, indeed you have to solve
 * $$\int \left(\frac{4 x}{\sqrt{16-x^2}}+4 x\right)\,dx$$
 * 4x is easy. For the second, you may try the substitution
 * $$u=16-x^2$$, which yields du=-2x dx. Pallida  Mors  05:39, 3 January 2008 (UTC)


 * I'd say that $$u^2=16-x^2$$ would be a better choice. --Taraborn (talk) 17:46, 6 January 2008 (UTC)

probability question
An accounting population consists of 100 accounts, 75 correct and 25 wrong. If 12 accounts are chosen at random, find the probability that at least one account is in error.

Would I be right in saying the answer to this question is

$$1-(88!/100!)$$136.206.1.17 (talk)
 * No. Note that this answer doesn't in any way make use of the relevant information regarding the amount of wrong accounts. You are correct, though, that you need to take 1 minus the probability that none will be wrong. So, think about this: What is the probability that the first account chosen is correct? Given that it is, what is the probability for the second? The third? And so on. -- Meni Rosenfeld (talk) 17:07, 3 January 2008 (UTC)


 * In case the OP is interested in reading more about such probability settings and analysis, Wikipedia has an article on hypergeometric distribution. Pallida  Mors  19:46, 3 January 2008 (UTC)

Okay than would it be

1-((25!/100!)*(88!/13!))

If its not could somone tell me the answer and why its like that. This isnt a homework question by the way, i'm trying to understand it for an exam. 136.206.1.17 (talk) 10:30, 4 January 2008 (UTC)
 * Almost. The expression $$\frac{25!}{100!}\frac{88!}{13!}$$ will give the probability that all of them will be wrong, but you are trying to find the probability that all of them will be correct (so 1 minus that will give you the probability that at least one is in error). -- Meni Rosenfeld (talk) 12:09, 4 January 2008 (UTC)

As I have just been told off by ConMan for volunteering to give you the answer I will just give you a BIG clue. The probabaility that the first one is correct is $$\frac{75}{100}$$, the probability that the second one is correct is $$\frac{74}{99}$$ etcetera. Multiply together all 12 of those, express it in factorials and subtract from one. You should now be able to see what is wrong with your expression.  Sp in ni ng  Spark  17:36, 5 January 2008 (UTC)
 * I'm not sure the OP requested this extra hint. I think he misunderstood you, and his plea to "put [him] out of [his] misory [sic]" referred to the new question. -- Meni Rosenfeld (talk) 17:40, 5 January 2008 (UTC)


 * Sorry for butting in in that case. But I thought you were going to bed?  Sp in ni  ng  Spark  17:55, 5 January 2008 (UTC)
 * Oops, wrong again, it was ConMan who said he was tired.  Sp in ni ng  Spark  17:57, 5 January 2008 (UTC)

Numeral system incorporating e
The article on numeral systems mentions that the only reason we use a system with ten bases is because we happen to have five fingers on each hand. Although the article mentions systems using different bases (five, eight, even sixty!), I didn't see one that uses the mathematical constant e. Does such a system exist? If not, would it be possible and/or useful?

Mikmd (talk) 22:18, 3 January 2008 (UTC)


 * Such a system does exist, though I know of no practical uses. Any numerical value can be used as a base. Think about this, what is "2" in base e??? what about "10" in base e?? or "20"? or "100"? (Answers: 2, e, 2e, and $$e^2$$.) Base e handles natural logarithms similar to how base 10 handles base 10 logarithms A math-wiki (talk) 22:32, 3 January 2008 (UTC)


 * I've always thought about a base e numerical system, but never had the will to explore it. Given e's useful properties as an exponentiation base, shouldn't a numerical system based on it have some interesting properties as well? &mdash; Kieff | Talk 22:40, 3 January 2008 (UTC)
 * We've had a similar discussion a while ago, it's worth checking it out. Ultimately, I don't think it turns out to be very useful or elegant. -- Meni Rosenfeld (talk) 23:01, 3 January 2008 (UTC)
 * I'll just say that Non-standard positional numeral systems has some interesting examples. I find $$\varphi$$ to be the most interesting irrational base (see Golden ratio base), since its rules are remarkably simple (the digits are 0 and 1, with the restriction that "11" doesn't appear anywhere). This easily generalizes to any algebraic number solving a polynomial with coefficients 0 and 1, and perhaps also to other algebraic numbers. For other numbers, there is a "brute force" construction which works but is very ugly. A somewhat irrelevant example is factoradic, which is interesting for integers but not seamless for fractions. -- Meni Rosenfeld (talk) 23:14, 3 January 2008 (UTC)
 * My personal favourite is the Knuth Quarter-imaginary base, where you can represent every complex number using only the digits 0,1,2,3 without a sign. 83.250.203.75 (talk) 10:07, 5 January 2008 (UTC)
 * I had some fun with base $$-1+i$$: use the high bit for green, the next bit for red ... The resulting fractal pattern looks like autumn leaves (of a species whose colors include blue). &mdash;Tamfang (talk) 00:16, 7 January 2008 (UTC)