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MidTerm / Re: MT, P1
« on: October 11, 2013, 10:05:54 AM »
thank u prof.
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MidTerm / Re: MT, P1
« on: October 10, 2013, 09:53:07 AM »And where this solution is defined? $\DeclareMathOperator{\arctanh}{arctanh}$
BTW, when collecting MT I have seen this problem and in many papers integration was atrocious.
As long as $|y|<1$ we have $\int \frac{dy}{1-y^2}=\arctanh (y)$ (inverse hyperbolic function $\tanh$) and then
$y=\tanh (t)=\frac{{e^t}-e^{-t}}{e^{t}+e^{-t}}$ (similar to $\int \frac{dy}{1+y^2}=\arctan(y)$).
hmm...yea...-1<y<1 T.T i feel i'm screwed...
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MidTerm / Re: MT, P4
« on: October 10, 2013, 09:43:00 AM »
ahh...omg, because the rectangle must contain the initial value point (0,1) in order to have an unique solution of the initial value problem. sin1>0 make the function increasing, so the retangle (a<0<b, 0<y<pi) which containing the initial point contains the unique solution function of (0,1) which is increasing?
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MidTerm / Re: MT, P2
« on: October 10, 2013, 09:27:59 AM »Xuewen Yang,
good job, for multiplication do not use * (it is a convolution, different operation) use either \cdot like in $a\cdot b$ or \times like $a\times b$.
Xiaozeng Yu, no point to post inferior solution (scan) after superior (typed) has been posted. This time I awarded "scan posted after scan", but not in the future.
sorry, it took me a while figuring out how to post the non-oversized version of my work...
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MidTerm / Re: MT, P1
« on: October 09, 2013, 10:52:39 PM »For Xiaozeng Yu,
Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?
no mistake. $d(\ln(1-y))/dt = -1/(1-y)$.
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