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  1. J

    Infinite Resistor Network

    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...
  2. J

    Have you noticed this?

    Don't worry, dex, it was intentional. tr.v. ef·fect·ed, ef·fect·ing, ef·fects 1. To bring into existence. --J
  3. J

    Have you noticed this?

    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
  4. J

    Solving a Differential Equation with a Fourier Series

    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
  5. J

    Solving a Differential Equation with a Fourier Series

    Looks good. Now go have another look at those b coefficients. --J Just one final left... Yay me!
  6. J

    Solving a Differential Equation with a Fourier Series

    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
  7. J

    Have you noticed this?

    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
  8. J

    Solving a Differential Equation with a Fourier Series

    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
  9. J

    Laplace Initial Value Problem

    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...
  10. J

    Double checking needed on 2 Differential Equations

    It's possible that your college has a license for it and will give it to you. Why don't you contact your IT department and ask? --J
  11. J

    Double checking needed on 2 Differential Equations

    Mathematica agrees with both of your solutions. Well done. --J
  12. J

    Laplace Initial Value Problem

    *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...
  13. J

    Series Diff EQ problem: (3 - x^2) y'' - (3x) y' - y = 0

    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
  14. J

    Series Diff EQ problem: (3 - x^2) y'' - (3x) y' - y = 0

    Your algebra is a little suspect. Try calculating those coefficients again. --J
  15. J

    Diff. EQ: How do I solve a 2nd order linear EQ with series?

    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...
  16. J

    Diff. EQ: How do I solve a 2nd order linear EQ with series?

    Now rewrite your factorials in factorial form and deduce an expression for a2n and a2n+1 in terms of a0 and a1, respectively. --J
  17. J

    Diff. EQ: How do I solve a 2nd order linear EQ with series?

    a2 is when n = 0. Calculate it for n = 0. You must choose two of the constants to be arbitrary. A most convenient choice would be a0 and a1. --J
  18. J

    Diff. EQ: How do I solve a 2nd order linear EQ with series?

    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
  19. J

    Healthy Breakfast

    What's this "breakfast" thing you speak of? --J
  20. J

    Solver and Excel

    Edit: Failure to read the original post... --J
  21. J

    How do you earn a living and still have a life?

    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...
  22. J

    Laplace Initial Value Problem

    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
  23. J

    Help really really really needed!

    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
  24. J

    Help with math puzzle

    Move the 8 and 9 to the right of the 1 and 2. Then move the 0 below the 6. You get 1289 345 67 0 --J
  25. J

    Laplace Initial Value Problem

    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
  26. J

    Integrals of Trig Functions

    You're integrating u to get du for some reason. You must differentiate u to get du. Stick with your original substitution. --J
  27. J

    Integrals of Trig Functions

    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
  28. J

    Integrals of Trig Functions

    Or just note that the derivative of tangent is the secant squared... u-substitution, anybody? --J
  29. J

    A conceptual problem regarding work

    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...
  30. J

    A conceptual problem regarding 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
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