last day (15 days later) » 

17:39
Hey, are you there?
17:54
Anyways, if you are interested, I leave here my solution to the previous problem (the one you solved is the one next to this one) mediafire.com/?ubbn4x8s7395fq0
18:54
Sorry, I was afk for a bit.
And now I have to leave again for a while. I will check when I get back.
19:28
Ok.
Let me know when you're back.
19:49
I got delayed, so I am still here. What's up?
@Peter: what's up?
I linked a similar problem for you to see how I solved it. It's the previous on the set of problems. You can see I exploited the fact that $$w = \frac{ds}{dt}$$
http://www.mediafire.com/?o67zpk1n2791a56
http://www.mediafire.com/?ubbn4x8s7395fq0
Those are the links.
I saw the PDF
However, I always called it the Pursuit Curve.
20:04
right, persecution is spanglish, =)
20:16
@Peter ah
Persecution has a pretty different meaning in English
But I guess pursuit could be considered a part of persecution :-)
Hahahaha yes I didn't think about it much when I posted.
If you have any spare time and interest I've got another question I'm getting tempted to make a bounty with sine it's freezed. Here it ishttp://math.stackexchange.com/questions/106389/recursive-solutions-to-linear-ode (it's quite interesting, I've made a pdf about that if you're interested)
3
Q: Recursive solutions to linear ODE.

PeterWhen finding the solutions to the simple ODE $$ y'- mxy= x^n \text{ ; } y(0) = 0$$ I found the following: Let $P_n$ be the particular solution for each integer exponent $n$. Then if we define $$P_0(x) = exp\left(\displaystyle \frac{mx^2}{2}\right)\int_0^x exp\left(-\displaystyle \frac{mt^...

20:30
$y=e^{mx^2/2}\int e^{-mx^2/2}x^n \;\mathrm{d}x$
for n odd, there are polynomial solutions.
for n even, you have erfs popping up
Yes, the general solutions are:
$${G_{2n}} = {c_1}{e^{\frac{{m{x^2}}}{2}}} + \frac{{\left( {2n - 1} \right)!!}}{{{m^n}}}{P_0} - \sum\limits_{k = 1}^n {\frac{{\left( {2n - 1} \right)!!}}{{\left( {2k - 1} \right)!!}}\frac{{{x^{2k - 1}}}}{{{m^{n - k + 1}}}}} $$

$${G_{2n + 1}} = {c_1}{e^{\frac{{m{x^2}}}{2}}} + \frac{{\left( {2n} \right)!!}}{{{m^{n + 1}}}}\left( {{e^{\frac{{m{x^2}}}{2}}} - 1} \right) - \sum\limits_{k = 1}^n {\frac{{\left( {2n} \right)!!}}{{\left( {2k} \right)!!}}\frac{{{x^{2k}}}}{{{m^{n - k + 1}}}}} $$
and $P_0$ has the erf in it.
I'm quite interested in the polynomial solutions.
and what would happen if we used an infinite sum of powers, for example if we used the solutions for each$$\frac{x^n}{n!}$$ to find the solution to $e^x$
Those do not render for me...
oops there they go :-)
20:54
quite an expression right?
 
1 hour later…
22:22
@Peter Just as I had said :-)
22:39
Hey rob so where are you from?
23:11
I live in the Los Angeles area

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