MAT244-2013F > MidTerm

MT, P1

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Xuewen Yang:

--- Quote from: Xiaozeng Yu on October 09, 2013, 10:52:39 PM ---
--- Quote from: Xuewen Yang on October 09, 2013, 10:47:45 PM ---For Xiaozeng Yu,

Is there a problem with the sign change? from "plus" (2nd row) to "minus" (3rd row) ?

--- End quote ---

no mistake. d(ln(1-y))/dt = -1/1-y

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You are right! My bad  :P

Victor Ivrii:
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 $\int \frac{dy}{1+y^2}=\arctan(y)$).

Xiaozeng Yu:

--- Quote from: Victor Ivrii on October 10, 2013, 06:03:59 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 $\int \frac{dy}{1+y^2}=\arctan(y)$).

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hmm...yea...-1<y<1  T.T i feel i'm screwed...

Victor Ivrii:

--- Quote from: Xiaozeng Yu on October 10, 2013, 09:53:07 AM ---
--- Quote from: Victor Ivrii on October 10, 2013, 06:03:59 AM ---And where this solution is defined?
hmm...yea...-1<y<1

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This is a range of $y$ (what values it takes), I asked about domain--for which $t$ it is defined.

So, my question about domain is pending

Xiaozeng Yu:
t is defined everywhere...