Wikipedia:Reference desk/Archives/Mathematics/2009 February 5

= February 5 =

Trig integral?
How do you do the integral:

$$\int {1 \over kv^2-g}\,dv.$$

I tried partial fractions and a substitution of v = sec(theta) but I can't get it. Thanks. Inasilentway (talk) 00:47, 5 February 2009 (UTC)
 * Try using tanh instead. Partial fractions should work fine, though. What went wrong? Algebraist 01:01, 5 February 2009 (UTC)

If k and g are positive, then partial fractions will do it without getting into imaginary numbers, etc.
 * $$ {1 \over kv^2-g} = {\text{something} \over \sqrt{k}\,v - \sqrt{g}} + {\text{something} \over \sqrt{k}\,v + \sqrt{g}}. $$

Do some algebra to figure out what the two "something"s are, and then you have
 * $$ \int {\text{something} \over \sqrt{k}\,v - \sqrt{g}}\,dv = {\text{something} \over \sqrt{k}}\cdot \ln(\sqrt{k}\,v - \sqrt{g}) + \text{constant} $$

etc.

If you use trigonometric substitutions, you should say
 * $$ \sqrt{k \over g}\,v = \sec\theta $$

so that
 * $$ \sqrt{k}\,v = \sqrt{g}\,\sec\theta $$

and then
 * $$ \sqrt{k}\,dv = \sqrt{g}\,\sec\theta\tan\theta\,d\theta $$

and
 * $$ kv^2 - g = g^2\sec^2\theta - g^2 = g^2\tan^2\theta.\, $$

That will lead you to the integtral of 1/sin. That's a hard one, but you don't need to do it since it's in the standard books and tables. Finally, you need to undo the substitution at the end, changing it back from a function of &theta; to a function of v.

Probably simpler just to go with partial fractions. Michael Hardy (talk) 01:14, 5 February 2009 (UTC)

Algebraist and Michael Hardy, thank you. Here's what went wrong when I tried using partial fractions. I set up the equation as Michael stated, using A and B as the somethings, respectively. I solved for A and B and one step in there was: $$kv^2+g = (A+B) \sqrt{k}V + (B-A) \sqrt{g}$$

which led to the system: $$B-A = \sqrt{g}$$

$$A+B = V*\sqrt{k}$$

and then $$ \int {\text{A} \over \sqrt{k}\,v - \sqrt{g}}\,dv + {\text{B} \over \sqrt{k}v + \sqrt{g}}  = \int 1 dv $$

which has got to be wrong... Inasilentway (talk) 01:38, 5 February 2009 (UTC)
 * However did you get $$kv^2+g = (A+B) \sqrt{k}V + (B-A) \sqrt{g}$$? That's not right at all. Algebraist 01:44, 5 February 2009 (UTC)
 * Bad algebra is how I did it. lol. So it should really be

$$1 = (A+B)v\sqrt{k} + (A-B)\sqrt{g}$$... which leaves me without a solution for A or B...

Can I just use this identity: http://www.integral-table.com/eq13.png and set a = k, b = 0, c = -g, x = v?
 * That's for cases where the discriminant b2 &minus; 4ac is negative. Michael Hardy (talk) 02:18, 5 February 2009 (UTC)


 * You have
 * $$ {1 \over kv^2-g} = {A \over \sqrt{k}\,v - \sqrt{g}} + {B \over \sqrt{k}\,v + \sqrt{g}}. $$
 * Multiply both sides by the common denominator and you get:
 * $$ 1 = A( \sqrt{k}\,v + \sqrt{g}\, ) + B( \sqrt{k}\,v - \sqrt{g} \,). $$
 * On the left side you have
 * $$ \blacksquare v + \blacksquare $$
 * and the first $$\blacksquare$$ is 0 and the second is 1.
 * One the left side you have
 * $$ \blacksquare v + \blacksquare $$
 * and the first $$\blacksquare$$ is
 * $$ (A+B)\sqrt{k}\, $$
 * and the second is
 * $$ (A - B)\sqrt{g}.\, $$
 * Therefore (since &radic;k &ne; 0) you have
 * $$ A + B = 0 \, $$
 * and
 * $$ (A - B)\sqrt{g} = 1. \, $$
 * From the first equation you get B = &minus;A. Then you can substitute  &minus;A for B in the second equation. Michael Hardy (talk) 02:15, 5 February 2009 (UTC)

A difficult algebra problem involving logarithms
I would like to express α as a function of f and p given the following:

0 = p * ln(1 - α + αf) + (1 - p) * ln(1 - α)

--Tigerthink (talk) 04:05, 5 February 2009 (UTC)


 * But can you specify what knd of numbers p and f are, and what kind of solution $$\alpha$$ are you looking for? (real/integer/positive/close to zero...?), for the behaviour/number of the solutions etc changes quite a lot. You are not happy with $$\alpha=0$$, I presume. So, (putting g:=1-f) you want to solve wrto $$\alpha$$ the equation
 * $$\textstyle(1-g\alpha)^p(1-\alpha)^{1-p}=1$$,
 * which at least looks more like an algebra problem; is it so? (By the way, if you use LaTeX format people will start answering earlier... it is in the human nature I guess) --pma (talk) 13:07, 5 February 2009 (UTC)


 * p is the probability of something (and therefore between 1 and 0.) f is a positive real number.  fp is greater than 1.  I would be surprised if α is ever less than 0 or greater than 1 if those two conditions hold.  Sorry, I don't know LaTeX.  And you might find it useful to know that the thing I'm really interested in is the limit of the value of α as f goes to infinity in terms of p.--Tigerthink (talk) 16:36, 5 February 2009 (UTC)
 * It looks to me like $$\alpha\rightarrow-\infty$$ as $$f\rightarrow\infty$$. The first term becomes (if you excuse the very informal notation) $$p\log(\infty)=\infty$$, so the only way the total could be 0 is for the second term to be $$-\infty$$. That can only happen if $$\alpha=-\infty$$. --Tango (talk) 16:54, 5 February 2009 (UTC)
 * But then the first term goes to log{-∞) or log(∞), depending on the exact rates at which f and α diverge, so it must be more complicated than that. α=0 always works, though. Algebraist 17:06, 5 February 2009 (UTC)
 * Good point, but it's irrelevant because I was answering the wrong question anyway (see below). α=0 is a rather boring answer... --Tango (talk) 22:47, 5 February 2009 (UTC)
 * If the second term $$\log(1-\alpha)$$ is $$-\infty$$, then actually $$\alpha=1$$, not $$\alpha=-\infty$$. — Emil J. 17:11, 5 February 2009 (UTC)
 * More precisely, if 0 < p < 1 is fixed, then I think $$\alpha=1-f^{-p/(1-p)}+O(f^{-2p/(1-p)})$$ as $$f\to\infty$$. — Emil J. 17:26, 5 February 2009 (UTC)
 * Wow, that's a pretty complicated formula. Can you give me any clues as to how you derived it?  And if you'll excuse my mathematical ignorance, what does O mean in this context?  Also, if it's the limit as f goes to infinity, then why is f present in the expression?--Tigerthink (talk) 20:52, 5 February 2009 (UTC)
 * See Big O notation. Algebraist 20:58, 5 February 2009 (UTC)


 * The limit as such is just 1, the more complicated expression provides info on the rate of convergence towards the limit. The big O notation also involves a sort of limit (limsup, actually), that's why it says $$f\to\infty$$. Now, the derivation is as follows.


 * First some observations. The expression is only well-defined for −1/(f − 1) < α < 1. Computing the derivative reveals that, as a function of α, it is increasing below p − (1 − p)/(f − 1), and decreasing above. The value for α = 0, where the function is increasing, is 0, and the limit for α → 1 is −∞, hence there exists exactly one nonzero solution, and it is in the interval (p − (1 − p)/(f − 1), 1).


 * We can write
 * $$\log(1-\alpha+\alpha f)=\log f+\log(\alpha+(1-\alpha)/f).\,$$
 * Since α > p − (1 − p)/(f − 1) > p/2 for large enough f, the term log(α + (1 − α)/f) is bounded, hence we have
 * $$p\log f+O(1)=-(1-p)\log(1-\alpha),\,$$
 * thus
 * $${\alpha=1-\exp\left(-\frac p{1-p}\log f+O(1)\right)=1-f^{-p/(1-p)}\Theta(1)=1+O(f^{-p/(1-p)}).\qquad(1)}$$
 * Now we repeat the process. By (1), we have
 * $$\log(\alpha+(1-\alpha)/f)=\log(1+O(f^{-p/(1-p)}))=O(f^{-p/(1-p)}),\,$$
 * thus
 * $$p\log f+O(f^{-p/(1-p)})=-(1-p)\log(1-\alpha),\,$$
 * $${\alpha=1-\exp\left(-\frac p{1-p}\log f+O(f^{-p/(1-p)})\right)=1-f^{-p/(1-p)}(1+O(f^{-p/(1-p)}))=1-f^{-p/(1-p)}+O(f^{-2p/(1-p)}).}$$
 * One could obtain even better estimates by the same method, but it gets messy. — Emil J. 12:35, 6 February 2009 (UTC)
 * "The value for α = 0, where the function is increasing, is 0, and the limit for α → 1 is −∞, hence there exists exactly one nonzero solution, and it is in the interval (p − (1 − p)/(f − 1), 1)." Are you sure you don't mean "zero" instead of "nonzero"?  Also, I don't think you've explained why f is present in the limit expression.  Your derivation is way beyond my current level of math ability.  Respect.--Tigerthink (talk) 04:52, 7 February 2009 (UTC)


 * Oh, yeah, for some reason I was thinking p>1, I'm not sure why... For the correct values of p, α=1 (in the limit) would work, you're right. --Tango (talk) 22:47, 5 February 2009 (UTC)

What's the best tool for signal intersection?
Say I have two signals, $$S_1$$ and $$S_2$$. Both contain a similar signal, with a particular noise each. Basically:
 * $$S_1 = S + N_1$$
 * $$S_2 = S + N_2$$

Now, what I can do to extract S? Given there are three variables and only two equations, I suppose it can't be solved so easily. In the case of signal processing, then, what tools are available to solve this kind of problem? What I really want here is some theoretical tool. Cheers! &mdash; Kieff | Talk 05:39, 5 February 2009 (UTC)
 * If you know nothing about what the signal might be, and nothing about how the noise might have occurred, then you can't really say anything. If you're able to experiment on the channel (measuring the output from a given input), you can get a very good picture of how the probability density function of the noise looks (it might be both white and Gaussian, for instance), and then it's easy to make an educated guess and also fairly easy to compute the reliability of that guess. The Statistical signal processing and Wiener filter articles might help. — JAO • T • C 17:21, 5 February 2009 (UTC)
 * I might have some sort of frequency/spatial mixup, but if this were a measurement of physical quantities, then a very sane approach is just to average the two signals, S3 = (S1 + S2)/2. Assuming the noises N1 and N2 have a mean of 0 (and possibly that they are a sum of a uniform (white) and a normal (gaussian) random variable with mean 0), then the noise of S3 should have smaller variance than the noise of either S1 or S2.  I tried reading the Wiener filter article, but it appeared to only look at one signal.  I might have missed something due to several implicit fourier transforms. JackSchmidt (talk) 17:41, 5 February 2009 (UTC)


 * There are a few things that occur to me, I'm sure a sound engineer would be much better.There are lot of different types of noise and the noise on both channels may have a different distribution. The fourier transforms of the signals can be compared and where a signal occurs on one but not the other that could be removed from that signal. After that if the two noises are independent then N1+N2 will have the same frequency distribution as N1-N2 = S1-S2, one could probably remove hum or suchlike that way. After that I guess just averaging would remove a bit more noise. Dmcq (talk) 21:25, 6 February 2009 (UTC)

Open interval as a disjoint union of closed intervals
Is it possible to write the open interval (0,1) as a disjoint union of closed positive-length intervals, or not?

131.111.8.96 (talk) 06:18, 5 February 2009 (UTC)BTS
 * No. Let T be the set of endpoints of the intervals. Then T ∪ {0,1} is a perfect set of reals (nonempty, closed and without isolated points), hence uncountable. But this is a contradiction since there clearly can be only countably many intervals. Joeldl (talk) 09:03, 5 February 2009 (UTC)

Sorry - why can there only be countably many intervals? Is it because each must uniquely enclose a rational?

131.111.8.97 (talk) 17:11, 5 February 2009 (UTC)BTS
 * Yes. Now go and do your own example sheet. Algebraist 17:14, 5 February 2009 (UTC)

Log normal distribution
Hello. Given the parameters (mu and sigma), how would I calculate the 99.5th percentile for a lognormal distribution? My calculator can calculate the percentiles for normal distributions. --Rekees Eht (talk) 08:36, 5 February 2009 (UTC)
 * The article lognormal distribution has formulas for computing the parameters for the corresponding normal distribution. Bo Jacoby (talk) 09:31, 5 February 2009 (UTC).

Sorry but I don't understand. Let's say I have a lognormal variable and I know its mean and variance. I can use the formulas to work out the corresponding normal parameters. But how would I work out the 99.5th percentile for the original lognormal variable? That article is hard to understand. --Rekees Eht (talk) 09:54, 5 February 2009 (UTC)

Okay I did something. Please let me know if it's right. Say I have a lognormal variable with mean 0.000325 and std dev 0.00900802 and I want to find the 99.5th percentile. Using the formulas you provided, I calculated that the corresponding parameters for the normal distribution is mu = -11 and sigma = 3. Then I used excel's "LOGNORMDIST" function with the parameters and did a goal seek on the "x" until the CDF equals 99.5%. The answer I got is 0.995566. Does this sound reasonable? --Rekees Eht (talk) 15:21, 5 February 2009 (UTC)


 * An easier approach is to use excel's inverse logdist function as follows: =LOGINV(0.995,0.000325,0.00900802). The solution this returns is 1.023807. Wikiant (talk) 15:49, 5 February 2009 (UTC)


 * Thanks but I think you are supposed to use the corresponding normal parameters (-11 and 3). When I do that I get an answer of 0.008967. Can you explain why that is so small? --Rekees Eht (talk) 16:15, 5 February 2009 (UTC)


 * If I am understanding you correctly, -11 and 3 are the parameters for the *normal* distribution. The LOGINV function is expecting parameters for the *lognormal* distribution. Wikiant (talk) 16:28, 5 February 2009 (UTC)

Countable series in the rationals
Is there any possible way to list the rationals Q as $$q_1,q_2,q_3....$$ such that $$\sum({q_n-q_{n+1}})^2$$ converges?

Thanks, 131.111.8.97 (talk) 16:42, 5 February 2009 (UTC)Mathmos6
 * Yes. For more details, consult your Analysis I supervisor, since he/she is paid to teach you this material, and we are not. Algebraist 16:54, 5 February 2009 (UTC)
 * An example would be $$q_n=(1/2)^n$$. Anythingapplied (talk) 19:43, 5 February 2009 (UTC)
 * I think the OP wants *all* the rationals to be included. If it's possible, it certainly wasn't taught in my first year analysis course... --Tango (talk) 19:46, 5 February 2009 (UTC)
 * It wasn't taught in mine, either, but it can be solved using results from the course. If it was taught, I might be willing to explain it, since it would be something the OP would have to know. It's supposed to be an interesting problem for students to attack, and it doesn't do someone any good to tell them the solution to an interesting problem. Attacking interesting problems is how one becomes better at mathematics. Algebraist 21:02, 5 February 2009 (UTC)
 * nb: this is a homework question. Algebraist 21:09, 5 February 2009 (UTC)
 * But how can we know if (s)he really have a paid supervisor... And in the case, what's wrong if (s)he wants our help? As a first hint for 131, I suggest to reflect on this fact: a finite sum of the form $$\sum_{j=1}^{n}({q_j-q_{j+1}})^2$$ can be made arbitrarily small choosing suitably the number $$n$$ and the $$q_1,q_2....,q_{n+1}$$, and one can do that even under the condition that the first and the last of the $$q_j$$ are two given numbers a and b, and there is still the freedom to choose the other$$q_j$$ avoiding some numbers, if one wants... I hope this will be of help, but not too much help. (Algebraist: I'll remove this post if it is not opportune) pma (talk) 21:30, 5 February 2009 (UTC)

I'm actually a Physics student rather than a Mathematics student - I simply do the maths example sheets too to try and improve my ability/as a matter of interest, since the physics isn't particularly challenging and I enjoy them, so I don't actually have an analysis supervisor. I try to read up as much as I have time to on whatever the topic is, in this case analysis, but this sort of problem hasn't been brushed upon in any of my reading so far - I wasn't expecting an explicit answer, because I was under the impression the reference desk was intended for guidance rather than that, but I apologize if I made it sound like I expected you to 'teach me' the material; I merely wanted a nudge in the right direction, if nothing else just so I could read up further on the appropriate topics in analysis, but any additional assistance was always welcomed. Thankyou for the help - I'll go away and have a think about it. 131.111.8.99 (talk) 21:38, 5 February 2009 (UTC)Mathmos6
 * We'll be happy to help you if you come back with your ideas after thinking about it. A hint: I doubt you'll find much useful in further reading; this requires thought, not knowledge. Algebraist 21:44, 5 February 2009 (UTC)

I do not know the answer, but I think that the answer is no. If the series $$\scriptstyle\sum_{n=1}^\infty({q_n-q_{n+1}})^2$$ converges, then for each positive real $$\scriptstyle\varepsilon $$ there exists a positive integer  $$\scriptstyle N_\varepsilon$$ such that $$\scriptstyle\sum_{n=N_\varepsilon}^\infty({q_n-q_{n+1}})^2<\varepsilon$$. So all, except a finite number, of the $$\scriptstyle q_n$$ are very close to one another. I do not expect these numbers to include all, except a finite number, of the rationals. Perhaps a proof can be made along this line. Good luck. Bo Jacoby (talk) 22:21, 5 February 2009 (UTC).
 * I already gave the answer. It is yes. Algebraist 22:23, 5 February 2009 (UTC)
 * I thought my hint was even too explicit. So, to start try to go from $$q_0=0$$ to $$q_{100}=1$$ with a corresponding sum =1/100 or so . --pma (talk) 22:48, 5 February 2009 (UTC)

So then, could you split the infinite sum up into an infinite series of finite sums, the values and bounds of which you 'choose' as pma suggested so that the size of the sums tends to 0 but the full collection of sums contains every single rational? —Preceding unsigned comment added by 131.111.8.104 (talk) 23:04, 5 February 2009 (UTC)
 * Well suppose you have an enumeration $$r_k$$ of the rationals; start your sum defining $$q_0:=r_0$$; your first task is to reach $$r_1$$; you can make as many steps as you want, till you get $$\scriptstyle r_1=q_{n_1}$$ for some $$ n_1$$ with a partial sum say < $$1/2^1$$. Then go on. Remember, you have to reach all $$r_k$$'s, each one only once... --pma (talk) 23:22, 5 February 2009 (UTC)
 * (PS: formally: define the $$q_k$$ by induction accordingly)

How about this: Let a1, a3, a5, ... list all the rationals. Then for n odd, let an+1 = an + 1/n. Then at least the sum of (an+1 &minus;an)2 over odd n would converge. Next try to figure out how to tweak that idea to make it work when the terms (an+2 &minus;an+1)2 for n odd are also added. Michael Hardy (talk) 23:38, 5 February 2009 (UTC)


 * But don't you have repetitions this way, since the a with odd indices already cover all rationals? I am also curious to see Algebraist's construction for he has some elegant trick for sure.... The OP is not homework! pma (talk) 00:22, 6 February 2009 (UTC)


 * Yes, you get repetitions. Does "list the rationals" mean no repetitions?  If so, we have still more work to do. Michael Hardy (talk) 23:58, 6 February 2009 (UTC)

So then - sorry if I haven't quite followed - you could take r0 to be say 1, reaching r0 from 0 with a partial sum of less than or equal to 1/2, then since you have more omitted rationals on either side of 1 you need to go 'both ways', so you could go from r0=1 to r1=-1 say, omitting the already counted rationals as 'pma' pointed out is possible earlier, with a new partial sum <1/4, then back to r2 = say, 2, omitting previously chosen rationals to obtain a partial sum <1/8, then to -2, 3, -3 and so on by the same process? As your steps must get 'smaller' in order to reduce the partial sum, would this ensure that you didn't miss out any rationals? I suppose going from 0->1 you could just take 1 step, 0->1/2->1, then on the way back, you could choose the path of previously unchosen 'thirds' to make sure you didn't omit any rationals (1->2/3->1/3->) and then choose halves (0->-1/2->-1) for our newly included interval, adding as many interstitial previously unchosen rationals as needed inbetween each rational to make sure your sum is less than 1/4? (Would this step need to be formalized more to avoid causing problems later on? You should be able to 'influence' the sum as necessary, since the closer the rationals you choose are together, the smaller the sum will be.) Continuing onwards with -1->-2/3->-1/3->0->1/4->...2 and then adding in more interstitial rationals to influence your sum - would this method work, since the rationals are dense and you could always ? Apologies for the lack of formatting, too tired to LaTeX! Thanks,

131.111.8.102 (talk) 03:29, 6 February 2009 (UTC)Mathmos6


 * Yes, you got the idea pretty well, but notice that you don't have to worry about which is the value of a given $$r_k$$, nor whether to go left or right. Let's just say that $$\textstyle \{r_j\}_{j\geq0}$$ is an enumeration (bijective) of $$\scriptstyle \Q$$. The situation at the kth stage is: you have already choosen rational numbers $$\textstyle q_0, q_1, q_2,..,q_{n_k} $$ which include among them all $$\textstyle r_0, r_1, r_2,..,r_{m_k} $$, but not  $$\textstyle  r_{m_k+1}$$, and you did it with a partial sum
 * $$\textstyle \sum_{j=1}^{n_k}({q_j-q_{j-1}})^2<\sum_{j=1}^k 2^{-j}.

$$. Now, you have to go from $$\textstyle q_{n_k}$$ to $$\textstyle r_{m_k+1}$$ choosing some not previously choosen rationals $$\textstyle q_{n_{k}+1},...,q_{n_{k+1}}:=r_{m_k+1}$$, and with increment of the sum :$$\textstyle \sum_{j=n_k+1}^{n_{k+1}}(q_j-q_{j-1})^2 < 2^{-k-1}  $$. To do so you just need a large number of little steps (precisely, $$\scriptstyle n_{k+1}-n_{k}>2^k(q_{n_k}-r_{m_k+1})^2$$ sufficies). The best should be making equal steps, but in order to avoid repetitions you can perturb them a little, keeping the sum less than $$2^{-k-1}$$. You can do that, of course, because the set where you pick the $$q_j$$ for $$\scriptstyle n_k<j<n_{k+1}$$, that is $$\scriptstyle \Q\setminus\{ q_0,q_1,q_2,..,q_{n_k}\}$$, is dense. The final sequence $$\textstyle \{q_j\}_{j\geq0}$$ is a re-ordering of  $$\textstyle \{r_j\}_{j\geq0}$$ for it contains each term of it only once, and gives a sum of the whole series less or equal than 1 (and by the way if you want it to be less than 1/100 just take the $$q_j/10$$, that are again an enumeration of the rationals) .--pma (talk) 11:12, 6 February 2009 (UTC)

Take any sequence of rationals that starts with 0, steps (using some number of upwards steps) up to 1, steps down to -1, steps up to 2, steps down to -2, and so on. Make the steps small enough that the total sum of squares is less than 1. This sequence steps across each rational that is not in the sequence infinitely often. Now insert each such missing rational x/y (written in lowest form) into the sequence during the y-th time it is stepped across. The new sequence includes all rationals and still has sum of squares less than 1. Does it work? McKay (talk) 07:06, 7 February 2009 (UTC)
 * Well it should, up to details to be fixed... but then I prefer to go from $$r_0$$ to $$ r_1$$, then to $$r_2$$ (if not already visited), then to $$r_3$$,...&c, so this way we are sure to be visiting all rationals. And since they are dense, we are able not to pass twice for the same one.
 * A further question ( expecially for whom skipped his homework ) is: can one list the rationals vectors $$\Q\times\Q$$ as $$q_1,q_2,q_3....$$ such that the sum of square norms $$\sum|{q_n-q_{n+1}}|^2$$ converges?? --pma (talk) 13:34, 7 February 2009 (UTC)