Results 1 to 8 of 8

Math Help - Hyperbolic Function Questions

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    92

    Hyperbolic Function Questions

    Hello

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

    How does sin(jy) = jsinh(y) ?

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

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    #1: \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.
    \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: f(u) = \ln u isn't defined for u < 0 ... And f(x) = \tanh x is differentiable. In fact, \frac{d}{dx} \tanh x = \text{sech}^2 x
    Last edited by o_O; January 15th 2009 at 03:25 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Nov 2008
    Posts
    92
    Thanks.

    Regarding #2, I got:

    <br />
e^{ix} - e^{-ix} = 2isin(x)

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

    Where to from here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Let x = iy

    Edit: Sorry, I had a typo in my first post. Should've been iy instead of ix using the sub here.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,388
    Thanks
    1476
    Awards
    1
    You should know \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: \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)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2008
    Posts
    92
    Thanks, I think I got the result:

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

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

    x = iy

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

    sinh(x) = isin(-ix)

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

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

    Also, trying to obtain cosh(x):

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

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

    Is that correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    No. Try reading over our posts again.

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

    Here, we let x = iy to get: \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)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2008
    Posts
    92
    Thanks I get it now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Hyperbolic Geometry Some Questions - 3
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 23rd 2011, 04:36 PM
  2. Hyperbolic Geometry Some Questions - 2
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 23rd 2011, 04:30 PM
  3. Hyperbolic Geometry Some Questions
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: April 23rd 2011, 04:23 PM
  4. more hyperbolic function questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 6th 2010, 04:12 PM
  5. Hyperbolic function questions
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 6th 2010, 04:26 AM

Search Tags


/mathhelpforum @mathhelpforum