Bug on Band (from Feynman Lectures on Physics)

CHECK THE COMMENT BEFORE YOU READ THIS! (It's pretty awesome.)

This is a problem that i found while on this website: http://feynmanlectures.info/ (all credit goes there). The problem is this (and there is a funny story behind it):

An infinitely stretchable rubber band has one end nailed to a wall, while the other end is pulled away from the wall at the rate of 1 m/s;  initially the band is 1 meter long. A bug on the rubber band, initially near the wall end, is crawling toward the other end at the rate of 0.001 cm/s. Will the bug ever reach the other end? If so, when?

Lev Okun  gave this problem to Andrei Sakharov to pass the time while they were being driven from Moscow to JINR in Dubna.  However, it didn't work (to pass the time) because immediately after being told the problem, Sakharov pulled out a pen, took Lev's magazine and wrote down the solution, without any hesitation whatsoever.


So it actually has a pretty nice history, and much respect should be given to Sakharov for his quick solution! Although, i see how, after having done the problem, that once you do one, you've done 'em all. I also want to mention another similar problem that i came across some time ago, and just remembered when i read this one. It is from a book called Professor Stewart's Cabinet of Mathematical Curiosities by Ian Stewart; the problem is as follows and is made slightly simpler than the one above (but trivially so, as it turns out. But it is actually kind of nice that way! - I'll explain more what I mean after I present my solutions below...):


The spaceship Indefensible is at the center of a spherical galaxy with radius 1000 lightyears (lyrs). The spaceship travels at a rate of one lightyear every year - the speed of light (assume it's made of photons...). At exactly one year intervals after the Indefensible starts its voyage, the universe expands instantly by 1000 lyrs, and the ship is carried along with the space in which it sits.


So these two problems are really one and the same, except that the bug is being carried along as the band expands constantly while the spaceship is only effected by the expanding space at discrete intervals.
We'll see how this effects the solution later.


At this point I kind of want to leave these problems up to you guys to solve, without giving away too many hints. One hint I will give for Stewart's problem (which he gives as well) is that the following link may be helpful: http://en.wikipedia.org/wiki/Harmonic_series_(mathematics) (particularly the section on Rate of Divergence...)

I'll upload the solutions sometime later, but don't peek! It's (They're) a quite rewarding problem(s) to work!

I'll note that the entire time I was working the Bug on Band problem I thought that there was absolutely no way that my solution was even remotely correct, until I got the right answer! Just goes to show you sometimes...

Y!A: Power series

This was a question i very much enjoyed answering because many people have trouble with - and actually absolutely hate! - power series and convergence and whatnot. but it is exactly the opposite! i LOVED the subject! (don't even get me started on Taylor Series!) and so i feel that this is an area in which i can do a lot of good and help clarify some subjects that some people have trouble with and dislike for no good reason at all!

and i'll also say that one of things that i think makes this subject hard for some people is that it requires the ability to really understand the mathematics and be able to draw conclusions and meaning from the equations. this is something A LOT of people lack, and it is not even something that is emphasized in schools! and this is a terrible circumstance, which i battle frequently, and i hope that i am doing my part to give people this familiarity with mathematics which is really useful and, above all, FUN!

it is also good to see that the asker was very pleased by my answer, and i really hope i was able to make a difference in the way he/she thinks about these subjects.

http://answers.yahoo.com/question/index;_ylt=AoITVMn2GywDSI2pg_R9oV3ty6IX;_ylv=3?qid=20100806112614AAmbpOX&show=7#profile-info-ewbG5Mazaa

Q:

Calculus: Power Series?

I thought that determining weather series converged or diverged was easy, but I seem to be having some problems figuring out how to do that with power series, so can you help me with this problem. Thanks :)

Find the interval of convergence for the given power series:

∑ (n=1 to ∞) [((x-9)^n) / (n(-7)^n)]

Additional Details [the asker posted this after someone else had posted a response with just the answer and nothing else. really pathetic in my opinion...]

I already knew that the answer was (2,16] so you haven't helped me at all. I want to see the steps taken to get to the answer because that is the part that I need help with.

A:

the key you need to solving this problem is to use the ratio test. most of the time, the ratio test will get you the answer you need to convergence problems.

if you recall, the ratio test is the limit (as n approaches ∞) of the absolute value of the ratio of the n+1th term to the nth term of the sequence. or in a different language:

lim(n-->∞) | (a_n+1) / (a_n) |

in this case, the nth term of your sequence is a_n = (x-9)ⁿ / n*(-7)ⁿ.

it follows that the next term (the n+1th term) is a_n+1 = (x-9)ⁿ⁺¹ / (n+1)*(-7)ⁿ⁺¹.

it also happens that our sequence is a function of another variable (x), but this is of no consequence when we consider a limit whose parameter is n (and n and x are in no way related).

so now we need to take the ratio of these as n goes to infinity. we write:

lim(n-->∞) | [ (x-9)ⁿ⁺¹ / (n+1)*(-7)ⁿ⁺¹ ] / [ (x-9)ⁿ / n*(-7)ⁿ ] |

this looks ugly, so we simplify the fractions:

lim(n-->∞) | [ n*(x-9)ⁿ⁺¹ (-7)ⁿ ] / [ (n+1) (x-9)ⁿ (-7)ⁿ⁺¹ ] |

as you can see, some nice things cancel here (since Aⁿ⁺¹ = Aⁿ A¹)

[at this point i highly suggest you write this out as you read it because it become much easier to read and see for yourself what i mean]

now that we've cancelled, our limit looks as follows:

lim(n-->∞) | n*(x-9) / (n+1)*(-7) |

now, if we wish (and we do), we can take things which do not depend on n out of the limit.

notice that x-9 does not depend on n (it only depends on x) and -7 doesn't really depend on anything (especially n!) so we can take these both out of the limit. but the absolute value signs stay on! (don't forget those!)

so, our limit now looks like this:

|(x-9)/(-7)| lim(n-->∞) | n / (n+1) |

the next step is to evaluate this rather simple limit. note also that you can just take those absolute value signs right off of the n/(n+1). this is because everything inside them is guaranteed to be positive, so the signs are superfluous. (but don't take them off of the (x-9)/(-7)! those aren't guaranteed to be positive!)

so, we know that n/(n+1) approaches 1 as n approaches infinity, and so the value of the limit is 1.

our expression now looks simply like:

|(x-9)/(-7)|

now, i've been neglecting it, but the point of the ratio test is to see of the value of the ratio is less than 1 because this is where the series converges (and don't forget that it is inconclusive when it equals 1!)

so we really have to solve this problem:

|(x-9)/(-7)| < 1

also note that that -7 on the bottom can be written as just a 7 and taken out of the absolute value signs. (because |-7| = 7)

so now let's solve

|x-9| / 7 < 1

|x-9| < 7

as you know, you can write this 2 ways and get rid of the absolute value signs as:

x-9 < 7, -(x-9) < 7

solving these (finally), we get:

x < 16, x > 2

(don't forget that when you multiply by a negative in an inequality, you flip the sign)

so we're done right? almost. remember that the ratio test inconclusive when it equals 1. so we have to manually check those possibilities. we do this by checking convergence of the original series at a particular value of x. recall the original series:

∑ [ (x-9)ⁿ / n*(-7)ⁿ ]

the 2 we have to check now are x = 16 and x = 2. notice that at x = 2, the (-7)ⁿs cancel on the top and bottom and you are left with 1/n. as you should know, ∑ 1/n diverges. so, x = 2 does not yield a series which converges.

at x = 16, you get a different series. i'll actually leave it to you to check this one for convergence (hint: use the ratio test again). but if you have problems with it, just let me know, and i'll walk you through it again.

i hope this has helped! and i'm always happy to answer more questions if you have any :]

Y!A: College Algebra Question

this is another question that i answered on Y!Answers, which can be found at this url:

http://answers.yahoo.com/question/index;_ylt=An4wC0WopGR0xxZZcW2Zts3ty6IX;_ylv=3?qid=20090926133641AASsdGx&show=7#profile-info-7y2dxI6Iaa

Q:

John owns a hotdog stand. He has found that his profit is represented by the function p(x)=-x^2+66x+81, with p(x) being profits and x the number of hotdogs sold. How many hotdogs must he sell to earn the most profit.

my answer would be a million bajillion hot dogs lol. but my answer key says 33. help please

A:

haha yes, it would be nice if he could sell a million bajillion hot dogs, but look at your equation, do you notice that -x^2 term? that means that when he sells a lot of hot dogs, he'll be subtracting that huge number squared from his profits! that's really really bad for his profits!

also, if you'll notice, the formula for his profits is the equation of a parabola! not just any parabola, but one that is opening downwards. this means that there is maximum height that his parabola reaches, corresponding to his maximum profit, which makes this a maximization/minimization problem!

now, all that analysis aside, here's how you solve it:

since the slope (derivative, in other words) of his profits curve (the parabola) is zero at its maximim, we can take the dervative of the profits curve and set that equal to zero to solve the number of hot dogs:

P(x) = -x^2 +66x +81

so the derivative, P'(x) is

P'(x) = -2x +66

now we set that equal to zero (for the reasons mentioned above),

0 = -2x + 66

and solve for x (the number of hot dogs),

2x = 66

x= 33

voila!

to solve these problems in the future, try to analyze the problem as i did in the first two paragraphs. this will really help you get a grasp of the mathematics, and make solving these kinds of problems a breeze

Y!A: What is the Taylor series of ln((x^2)+1) centered at c=0?

this is a question i answered on Y!Answers and can be found here:

http://answers.yahoo.com/question/index;_ylt=AmsLM3qKuCKKGf8_wa161Zfty6IX;_ylv=3?qid=20090426105041AA1f5lh&show=7#profile-info-2AVjODBNaa

there are two ways you can do this:

the first is by deriving the taylor series directly. however that involves a lot of messy product rules for derivatives in this case (but i will start it for you anyway).

the second is to use substitution and integration.

THE FIRST METHOD:
you would write normally, as if you were deriving a taylor series:
f(x) = a ₀ + a₁x + a₂x² + ... = ln(x² + 1)

but you can see that when we differentiate both sides, ln(x² +1) yields confusing derivatives. so, i will use a different, smarter method.

THE SECOND METHOD:
the first thing i will do is take a known taylor series, and later manipulate it:
the taylor series for f(x) = 1/(1-x) is known to be 1 + x + x² + x³ + x⁴ + ...
so i will take f(-x) = 1 / (1 - (-x) ) = 1 / (1+x)
for which the taylor series becomes: 1 - x + x² - x³ + x⁴ - ... (plugging in -x everywhere for x)

next, i integrate 1 / (1+x) and its taylor series:
ln(1+x) = C + x - (x²)/2 + (x³)/3 - (x⁴)/4 + (x⁵)/5 - ...
(and we can eliminate C now by plugging in 0 for x and getting ln(1) = C = 0)

now, using some substitution, take our new function g(x) = ln(1+x) and its taylor series, and substitute x² for x. that is, g(x²) = ln(1 + x²), and its taylor series becomes (x²) - ((x²)²)/2 + ((x²)³)/3 - ((x²)⁴)/4 + ((x²)⁵)/5 - ... (placing x²everywhere that we see an x), which equals
x² - (x⁴)/2 + (x⁶)/3 - (x⁸)/4 + ...

thus ln( 1 + x² ) = x² - (x⁴)/2 + (x^6)/3 - (x⁸)/4 + (x^10)/5 - ...
and in general form:
∑ (-1)ⁿ⁻¹ * x²ⁿ/n

[from n=1 to ∞]

though this method seems very confusing, it will get easier and more natural with practice. the trick to picking the correct initial function will also come with practice and your familiarity with taylor series.

i hope this wasn't too confusing!