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Thread: Hyperbolic Function Questions

  1. #1
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    Hyperbolic Function Questions

    Hello

    How does $\displaystyle (exp[A] - exp[-A])/(exp[A] + exp[-A]) = (exp[2A] - 1)/(exp[2A] + 1)$ ?

    How does $\displaystyle sin(jy) = jsinh(y)$ ?

    If $\displaystyle -1 < tanh(x) < 1$, why is $\displaystyle ln[tanh(x)] < 0$ ?
    Also, why does this mean $\displaystyle tanh(x)$ can not be differentiated? (without using the 'multi-valued' approach and differentiating it).

    Thanks
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  2. #2
    o_O
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    #1: $\displaystyle \frac{e^a - e^{-a}}{e^a +e^{-a}} = \frac{e^a - \frac{1}{e^a}}{e^a + \frac{1}{e^a}} \cdot {\color{blue}\frac{e^a}{e^a}} = \cdots$

    #2: Use Euler's formula.
    $\displaystyle \begin{aligned} e^{ix} & = \cos x + i\sin x \\ e^{-ix} & = \cos x - i\sin x \end{aligned} \ \ \overbrace{\Rightarrow}^{x \ = \ iy} \ \sin iy = \frac{1}{2i}\left(e^{-iy} - e^{iy}\right)$ (Subtracted the second line from the first and divided both sides by i)

    Can you carry on from here?

    #3: $\displaystyle f(u) = \ln u$ isn't defined for $\displaystyle u < 0$ ... And $\displaystyle f(x) = \tanh x$ is differentiable. In fact, $\displaystyle \frac{d}{dx} \tanh x = \text{sech}^2 x$
    Last edited by o_O; Jan 15th 2009 at 03:25 PM.
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  3. #3
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    Thanks.

    Regarding #2, I got:

    $\displaystyle
    e^{ix} - e^{-ix} = 2isin(x)$

    $\displaystyle \frac{e^{ix} - e^{-ix}}{2}\ = isin(x)$

    Where to from here?
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  4. #4
    o_O
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    Let $\displaystyle x = iy$

    Edit: Sorry, I had a typo in my first post. Should've been iy instead of ix using the sub here.
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    You should know $\displaystyle \sin (z) = \frac{1}{{2i}}\left( {e^{iz} - e^{ - iz} } \right)$. That is the standard definition for the complex sine function.
    Then it is easy to see: $\displaystyle \sin (iy) = \frac{1}{{2i}}\left( {e^{i\left( {iy} \right)} - e^{ - i\left( {iy} \right)} } \right) = - \frac{i}{2}\left( {e^{ - y} - e^y } \right) = i\left( {\frac{{e^y - e^{ - y} }}{2}} \right) = i\sinh (y)$
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  6. #6
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    Thanks, I think I got the result:

    $\displaystyle e^{iy} - e^{-iy} = 2isin(y)$

    $\displaystyle \frac{e^{iy} - e^{-iy}}{2}\ = isin(y)$

    $\displaystyle x = iy$

    $\displaystyle \frac{e^{x} - e^{-x}}{2}\ = isin(x/i)$ (1)

    $\displaystyle sinh(x) = isin(-ix)$

    $\displaystyle \Rightarrow\ isinh(x) = sin(ix)$

    But...for (1), how does that equate to $\displaystyle sinh(x)$ if it contains the imaginary part $\displaystyle i$? (I thought only $\displaystyle sin(x)$ dealt with imaginary parts, given the $\displaystyle i$ 's in its definition).

    Also, trying to obtain $\displaystyle cosh(x)$:

    $\displaystyle \frac{e^{x} + e^{-x}}{2}\ = cos(x/i)$

    $\displaystyle \Rightarrow \ cosh(x) = cos(ix)$

    Is that correct?
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  7. #7
    o_O
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    No. Try reading over our posts again.

    $\displaystyle \sin x = \frac{1}{2i} \left(e^{ix} - e^{-ix}\right)$

    Here, we let $\displaystyle x = iy$ to get: $\displaystyle \sin (iy) = \frac{1}{2i}\left(e^{i(iy)} - e^{-i(iy)}\right) = \frac{1}{2i} \left(e^{-y} - e^{y}\right) = -\frac{1}{i} \sinh (y)$
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  8. #8
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    Thanks I get it now.
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