Wikipedia:Reference desk/Archives/Mathematics/2009 February 6

= February 6 =

Series
Is there a determine whether the infinite series
 * $$\sum_{n=3}^\infty \frac{1}{n\ln(n)\ln(\ln(n))}$$

converges or diverges? I've tries several tests repeatedly, but to no avail. —Preceding unsigned comment added by 70.52.46.213 (talk) 02:14, 6 February 2009 (UTC)

Consider your function $$\frac{1}{x\ln(x)\ln(\ln(x))}$$ over all x rather than just the natural numbers - is it the derivative of something similar to part of your denominator perhaps? See if you can work out the integral of $$\frac{1}{x\ln(x)}$$ first, in terms of a single function. In which case, do you know a series convergence test related to the derivative of a function? 131.111.8.102 (talk) 02:57, 6 February 2009 (UTC)Mathmos6


 * Well, I know that the integral of 1/nln(n) is ln(ln(n)), so by the integral test it diverges...but how would I go from there? —Preceding unsigned comment added by 70.52.46.213 (talk) 03:01, 6 February 2009 (UTC)

The integral of 1/n is ln(n), the ingegral of 1/nln(n) is ln(ln(n)), the integral of 1/nln(n)ln(ln(n)) is ___? Can you spot the pattern? 131.111.8.102 (talk) 03:04, 6 February 2009 (UTC)Mathmos6

Ah thank you very much. —Preceding unsigned comment added by 70.52.46.213 (talk) 03:16, 6 February 2009 (UTC)

Question on roots
This has left me totally flummoxed: how do I prove that the equation:
 * $$x^5 + x = 10\ $$

has only one root, lying between 1 & 2, which is irrational? While the 1st and 2nd parts can be solved by plotting the graph for the function $$ f(x) = x^5 + x- 10$$, the 3rd part stumps me. A helping hand please? I know the policy of not asking homework questions full well and respect it. This is just a practice question from one of the tougher Indian high school level books.--Leif edling (talk) 03:31, 6 February 2009 (UTC)


 * As f(1) = &minus;8 and f(2) = 24, there is clearly at least one real root between 1 and 2 (by the intermediate value theorem). I think you have to work a bit harder to show that there is only one real root - how do you know that there are not 3 or 5 real roots ? Hint: if a > b, what can you say about a5 and b5 ? So what can you say about a5 + a and b5 + b ?
 * To show that the real root is irrational, you can use the rational roots theorem to show that there are only a few possible values for rational roots, and they happen to all be integers. But you know that the only real root is not an integer (because it is greater than 1 and less than 2) so ... Gandalf61 (talk) 10:18, 6 February 2009 (UTC)


 * Why not just show that the derivative of $$y=x^5+x-10$$ is greater than zero for all values of x? 92.4.224.144 (talk) 22:54, 6 February 2009 (UTC)
 * That would show that there cannot be more than one root. It would NOT show that there is at least one root, and it would NOT show that the root is irrational. Michael Hardy (talk) 01:54, 7 February 2009 (UTC)
 * Why would you want to show there 'is at least one root' when you can show there is exactly one root, especially when the originally questions says 'has only one root'? 92.4.224.144 (talk) 18:46, 7 February 2009 (UTC)
 * Because to show that there is exactly one root requires showing that there is at least one root? Algebraist 18:48, 7 February 2009 (UTC)


 * To 99.2: it isn't true that a function whose derivative is positive for all values of x must have exactly one root.  Consider a function that has a horizontal asymptote above the x-axis, which it approaches as x approaches negative infinity.  Such a function wouldn't have a root, even if its derivative is, say, x^2+1.  --Bowlhover (talk) 22:10, 8 February 2009 (UTC)
 * But, of course, such a function couldn't have derivative x^2+1 since the only functions with that derivative are of the form 1/3 x^3+x+C, which tends to negative infinity as x tends to negative infinity. How about e^x+1, which has derivative e^x? --Tango (talk) 01:43, 10 February 2009 (UTC)

Nice example by Tango. But my MAIN problem was with the last part which has only been touched by Gandalf61. The root may not be an integer, but would that imply that it must be irrational (e.g. it could be a recurring decimal)? Everyone else has just discussed the methods to solve the first two parts, which weren't much of a struggle. Too late for a reply now, I guess. Thanks, though. --Leif edling (talk) 08:00, 11 February 2009 (UTC)


 * You should read the article on the rational roots theorem that Gandalf linked to. This theorem can be used to prove that your root is irrational:  the theorem provides an algorithm to find all rational roots, so any roots that are not found with this algorithm must be irrational.  If you are stuck ask for help again.  Eric.  131.215.158.184 (talk) 10:32, 11 February 2009 (UTC)

Beat Frequency for Pulses
Hello,

I was looking at the wikipedia article for beat frequency and there it lists this formula for calculating the beat frequency of two sin waves:


 * $${ \sin(2\pi f_1t)+\sin(2\pi f_2t) } = { 2\cos\left(2\pi\frac{f_1-f_2}{2}t\right)\sin\left(2\pi\frac{f_1+f_2}{2}t\right) }$$

It doesn't, however, say how to calculate the beat frequency of binary pulses. In a simple example, I'm thinking of two blinking lights that are blinking at slightly different frequencies. Or in a more complicated way, pulses of electricity that are very brief, with longer times of no pulse.

Does anyone know the formulas for these examples? Thank you in advance.

--Grey1618 (talk) 06:08, 6 February 2009 (UTC)


 * The same formula should work, shouldn't it? Just think of a sine wave with the light only being on at the peak of each wave. You won't get beats in the same way, though, since the blinking lights can't destructively interfere (cancel each other out) - you'll just see them coming into synch and then going out of synch again, but the frequency will be the same as with sine waves. --Tango (talk) 10:18, 6 February 2009 (UTC)

If one of them blinks at times t, 2t, 3t, 4t, ..., and the other at s, 2s, 3s, 4s, ..., then maybe you're looking for the least common multiple of t and s, which exists if t/s is rational. For example, if t/s = 3/8, then 8t = 3s = the least common multiple = the time between "common beats" ("common beats" = times when BOTH lights blink). The times when both lights blink simultaneously will be 8t, 16t, 24t,... and so on, which is the same as 3s, 6s, 9s,... .

If t/s is irrational, then they can blink simultaneously at most once. Then you may want to know when the blink at approximately the same time, and define "approximately" to mean a time difference of less than some amount you specify. Then you've got a problem in Diophantine approximations, which is more involved. Michael Hardy (talk) 23:37, 6 February 2009 (UTC)

Integers.
Two integers are related if their sum is a multiple of two. Show that this is an equivalence relation on the set of integers. —Preceding unsigned comment added by 196.201.151.5 (talk) 15:24, 6 February 2009 (UTC)


 * We're not going to do your homework for you. Do you know the definition of an equivalence relation? Just go through each of the conditions checking them one at a time. If you get stuck, write out what you've got so far here and we'll help you work out the next step. --Tango (talk) 15:28, 6 February 2009 (UTC)

Time in relation to Travel
A passenger is traelling from London to Los Angeles. The flight leaves London on Thursday 5th October at 1615 and will take 11 hours and 15 minutes, arriving in Los Angeles at 1930 5th October. Ignoring any potential Daylight Saving Time changes, can some one please answer the following qustions.

a) How many hours behind London is Los Angeles?

b) How many hours behind Auckland is Los Angeles?

c) What is the date and time in Los Angeles when the flight departs London?

d) What is the date and time in London when the flight arrives in Los Angeles?

e) What is the date and time in Auckland New Zealand whne the flight departs London?

f) What is the date and time in Auckland New Zealand when the flight arrives in Los Angeles?

Thanking any one who can help me. —Preceding unsigned comment added by 222.154.40.109 (talk) 20:35, 6 February 2009 (UTC)


 * Please show us your thinking on homework problems before asking for help. You can answer a, c, and d with the info given, but, since you included no info about the time zone for Auckland, we couldn't answer that without looking up the time zones first (which might be rather Auckward).  :-) StuRat (talk) 21:05, 6 February 2009 (UTC)


 * To help you get started, add the length of time for the flight to the departure time. This should give you the arrival time in the time zone of the departure city.  Now take the difference between that time and the arrival time in the arrival city, and that gives you the time difference between the two time zones (ignoring daylight savings time). StuRat (talk) 21:11, 6 February 2009 (UTC)


 * This is the strangest homework question I've seen here on the Reference Desk. Somebody who is learning basic arithmetic posting here on Wikipedia?  Weird.  --99.237.96.81 (talk) 04:35, 7 February 2009 (UTC)


 * I see nothing weird about it. We handle all ranges of questions, from the elementary to the post-doctoral level. StuRat (talk) 08:55, 7 February 2009 (UTC)
 * And, interestingly, the easier it is, the more likely it is to have "hard" in the title... --Tango (talk) 18:56, 7 February 2009 (UTC)

Ackland is 20 hours behind LA