Since the network is infinite, if you extend the network by one more copy of the pattern, the equivalent resistance shouldn't change. So represent the network's resistance with a single resistor with Req resistance. Now, can you add 3 resistors to the Req resistor to make it back into the...
Well, the next time you start examining the effects of your particle accelerator bunny exterminator on cute pink flying bunny rabbits and wish it could effect them more, don't come complaining to me when you drown in your own pool of rabbits.
--J
Looks like you got it.
And just in case you were interested, here's some plots of the fourier series truncated to 5, 15, 50, and 500 terms. Can't even tell it's not the function itself in the 500 one, can ya'?
Good job.
--J
You got an extra negative. \cos{n\pi} = (-1)^n, which we can verify by noting that when n = 0, n\pi = 0, so \cos{n\pi} = 1. Additionally, when n = 1, we just have cosine of pi, which is -1. So we must have (-1)n instead of (-1)n-1.
--J
In what instances? Sometimes "affect" is the correct word, and sometimes "effect" is the correct word. Naturally, most of the English speakers in the world have no idea that they are not the same word, so it can be pretty irritating if it's one of your pet peeves.
--J
You need to revisit your bn with your newly refound knowledge of the behavior of sine and cosine.
The sign of your an term is indeed incorrect.
Why do I let myself get dragged into other people's problems? *sighs* Back to finals! Work, damn me!
--J
Unfortunately, it's a 30 hour drive that must be completed in one weekend. Sometimes I wonder about the things I get myself into... But there's a bubble bath waiting for me at the end of it, so it'll all be worth it.
And while I'm driving for two days straight, you get to wander around in...
*sighs* If only I didn't have to write an essay and do a few more finals, move out of my room, then drive for 30-some-odd hours, I could look into it more. But alas, I've been screwed by finals week. :frown:
So don't bust your butt on my account, salty, because I won't be able to do...
Also, you shouldn't include the xn in your coefficients. Remember that these are the coefficients of the powers of x! They don't include the power of x themselves. You must multiply them by the appropriate power of x to get your solution. Otherwise, it looks like you're set. Good job.
--J
You know that there must be two arbitrary constants because the differential equation is of second order. You select the 0 and 1 constants because they are generally the most convenient when you have these problems that are designed to work out nicely. If your differential equation were instead...
You could start with a0 = a0 and a1 = a1. Or a0 = c1 and a1 = c2. Just eliminate as many of the an constants as you can.
Your denominator looks awfully similar to a factorial... And how did you get a4 = a2 = a0?
--J
Give up sleep. And consequently your sanity. 6 all-nighters in 2 weeks and you get used to sleeping every other day. You know the saying. Sleep, social, or school: pick two. (Disclaimer: Don't do it.)
And how did you do 7 hours of homework every day in high school...? Even if I wanted to, I...
In the brief intermission from me being overworked that I'm stealing from myself, did we figure out what the story with the c was? I noted that the numerical solution Mathematica found set c to zero, as saltydog said, but I've had little time to look into it more than that.
--J
He didn't copy the problem right in this post. The other post had a 0 above the 1 and looks like
0
1 2
3 4 5
6 7 8 9
Just in case you were interested in figuring it out for yourself.
--J
It seems to me if you apply parts enough times, the only hitch is in evaluating
\int_0^{\infty} te^{-st}ydt
But both ty'' and ty' reduce to that and known transforms, if I'm not mistaken.
Then again, small comfort, since that's not so easy to evaluate anyway.
--J
Are you familiar with u-substitution? You're going to have to make one to evaluate the integral. You have two choices for what to let u equal, both work. What do you think u might equal?
--J
Actually, I was thinking more along the lines of pushing the sofa out of the earth's field, letting the moon do a bit of stopping via the crater method, then pushing the sofa out of the moon's field and then letting the earth do the stopping. And then you have done work (No guarantees! All work...
You're right. The work done on the sofa is zero, but the work that you do on the sofa is not. I'm not (and wasn't) in much condition to post anything very sensible, so I'll just shut up until I have more rest. Until then, probably best to ignore me.
--J